# Hilbert Spaces Without The Countable Axiom of Choice

Bruce Blackadar, Ilijas Farah, and Asaf Karagila  Department of Mathematics
and Statistics, University of Nevada, Reno
Reno, NV, USA 89557<EMAIL_ADDRESS>Department of Mathematics and Statistics,
York University, 4700 Keele Street, Toronto, Ontario, Canada, M3J 1P3
Matematički Institut SANU
Kneza Mihaila 36
11 000 Beograd, p.p. 367
Serbia<EMAIL_ADDRESS>School of Mathematics
University of Leeds
Leeds, LS2 9JT, UK<EMAIL_ADDRESS>
I.F. was partially supported by NSERC

A.K. was supported by a UKRI Future Leaders Fellowship [MR/T021705/2]

We would like to thank the anonymous referee for a very useful report.

To the memory of Eberhard Kirchberg

This article examines Hilbert spaces constructed from sets whose existence is
incompatible with the Countable Axiom of Choice (CC). Our point of view is
twofold:

(1) We examine what can and cannot be said about Hilbert spaces and operators
on them in ZF set theory without any assumptions of Choice axioms, even the
CC. It turns out that quite a bit can be said in this context, some of which
seems a little counterintuitive. We are especially interested in Hilbert
spaces whose dimension is “finite” in a sense incompatible with the CC, where
strange properties sometimes hold. One philosophical consequence of not
assuming CC is that Hilbert spaces come in a great many more varieties and are
not nearly as homogeneous as normally thought (and are much more interesting!)

(2) We view Hilbert spaces as “quantized” sets and obtain some set-theoretic
results from associated Hilbert spaces.

The article is written in a partly expository style. Some of the results were
previously obtained, primarily by N. Brunner and coauthors in [Bru86],
[Bru84], and [BSB96] (although these papers have a somewhat different point of
view and contain some errors; see the discussion in §6.1.2). Many of our basic
results are modest and qualify as “low-hanging fruit”; we have certainly not
picked all the fruit and invite other interested readers to join the harvest.

A subsequent article [BF23] will discuss C*-algebras without AC, including the
validity of the two Gelfand-Naimark representation theorems.

Caution: Without CC one must be very careful since many familiar techniques we
use, sometimes without much thought, do not work. The principal tools we
cannot use are Zorn’s Lemma, Tikhonov’s Theorem, the Hahn-Banach Theorem, and
the Baire Category Theorem, all of which require some version of the AC.
(Note, however, that we can still make finitely many choices.) Thus we will
have to prove or reprove some “standard” or “well-known” properties of Hilbert
spaces. So we cannot assume we know anything about Hilbert spaces beyond some
basic facts obviously independent of Choice, such as the CBS inequality and
Parallelogram Law. [Bla23] is a convenient reference for some standard
arguments, which we will use as standard reference, but they can also be found
in many other references. We use the capitalized word Choice to generically
refer to versions of the Axiom of Choice (e.g. “can be proved without any
Choice” means “theorem of ZF”), and we will not assume any Choice without
explicit mention.

###### Contents

  1. 1 Completeness and its Consequences
  2. 2 Hilbert Spaces
  3. 3 Orthonormal Bases and $\ell^{2}$-Spaces
  4. 4 Finite-Dimensional and Separable Hilbert Spaces
    1. 4.1 Separable Hilbert Spaces
  5. 5 Dedekind-Finite Sets of Various Flavors
    1. 5.1 Vanilla Flavor
    2. 5.2 Other Flavors
    3. 5.3 Amorphous Sets
    4. 5.4 Permanence Properties of CF Sets
    5. 5.5 Examples and Constructions
    6. 5.6 Rigid Sets
    7. 5.7 Russell Sets and Russell Cardinals
  6. 6 $\ell^{2}(X)$ For DF Sets
    1. 6.1 $\ell^{2}(X)$ for Russell Sets
    2. 6.2 Bases in Hilbert Spaces
    3. 6.3 Uniqueness of Orthogonal Dimension
    4. 6.4 Universality of Hilbert Spaces With a Basis?
  7. 7 Bounded Operators
    1. 7.1 Compact Operators
  8. 8 ${\mathcal{B}}(\ell^{2}(X))$ For DF Sets
    1. 8.1 ${\mathcal{B}}(\ell^{2}(X))$ for CF Sets
    2. 8.2 ${\mathcal{K}}(\ell^{2}(X))$ for CF Sets
    3. 8.3 How Big Is ${\mathcal{Q}}(\ell^{2}(X))$?
    4. 8.4 Stably Finite $\mathrm{C}^{*}$-algebras With no Quasitraces
    5. 8.5 Spectrum of Operators
  9. 9 New Size and Comparison Relations on Hilbert Spaces and Sets
    1. 9.1 Hilbert Dedekind-finiteness
    2. 9.2 Quantum Cardinals
  10. 10 Models of ZF
  11. 11 Problems

### Brief Summary

In Section 1 we introduce the notion of $\sigma$-completeness of metric spaces
and argue that it is the correct notion of completeness in a context where CC
fails (see also Example 6.0.6). In Secion 2 we establish the theory of Hilbert
spaces in this context. Spaces $\ell^{2}(X)$ are introduced in Section 3, and
in Section 4 we show that the analysis of finite-dimensional and separable
Hilbert spaces is unaffected even if CC fails. Dedekind-finite sets and their
variations are introduced in Section 5. The most important notion for us is
that of a Cohen-finite set (Definition 5.2.3), and a CF set is a Cohen-finite,
infinite set. Simple permanence properties of CF sets are established in
Subsection 5.4. This includes Lemma 5.4.2, used throughout. Fig. 1 represents
implications between different flavors of Dedekind-finite sets. In Section 6
we establish basic results on $\ell^{2}(X)$ spaces for a DF (Dedekind-finite
and infinite) set $X$. In Proposition 6.0.3 we prove, among other
equivalences, that $X$ is Cohen-finite if and only if the standard basis is a
Hamel basis. In Theorem 6.2.3 we find (using a rather peculiar variant of a
Dedekind-finite set, proved consistent with ZF by J. Truss) a Hilbert space
without an orthonormal basis, and even without an infinite orthonormal
sequence, and in Example 6.3.1 we prove that if $X$ is a DF set which is not
Cohen-finite then $\ell^{2}(X)$ has orthonormal bases of different
cardinalities. The question whether every Hilbert space is (provably in ZF)
isometric to a subspace of one with an orthonormal basis is discussed in
Subsection 6.4. In Section 7 we turn to the study of bounded linear operators,
and in Section 8 we consider the algebra of bounded linear operators on
$\ell^{2}(X)$ for a DF set $X$. If $X$ is CF then this algebra is non-
separable, but it is stably finite (Corollary 8.1.3) and every compact
operator has finite rank (Proposition 8.2.1). In Section 8.3 we study the
Calkin algebra ${\mathcal{Q}}(\ell^{2}(X))$ and prove that it is one-
dimensional if $X$ is strongly amorphous (Proposition 8.3.3) and that it can
be non-separable and abelian (Proposition 8.3.8) or nonabelian and stably
finite (Proposition 8.3.9) for specific choices of a CF set $X$. In Corollary
8.4.3 we prove that in Solovay’s model there is a stably finite
$\mathrm{C}^{*}$-algebra with no tracial states. This example is commutative
and it has no nonzero representation on a Hilbert space. In Proposition 8.4.4
we show that if there is a CF set $X$ such that its power set has no finitely
additive probability measure that vanishes on singletons, then
${\mathcal{B}}(\ell^{2}(X))$ is stably finite and has no tracial states, and
is therefore not satisfactory (this is an immediate consequence of a well-
known fact that if all sets of reals have the Property of Baire then there is
no probability measure on ${\mathcal{P}}(\mathbb{N})$ that vanishes on the
singletons). In Section 8.5 we prove that for a power Dedekind-finite set $X$,
the spectrum of every bounded linear operator $T$ on $\ell^{2}(X)$ is finite
and every element of the spectrum is an eigenvalue. In Section 9 we introduce
Hilbert space analogs of principal flavors of Dedekind-finite sets (see Fig.
2). Models of ZF used in the previous sections are constructed in Section 10,
and we conclude with a short list of open problems in Section 11.

##  1\. Completeness and its Consequences

The first technicality concerns the “right” notion of completeness for Hilbert
spaces and, more generally, metric spaces. The Cauchy sequence definition
(which we will call Cauchy completeness) turns out not to be useful in the
absence of CC. Instead, the following version of completeness should be used.

##### 1.0.1

Definition. Let $(X,\rho)$ be a metric space.

(i) $(X,\rho)$ is absolutely closed if, whenever $\phi$ is an isometric
embedding of $(X,\rho)$ into a metric space $(Y,\rho^{\prime})$, $\phi(X)$ is
closed in $(Y,\rho^{\prime})$.

(ii) $(X,\rho)$ is $\sigma$-complete if, whenever $(A_{n})$ is a decreasing
sequence of nonempty closed subsets of $X$ with $\operatorname{diam}(A_{n})\to
0$, then $\cap_{n}A_{n}$ is nonempty (it is then necessarily a singleton). In
[BSB96] spaces with this property were called _Cantor complete_.

(iii) $(X,\rho)$ is totally complete if, whenever $(A_{i})$ is an indexed set
of nonempty closed subsets of $X$, directed by reverse inclusion (i.e.
$A_{j}\subseteq A_{i}$ if and only if $i\leq j$, and for every $i,j$ there is
a $k$ with $A_{k}\subseteq A_{i}\cap A_{j}$; the indices run over an arbitrary
directed set) with $\operatorname{diam}(A_{i})\to 0$ (i.e. for every
$\epsilon>0$ there is an $i_{0}$ with
$\operatorname{diam}(A_{i_{0}})<\epsilon$ and hence
$\operatorname{diam}(A_{i})<\epsilon$ for all $i>i_{0}$), then $\cap_{i}A_{i}$
is nonempty (it is then necessarily a singleton).

(iv) $(X,\rho)$ is uniformly complete if the uniform space defined by the
metric is complete, in the sense that every Cauchy net is convergent.

In the context of the Axiom of Choice, a metric space is uniformly complete if
and only if it is Cauchy complete. However, without CC in general a non-
convergent Cauchy sequence cannot be extracted from a non-convergent Cauchy
net (cf. 6.0.3, 6.2.7). The following is a theorem of ZF, i.e. no form of
Choice is used. If the CC is assumed, all five conditions are equivalent. See
[Bla23, XII.16.9.5] for a proof.

##### 1.0.2

Theorem. Let $(X,\rho)$ be a metric space. The following are equivalent:

  1. (i)

$(X,\rho)$ is absolutely closed.

  2. (ii)

$(X,\rho)$ is $\sigma$-complete.

  3. (iii)

$(X,\rho)$ is totally complete.

  4. (iv)

$(X,\rho)$ is uniformly complete.

These conditions imply

  1. (v)

$(X,\rho)$ is Cauchy complete.

We will use the term $\sigma$-complete generically for a metric space
satisfying (i)–(iv). A closed subset of a $\sigma$-complete metric space is
obviously $\sigma$-complete. A compact metric space is $\sigma$-complete.

Since every uniform space has an essentially unique completion (cf. e.g.
[Bla23, §XII.16.8], and note that no Choice is used there), we obtain:

##### 1.0.3

Corollary. If $(X,\rho)$ is a metric space, there is a canonical
$\sigma$-complete metric space $(\bar{X},\bar{\rho})$ containing $(X,\rho)$.
If $\phi:X\to Y$ is an isometry from $X$ to a $\sigma$-complete metric space
$(Y,\tau)$, then $\phi$ extends uniquely to an isometry from $\bar{X}$ to the
closure of $\phi(X)$ in $Y$. Thus the completion of $(X,\rho)$ is unique up to
isometry which is the identity on $X$.

##  2\. Hilbert Spaces

##### 2.0.1

Definition. A Hilbert space is a complex inner product space which is
$\sigma$-complete under the induced norm (almost everything we do works at
least as well for real Hilbert spaces). More generally, a (real or complex)
Banach space is a $\sigma$-complete normed (real or complex) vector space.

In the presence of CC, this agrees with the usual definition, and we believe
it is the “right” definition in the general case. Note that a Hilbert space by
this definition is Cauchy-complete. Every (complex) inner product space has a
canonical $\sigma$-completion (1.0.3). Since closed bounded subsets of a
finite-dimensional normed vector space are compact (no Choice is needed to
prove this), any finite-dimensional inner product space is a Hilbert space.
The problem with the example in [BSB96] (cf. §6.0.6) is that, while it is
Cauchy complete, it is not $\sigma$-complete (and thus not a Hilbert space by
our definition).

##### 2.0.2

The CBS inequality and the Parallelogram Law hold in any inner product space
(no Choice required), and the following fact (in ZF) is a fundamental
consequence. See e.g. [Bla23, XVI.9.4.2] for a proof.

##### 2.0.3

Lemma. Let $C$ be a nonempty convex set in an inner product space
${\mathcal{V}}$, $\xi\in{\mathcal{V}}$, and $\epsilon>0$.

If $\eta,\zeta\in C$ with111We write
$\rho(\xi,C)=\inf\\{\|\xi-\eta\|\mid\eta\in C\\}$.
$\|\xi-\eta\|^{2}<\rho(\xi,C)^{2}+\epsilon$ and
$\|\xi-\zeta\|^{2}<\rho(\xi,C)^{2}+\epsilon$, then
$\|\eta-\zeta\|^{2}<4\epsilon$.

The most important consequence of $\sigma$-completeness in inner product
spaces is that the following fundamental fact holds (in ZF!). The usual proof
uses Cauchy sequences and requires CC, but there is an alternate proof from
$\sigma$-completeness not requiring CC. This proof can be found in some
references, but we give it here to illustrate the use of
$\sigma$-completeness.

##### 2.0.4

Theorem. [Closest Vector Property] Let ${\mathcal{H}}$ be a Hilbert space and
$C$ a nonempty closed convex subset of ${\mathcal{H}}$. For any
$\xi\in{\mathcal{H}}$, there is a unique $\eta\in C$ such that

$\|\xi-\eta\|=\rho(\xi,\eta)=\rho(\xi,C)=\inf_{\zeta\in
C}\|\xi-\zeta\|=\min_{\zeta\in C}\|\xi-\zeta\|\ .$

Proof: Let $r=\rho(\xi,C)$. For each $n\in\mathbb{N}$ let $A_{n}$ be the
intersection of $C$ with the closed ball of radius $r+\frac{1}{n}$ around
$\xi$. Then each $A_{n}$ is a nonempty closed convex set in ${\mathcal{H}}$,
with $A_{n+1}\subseteq A_{n}$, and $\rho(\xi,A_{n})=\rho(\xi,C)$. Lemma 2.0.3
implies that $\operatorname{diam}(A_{n})\to 0$ as $n\to\infty$. Thus by
$\sigma$-completeness $\cap_{n}A_{n}=\\{\eta\\}$ for some $\eta\in C$. Clearly
$\|\xi-\eta\|=r$ and $\eta$ is the unique vector in $C$ with this property.

A key consequence of 2.0.4 is the existence of orthogonal complements:

##### 2.0.5

Corollary. Let ${\mathcal{H}}$ be a Hilbert space, and ${\mathcal{Y}}$ a
closed subspace of ${\mathcal{H}}$. Then every $\xi\in{\mathcal{H}}$ can be
uniquely written as $\eta+\zeta$, where $\eta\in{\mathcal{Y}}$ and
$\zeta\in{\mathcal{Y}}^{\perp}$. This $\eta$ is the closest vector in
${\mathcal{Y}}$ to $\xi$. So ${\mathcal{Y}}$ and ${\mathcal{Y}}^{\perp}$ are
complementary closed subspaces of ${\mathcal{H}}$, and in particular
${\mathcal{Y}}$ has an orthogonal complement, and
$({\mathcal{Y}}^{\perp})^{\perp}={\mathcal{Y}}$. There is an orthogonal
projection $P_{{\mathcal{Y}}}$ from ${\mathcal{H}}$ onto ${\mathcal{Y}}$ with
null space ${\mathcal{Y}}^{\perp}$.

##### 2.0.6

Theorem. [Riesz Representation Theorem] Let ${\mathcal{H}}$ be a Hilbert
space, and $\phi$ a bounded linear functional on ${\mathcal{H}}$. Then there
is a unique vector $\eta\in{\mathcal{H}}$ with
$\phi(\xi)=\langle\xi,\eta\rangle$ for all $\xi\in{\mathcal{H}}$. The map
$\phi\mapsto\eta$ is a (conjugate-linear) isometry from ${\mathcal{H}}^{*}$
onto ${\mathcal{H}}$. In particular, ${\mathcal{H}}$ is reflexive.

The standard proofs work verbatim.

We also note the following immediate (actually somewhat nontrivial)
consequence of 1.0.3:

##### 2.0.7

Proposition. Let ${\mathcal{H}}$ and ${\mathcal{H}}^{\prime}$ be Hilbert
spaces, and ${\mathcal{H}}_{0}$ a dense subspace of ${\mathcal{H}}$. An
isometric linear map $T$ from ${\mathcal{H}}_{0}$ to ${\mathcal{H}}^{\prime}$
extends uniquely to an isometric linear map $\bar{T}$ from ${\mathcal{H}}$ to
${\mathcal{H}}^{\prime}$. If $T({\mathcal{H}}_{0})$ is dense in
${\mathcal{H}}^{\prime}$, then $\bar{T}$ is surjective.

More generally, since a bounded linear operator is uniformly continuous, any
bounded $T:{\mathcal{H}}_{0}\to{\mathcal{H}}^{\prime}$ extends uniquely to
${\mathcal{H}}$.

##  3\. Orthonormal Bases and $\ell^{2}$-Spaces

One standard property of Hilbert spaces is dramatically absent in our non-
Choice setting: existence of orthonormal bases. Every Hilbert space that has a
well-ordered dense subset has an orthonormal basis, even a well-ordered one
(Theorem 4.1.3). In particular, every separable Hilbert space has an
orthonormal basis. A Hilbert space with no well-ordered dense subset is non-
separable, and without the Axiom of Choice (or at least the axiom of Dependent
Choices) it does not necessarily have a separable infinite-dimensional
subspace (cf. 6.0.3, 6.2.7). Because of this, in the non-Choice setting a non-
separable Hilbert space is not necessarily either “larger” or “smaller” than a
separable one). Not every Hilbert space has an orthonormal basis without some
additional assumptions (6.2.3). In this section, we will examine Hilbert
spaces which come equipped with a natural orthonormal basis, which might be
better behaved.

##### 3.0.1

The first thing we need to make precise is the notion of the sum of a function
over a set. We will take it to mean the integral with respect to counting
measure; thus the sum of a nonnegative real-valued function on a set $X$ is
the supremum of the sums over finite subsets of $X$.

##### 3.0.2

Definition. Let $X$ be a set. Define

$\ell^{2}(X)=\left\\{\eta:X\to\mathbb{C}:\sum_{x\in
X}|\eta(x)|^{2}<\infty\right\\}\ .$

##### 3.0.3

The usual proofs show that $\ell^{2}(X)$ is a complex vector space under
pointwise operations, and the CBS inequality shows that if
$\eta,\zeta\in\ell^{2}(X)$, then the net

$\langle\eta,\zeta\rangle_{F}=\sum_{x\in F}\eta(x)\overline{\zeta(x)}$

in $\mathbb{C}$ indexed by finite subsets $F$ of $X$, directed by inclusion,
converges to a complex number we call $\langle\eta,\zeta\rangle$, and
$\langle\cdot,\cdot\rangle$ is an inner product on $\ell^{2}(X)$.

##### 3.0.4

Proposition. Let $X$ be a set. Then $\ell^{2}(X)$ is $\sigma$-complete, hence
a Hilbert space.

Proof: The proof follows the pattern of the usual proof of completeness. Let
$(A_{n})$ be a decreasing sequence of closed bounded subsets of $\ell^{2}(X)$
whose diameters go to zero. For each $x\in X$, let $A_{n}^{x}$ be the set of
$x$’th coordinates of elements of $A_{n}$, i.e. $A_{n}^{x}=P_{x}(A_{n})$,
where $P_{x}$ is the orthogonal projection of ${\mathcal{H}}$ onto the one-
dimensional subspace spanned by $\xi_{x}$. Then $(\overline{A_{n}^{x}})$ is a
decreasing sequence of closed bounded subsets of $\mathbb{C}$ whose diameters
go to 0; hence $\cap_{n}\overline{A_{n}^{x}}=\\{c_{x}\\}$ for some
$c_{x}\in\mathbb{C}$. We claim

$\sum_{x\in X}c_{x}\xi_{x}\in\bigcap_{n}A_{n}\ .$

If $F$ is a finite subset of $X$, set $A_{n}^{F}=P_{F}(A_{n})$, where $P_{F}$
is the orthogonal projection of $\ell^{2}(X)$ onto the span
${\mathcal{Y}}_{F}$ of $\\{\xi_{x}:x\in F\\}$. Then $(\overline{A_{n}^{F}})$
is a decreasing sequence of closed subsets of ${\mathcal{Y}}_{F}$ whose
diameters go to zero, so $\cap_{n}\overline{A_{n}^{F}}=\\{\eta_{F}\\}$ for
some $\eta_{F}\in{\mathcal{Y}}_{F}$ since ${\mathcal{Y}}_{F}$ is finite-
dimensional; we must have $\eta_{F}=\sum_{x\in F}c_{x}\xi_{x}$.

Suppose $\|\zeta\|\leq M$ for all $\zeta\in A_{1}$. It follows that for each
finite subset $F$ of $X$, $\|\zeta\|\leq M$ for all $\zeta\in A_{1}^{F}$; so
$\sum_{x\in F}|c_{x}|^{2}\leq M^{2}$ for any $F$. Thus $\sum_{x\in
X}|c_{x}|^{2}\leq M^{2}$, so the $c_{x}$ define a vector

$\eta=\sum_{x\in X}c_{x}\xi_{x}\in\ell^{2}(X)\ .$

To show $\eta\in\cap_{n}A_{n}$, let $\epsilon>0$, and fix $m$ with
$\operatorname{diam}(A_{m})\leq\epsilon$. Let $n\geq m$. If $F$ is a finite
subset of $X$, we have
$\operatorname{diam}(\overline{A_{n}^{F}})=\operatorname{diam}(A_{n}^{F})\leq\epsilon$,
so, for any $\zeta\in A_{n}$,

$\|P_{F}(\zeta)-\eta_{F}\|^{2}=\sum_{x\in
F}|\langle\zeta,\xi_{x}\rangle-\langle\eta,\xi_{x}\rangle|^{2}=\sum_{x\in
F}|\langle\zeta-\eta,\xi_{x}\rangle|^{2}\leq\epsilon^{2}\ .$

Thus $\|\zeta-\eta\|\leq\epsilon$ for all $\zeta\in A_{n}$. This is true for
every $\epsilon>0$, for all sufficiently large $n$. Since the $A_{n}$ are
decreasing and closed, $\eta\in\cap_{n}A_{n}$.

##### 3.0.5

If $X$ is a set, $\ell^{2}(X)$ has a canonical orthonormal basis
$\\{\xi_{x}:x\in X\\}$, where $\xi_{x}$ is the characteristic (indicator)
function of $\\{x\\}$. If $\eta\in\ell^{2}(X)$, then we have

$\eta=\sum_{x\in X}\eta(x)\xi_{x}$

in the sense that the sums over finite subsets converge in norm to $\eta$.
Every square-summable series occurs: if $\\{c_{x}:x\in X\\}$ is a square-
summable set of complex numbers, there is a unique $\eta\in\ell^{2}(X)$ with
$c_{x}=\langle\eta,\xi_{x}\rangle$ for all $x\in X$.

Conversely:

##### 3.0.6

Proposition. Let ${\mathcal{H}}$ be a Hilbert space with an orthonormal basis
$\\{\zeta_{x}:x\in X\\}$ for some set $X$. Then the map
$\zeta_{x}\mapsto\xi_{x}$ gives an isometric isomorphism from ${\mathcal{H}}$
onto $\ell^{2}(X)$.

Proof: Let ${\mathcal{H}}_{0}$ be the dense subspace of ${\mathcal{H}}$ of
finite linear combinations of the $\zeta_{x}$. The map
$\zeta_{x}\mapsto\xi_{x}$ defines an isometric linear map of
${\mathcal{H}}_{0}$ onto a dense subspace of $\ell^{2}(X)$. Since
$\ell^{2}(X)$ is $\sigma$-complete, this map extends (uniquely) to an
isometric linear map $T$ from ${\mathcal{H}}$ onto $\ell^{2}(X)$ by 2.0.7.

Thus the $\ell^{2}(X)$ for various $X$ are universal models of Hilbert spaces
with orthonormal bases.

##  4\. Finite-Dimensional and Separable Hilbert Spaces

##### 4.0.1

No Choice is needed to prove the standard results of finite-dimensional linear
algebra. If ${\mathcal{H}}$ is a finite-dimensional inner product space
(finite-dimensional means spanned by a finite set of vectors), then:

  1. (i)

Every set of orthonormal vectors can be expanded to an orthonormal basis. In
particular, ${\mathcal{H}}$ has an orthonormal basis.

  2. (ii)

${\mathcal{H}}$ has a well-defined orthogonal dimension $n$, which coincides
with its linear dimension.

  3. (iii)

Every linearly independent set can be orthonormalized (Gram-Schmidt).

In addition, we have (in ZF):

  1. (iv)

${\mathcal{H}}$ is $\sigma$-complete.

  2. (v)

The closed unit ball of ${\mathcal{H}}$ is compact.

##### 4.0.2

There is a converse to (v): if ${\mathcal{H}}$ is an infinite-dimensional
inner product space, then for any $n$ there is an orthonormal set of vectors
in ${\mathcal{H}}$ of cardinality $n+1$ made by Gram-Schmidt, so the closed
unit ball cannot be covered by $n$ open balls of radius $\frac{1}{\sqrt{2}}$
or less, and thus the closed unit ball of ${\mathcal{H}}$ is not totally
bounded.

(No Choice is needed to show that a compact metric space is $\sigma$-complete
and totally bounded, but the converse requires some Choice, cf. [Bla23,
XII.16.12.9].)

###  4.1. Separable Hilbert Spaces

##### 4.1.1

Most of the results about finite-dimensional Hilbert spaces extend to
separable Hilbert spaces (every finite-dimensional Hilbert space is
separable), due to the Gram-Schmidt orthogonalization process where any
sequence of vectors in an inner product space can be orthogonalized into an
orthonormal sequence (or finite sequence) with the same span. The procedure is
inductive (recursive) and requires no Choice and, indeed, no completeness.
Using Gram-Schmidt plus previous results, we have:

##### 4.1.2

Theorem. Let ${\mathcal{H}}$ be a separable Hilbert space. Then

  1. (i)

${\mathcal{H}}$ has a countable orthonormal basis.

  2. (ii)

Every closed subspace of ${\mathcal{H}}$ is separable (hence has a countable
orthonormal basis).

  3. (iii)

Every orthonormal set in ${\mathcal{H}}$ can be expanded into an orthonormal
basis.

  4. (iv)

Any two orthonormal bases for ${\mathcal{H}}$ have the same cardinality
(finite or $\aleph_{0}$).

Conversely, any Hilbert space with a countable orthonormal basis is separable.

Proof: (i) is exactly the Gram-Schmidt process. For (ii), if ${\mathcal{Y}}$
is a closed subspace of ${\mathcal{H}}$, then there is a projection of norm
one from ${\mathcal{H}}$ onto ${\mathcal{Y}}$ (2.0.5), and a continuous image
of a separable space is separable. For (iii), let
${\mathcal{S}}=\\{\xi_{x}:x\in X\\}$ be an orthonormal set in ${\mathcal{H}}$,
and let ${\mathcal{Y}}$ be the closed linear span of ${\mathcal{S}}$. Then
${\mathcal{Y}}^{\perp}$ is a separable Hilbert space, hence has a (countable)
orthonormal basis ${\mathcal{B}}$, and ${\mathcal{B}}\cup{\mathcal{S}}$ is an
orthonormal basis for ${\mathcal{H}}$. For the converse, finite linear
combinations of the basis vectors with coefficients in
$\mathbb{Q}+\mathbb{Q}i$ are dense, and can be effectively enumerated without
any Choice.

(iv) Let ${\mathcal{H}}$ be a separable Hilbert space, and $\\{\xi_{x}:x\in
X\\}$ be an orthonormal set in ${\mathcal{H}}$. Let
$\\{\eta_{n}:n\in\mathbb{N}\\}$ be a countable dense set in ${\mathcal{H}}$.
For each $x\in X$, let $n(x)$ be the smallest $n$ for which
$\|\eta_{n}-\xi_{x}\|<\frac{1}{2}$. Then $x\to n(x)$ is injective, so $X$ is
countable. If ${\mathcal{H}}$ is finite-dimensional, an orthonormal basis for
${\mathcal{H}}$ is a Hamel (vector space) basis for ${\mathcal{H}}$, and the
vector space dimension of a finite-dimensional vector space is well defined by
elementary linear algebra (no Choice needed). If ${\mathcal{H}}$ is infinite-
dimensional, then every orthonormal basis for ${\mathcal{H}}$ is countably
infinite.

Theorem 4.1.2 is about the existence of a well-ordered basis rather than
separability.

##### 4.1.3

Theorem. Suppose that $\kappa$ is a well-ordered cardinal. Let ${\mathcal{H}}$
be a Hilbert space with a dense subset of cardinality $\kappa$. Then

  1. (i)

${\mathcal{H}}$ has an orthonormal basis of cardinality $\lambda$, for some
$\lambda\leq\kappa$.

  2. (ii)

Every closed subspace of ${\mathcal{H}}$ has orthonormal basis of cardinality
$\leq\kappa$.

  3. (iii)

Every orthonormal set in ${\mathcal{H}}$ can be expanded into an orthonormal
basis.

  4. (iv)

Any two orthonormal bases for ${\mathcal{H}}$ have the same cardinality.

The proof of this theorem is given below after a definition and a lemma.

##### 4.1.4

Definition.The following describes the _transfinite Gram–Schmidt process_.
Suppose that $\kappa$ is an ordinal and $\eta_{\alpha}$, for $\alpha<\kappa$,
are nonzero vectors in a Hilbert space. Define vectors $\zeta_{\alpha}$, for
$\alpha<\kappa$, by transfinite recursion as follows.

First let $\zeta_{0}=\eta_{0}\|\eta_{0}\|^{-1}$. If $\zeta_{\alpha}$, for
$\alpha<\beta$, have been determined, then consider the projection of
$\xi_{\beta}$ to the closed linear span of $\zeta_{\alpha}$, for
$\alpha<\beta$:

(1)
$\xi_{\beta}=\sum_{\alpha<\beta}\langle\eta_{\beta},\zeta_{\alpha}\rangle\zeta_{\alpha}.$

If $\eta_{\beta}-\xi_{\beta}$ is nonzero, let
$\zeta_{\beta}=(\eta_{\beta}-\xi_{\beta})\|\eta_{\beta}-\zeta_{\beta}\|^{-1}$.
Otherwise let $\zeta_{\beta}=0$.

##### 4.1.5

Lemma. If $\zeta_{\alpha}$, for $\alpha<\kappa$, are vectors in a Hilbert
space indexed by an ordinal, then the transfinite Gram–Schmidt process results
in a well-ordered orthonormal basis for the closed linear span of these
vectors.

Proof: The proof is analogous to one in the finitary case. At the limit
stages, Bessel’s inequality implies that the finite partial sums of the right-
hand side of (1) converge. In the resulting sequence $(\zeta_{\alpha})$, the
non-zero vectors are well-ordered by some ordinal $\kappa^{\prime}\leq\kappa$.

Notably, the use of completeness at limit stages of the transfinite
Gram–Schmidt process is necessary even if the AC holds, if $H$ is a non-
separable (not necessarily $\sigma$-complete) inner product space then it may
not contain an orthonormal basis (see [Dix53]).

Proof of Theorem 4.1.3: The proof is analogous to the proof of Theorem 4.1.2.

(i) Let $\eta_{\alpha}$, for $\alpha<\kappa$, be an enumeration of a dense
subset of ${\mathcal{H}}$. By applying the transfinite Gram–Schmidt process
and using Lemma 4.1.5 we obtain an ordinal $\lambda\leq\kappa$ such that
${\mathcal{H}}\cong\ell^{2}(\lambda)$. By re-enumerating, we can assume
$\lambda$ is a cardinal.

(ii) If ${\mathcal{H}}_{0}$ is a closed subspace of ${\mathcal{H}}$, consider
the image of a dense subset of ${\mathcal{H}}$ under the projection to
${\mathcal{H}}_{0}$ and apply (i).

(iii) Apply (ii) to the orthogonal complement of the span of the given
orthonormal set.

(iv) By (ii), we need to prove that if $\kappa$ and $\lambda$ are well-ordered
cardinals and $\ell^{2}(\kappa)$ and $\ell^{2}(\lambda)$ are isomorphic, then
$\kappa=\lambda$. This is elementary if $\kappa$ is finite. For the infinite
case, by expanding each basis vector in terms of the other basis and using
that the support of a vector is countable (6.0.2, and we are working in a
well-ordered set), we obtain $\kappa\leq\aleph_{0}\cdot\lambda$ and
$\lambda\leq\aleph_{0}\cdot\kappa$. But if $\kappa$ is a well-ordered infinite
cardinal, then $\kappa=\aleph_{0}\cdot\kappa$.

##  5\. Dedekind-Finite Sets of Various Flavors

In the absence of CC, there can be bizarre sets which are infinite but “almost
finite.” In this section we survey various types of such sets which exist in
some models of ZF. For more details see e.g. [Jec08], also §10.

We start by making precise the notion of cardinality in the context of ZF.

###  5.1. Vanilla Flavor

##### 5.1.1

Definition. The _cardinality_ of a set $X$, denoted $|X|$, is the equivalence
class of all $Y$ such that there is a bijection between $X$ and $Y$. (This is
a proper class, but by Scott’s trick it can be identified with a set.) AC is
equivalent to the assertion that every set can be well-ordered, and one writes
$\kappa=|X|$ where $\kappa$ is the least ordinal for which such bijection
exists. We take such $\kappa$ (if it exists) as the canonical representative
of $|X|$, and refer to it as a _well-ordered cardinal_.

We will use symbols $\kappa,\lambda$ to denote cardinals, with understanding
that they are not necessarily well-orderable.

The cardinality of $X$ is less or equal than the cardinality of $Y$, in
symbols $|X|\leq|Y|$, if there is an injection from $X$ to $Y$. If $Y$ is
nonempty, then $|Y|\leq|X|$ implies that there is a surjection from $X$ onto
$Y$. (To see this, fix an injection $f\colon Y\to X$ and $y_{0}\in Y$. Define
$g\colon X\to Y$ by $g(f(y))=y$ and $g(x)=y_{0}$ if $x$ is not in the range of
$g$.)

##### 5.1.2

We use the following notation for a set $X$. Denote by ${\mathcal{P}}(X)$ the
power set of $X$, the set of all subsets of $X$. Let $\operatorname{Fin}(X)$
be the subset of ${\mathcal{P}}(X)$ consisting of all finite subsets of $X$
(set theorists often denote this by $[X]^{<\omega}$). $\operatorname{Seq}(X)$
is the set of all finite sequences of distinct elements of $X$ (including the
“empty sequence”).

We have $|X|\leq|\operatorname{Fin}(X)|\leq|{\mathcal{P}}(X)|$ and, of course,
$|X|<|{\mathcal{P}}(X)|=2^{|X|}$; and $|X|\leq|\operatorname{Seq}(X)|$. There
is an obvious surjective map from $\operatorname{Seq}(X)$ to
$\operatorname{Fin}(X)$. We will have use for the following well-known fact, a
theorem of ZF (i.e. no Choice is needed):

##### 5.1.3

Proposition. Let $\mathbb{N}$ be the set of natural numbers. Then
$\operatorname{Seq}(\mathbb{N})$ and $\operatorname{Fin}(\mathbb{N})$ are
countably infinite (equipotent with $\mathbb{N}$).

##### 5.1.4

We write $|Y|\leq^{*}|X|$ if there is a surjective function from $X$ to $Y$
(or if $Y=\emptyset$). We have $|Y|\leq|X|\Rightarrow|Y|\leq^{*}|X|$, and the
converse is true under AC but not otherwise (whether this is equivalent to AC
is an old open problem).

The following well-known notion is central to our study.

##### 5.1.5

Definition. A set $X$ is Dedekind-finite if there is no bijection from $X$ to
a proper subset of $X$, i.e. every injective function from $X$ to $X$ is
surjective. Otherwise $X$ is Dedekind-infinite. A set is _finite_ if its
cardinality is equal to the cardinality of some natural number. A cardinal
$\kappa$ is Dedekind-finite if it is the cardinal of a Dedekind-finite set.
Say a set $X$ or a cardinal $\kappa$ is DF if it is infinite and Dedekind-
finite.

While every finite set is Dedekind-finite, the converse fails in some models
of ZF (Proposition 10.0.1). Whether a set $X$ is Dedekind-finite or Dedekind-
infinite depends only on the cardinal of $X$: if $X$ and $Y$ are equipotent
and one is Dedekind-finite, so is the other. Any subset of a Dedekind-finite
set is Dedekind-finite.

The next result is fundamental:

##### 5.1.6

Proposition. Let $X$ be a set. Then $X$ is Dedekind-infinite if and only if
there is an injective function from $\mathbb{N}$ to $X$ (i.e. there is a
sequence of distinct elements of $X$). A cardinal $\kappa$ is Dedekind-
infinite if and only if $\aleph_{0}\leq\kappa$.

Proof: If $|Y|\leq|X|$ and $Y$ is Dedekind-infinite, then so is $X$; in
particular, if there is an injection from $\mathbb{N}$ into $X$ then $X$ is
Dedekind-infinite. Suppose that $X$ is Dedekind-infinite and let $f\colon X\to
X$ be an injection that is not a surjection. Choose $x\in X\setminus f[X]$.
Writing $f^{n}$ for the $n$-fold iteration of $f$ and $f^{0}$ for the identity
map, $g\colon\mathbb{N}\to X$ defined by $g(n)=f^{n}(x)$ is an injection from
$\mathbb{N}$ into $X$.

The second sentence is a restatement of the first.

##### 5.1.7

Corollary. If $X$ is a proper subset of $Y$ and $X$ is DF, then $|X|<|Y|$.

If the CC is assumed, a set is Dedekind-infinite if and only if it is
infinite, so a set is Dedekind-finite if and only if it is finite; thus DF
sets are incompatible with the CC.

The next proposition gives some alternate characterizations of Dedekind-
infiniteness:

##### 5.1.8

Proposition. Let $X$ be a set, and $\kappa=|X|$. The following are equivalent:

  1. (i)

$X$ is Dedekind-infinite, i.e. there is a proper subset $Y$ of $X$ which is
equipotent with $X$.

  2. (ii)

$\kappa=\kappa+\aleph_{0}$.

  3. (iii)

$\kappa=\kappa+1$.

  4. (iv)

If $\kappa=\lambda+1$, then $\lambda=\kappa$.

  5. (v)

If $Y$ is any set obtained from $X$ by adding or removing finitely many
points, then $|Y|=\kappa$.

##### 5.1.9

If we have one DF cardinal, we have many. If $X$ is a DF set and
$n\in\mathbb{N}$, let $F_{n}$ be an $n$-element subset of $X$, and
$\kappa_{n}=|X\setminus F_{n}|$ (note that we can choose $F_{n}$ for any
particular $n$, or for finitely many $n$, but we may not choose $F_{n}$
simultaneously for all $n$, and in particular we cannot choose the $F_{n}$ so
that $F_{n}\subseteq F_{n+1}$ for all $n$). It is easily seen that the
cardinality of $X\setminus F_{n}$ depends only on $n$ and not on the choice of
$F_{n}$, so $\kappa_{n}$ is well defined for each $n$. We clearly have
$\kappa_{n+1}\leq\kappa_{n}$ for each $n$, and we have
$\kappa_{n+1}\neq\kappa_{n}$ since $X\setminus F_{n}$ is DF. Thus there is a
strictly decreasing sequence of DF cardinals smaller than $|X|$, which
dramatically contradicts the Well-Ordering Principle (there is even a
collection of DF cardinals order-isomorphic to $\mathbb{R}$, cf. [Bla23,
II.9.5.21]). More generally, if $Y$ is any proper subset of $X$, then
$|Y|<|X|$. DF cardinals do not even form a set in general: in fact, there are
models of ZF in which every set is the image of a DF set. This holds in the
‘improved Morris model’ of [Kar19, §5.8], and we briefly explain why. It
suffices to prove that every rank-initial segment of the universe,
$V_{\alpha}$ is the image of a Dedekind-finite set. [Kar19, Proposition 4.2]
shows that the set of generic branches is Dedekind-finite, and by construction
it can be mapped onto $V_{\alpha}$; as the forcing described in [Mon75] is
iterated through the universe, in the final model each $V_{\alpha}$ is the
image of some Dedekind-finite set.

##### 5.1.10

Proposition. A finite Cartesian product of Dedekind-finite sets is Dedekind-
finite. A finite union of Dedekind-finite sets is Dedekind-finite.

Proof: If $(x_{j})$ is a sequence of distinct elements of
$X_{1}\times\cdots\times X_{n}$, then for some $k$, $1\leq k\leq n$, the
sequence of $k$’th coordinates of the $x_{j}$ must contain infinitely many
distinct elements of $X_{k}$, hence $X_{k}$ is not Dedekind-finite. Similarly,
if $X=\cup_{k=1}^{n}X_{k}$, and $(x_{j})$ is a sequence of distinct elements
of $X$, then infinitely many of the $x_{j}$ must be in some $X_{k}$, so
$X_{k}$ is not Dedekind-finite.

In particular, if $X$ is Dedekind-finite, then $X^{n}$ and $X\times F$ are
Dedekind-finite for any $n$ and any finite set $F$. These sets all have
different cardinalities by 5.1.9. By repeating these constructions and taking
subsets, a very large collection of Dedekind-finite cardinals can be obtained.

###  5.2. Other Flavors

Here are some important variations on Dedekind-infiniteness:

##### 5.2.1

Definition. Let $X$ be a set. Then

  1. (i)

$X$ is Cohen-infinite if $\operatorname{Fin}(X)$ is Dedekind-infinite, i.e.
there is a sequence of distinct finite subsets of $X$.

  2. (ii)

$X$ is weakly Dedekind-infinite ${\mathcal{P}}(X)$ is Dedekind-infinite, i.e.
there is a sequence of distinct subsets of $X$.

  3. (iii)

$X$ is dually Dedekind-infinite if there is a surjective function from $X$ to
$X$ which is not injective.

##### 5.2.2

We have ($X$ Dedekind-infinite) $\Rightarrow$ ($X$ Cohen-infinite)
$\Rightarrow$ (X weakly Dedekind-infinite) $\Rightarrow$ (X infinite) and ($X$
Dedekind-infinite) $\Rightarrow$ ($X$ dually Dedekind-infinite).

##### 5.2.3

Definition. Let $X$ be a set. Then

  1. (i)

$X$ is Cohen-finite if $\operatorname{Fin}(X)$ is Dedekind-finite.

  2. (ii)

$X$ is power Dedekind-finite if ${\mathcal{P}}(X)$ is Dedekind-finite, i.e. if
$X$ is not weakly Dedekind-infinite.

  3. (iii)

$X$ is dually Dedekind-finite if it is not dually Dedekind-infinite.

We use the following abbreviations:

  1. CF means “infinite and Cohen-finite.”

  2. PF means “infinite and power Dedekind-finite.”

  3. DDF means “infinite and dually Dedekind-finite.”

We use the term power Dedekind-finite instead of weakly Dedekind-finite for
the negation of weakly Dedekind-infinite since the condition of being power
Dedekind-finite is considerably stronger than being Dedekind-finite (5.2.14).

The terminology Cohen-finite is justified since by [KS20, §6], a set $X$ is CF
if and only if the forcing $\operatorname{Add}(X,1)$ for adding a subset of
$X$ by finite conditions (‘adding a Cohen subset of $X$’) has the property
that every statement in the language of forcing is decided by a finite
predense set. This property is also equivalent to the assertion that $X$
remains DF in the forcing extension by $\operatorname{Add}(X,1)$.

Any subset of a Cohen-finite [resp. power Dedekind-finite, dually Dedekind-
finite] is Cohen-finite [resp. power Dedekind-finite, dually Dedekind-finite].
The definitions easily imply that ($X$ PF) $\Rightarrow$ ($X$ CF)
$\Rightarrow$ (X DF) and ($X$ DDF) $\Rightarrow$ ($X$ DF), and 5.2.10 implies
($X$ PF) $\Rightarrow$ ($X$ DDF). None of these implications can be reversed
(5.2.11 shows that DF sets are not necessarily DDF, 5.7.2 shows that some DF
sets are not CF, and combining 5.2.5 and 5.5.2 shows that CF sets are not
always PF). In addition to that, ($X$ DDF) does not imply ($X$ CF), as shown
in 5.7.3, nor ($X$ CF) implies ($X$ DDF), as shown by 5.2.11 when considered
in Cohen’s model.

The next result, due to Kuratowski [Tar24, pp.94–95], gives alternate
characterizations of weak Dedekind-infiniteness. See e.g. [Bla23, II.9.5.7]
for a proof.

##### 5.2.4

Theorem. Let $\kappa$ be a cardinal, and $X$ a set with $|X|=\kappa$. The
following are equivalent:

  1. (i)

$\aleph_{0}\leq^{*}\kappa$, i.e. there is a sequence of pairwise disjoint
nonempty subsets of $X$.

  2. (ii)

$\aleph_{0}\leq 2^{\kappa}$, i.e. $X$ is weakly Dedekind-infinite.

  3. (iii)

$2^{\aleph_{0}}\leq 2^{\kappa}$.

##### 5.2.5

Proposition. Let $X$ be an infinite second-countable Hausdorff space (e.g. an
infinite subset of $\mathbb{R}$). Then $X$ is weakly Dedekind-infinite.

Proof: We will prove that $X$ maps onto $\mathbb{N}$. Fix an enumeration
$U_{n}$, for $n\in\mathbb{N}$, of a base for the topology of $X$ with
$U_{0}=X$. Suppose for a moment that all points of $X$ are isolated. Then
$f(x)=\min\\{n\mid U_{n}\cap X=\\{x\\}\\}$ defines a surjection from $X$ onto
an infinite subset of $\mathbb{N}$.

We may therefore assume that $X$ has an accumulation point, $x$. Recursively
define $m(n)\in\mathbb{N}$, for $n$ in $\mathbb{N}$, by $m(0)=0$, and if
$m(n)$ has been defined let

$m(n+1)=\min\\{j\mid U_{j}\subseteq U_{m(n)},\text{ $x\in U_{j}$, and
}(U_{m(n)}\setminus U_{j})\cap X\neq\emptyset\\}.$

Then the function that sends $X\cap(U_{m(n)}\setminus U_{m(n+1)})$ to $n$ is a
surjection.

See [Sie65] or [Bla23, II.9.5.6] for another proof. Subsets of $\mathbb{R}$
are special among all sets (in the absence of AC) in part because they can be
totally ordered.

##### 5.2.6

Proposition. Let $X$ be a CF set. Then every subset of $X$ that can be covered
by a countable union of finite sets is finite.

Proof: Let $A=\cup_{n=1}^{\infty}A_{n}\subseteq X$, with each $A_{n}$ finite.
If $A$ is infinite, then there must be infinitely many distinct $A_{n}$, hence
a sequence of distinct $A_{n}$, so $\operatorname{Fin}(X)$ is not Dedekind-
finite.

##### 5.2.7

The analog of 5.2.4 (ii) $\Rightarrow$ (i) (which is quite subtle) for finite
subsets is easily proved by induction. Thus a set is Cohen-finite if and only
it it does not contain a sequence of pairwise disjoint nonempty finite
subsets.

##### 5.2.8

Proposition. If $X$ can be linearly ordered, then it is DF if and only if it
is CF.

Proof: Only the direct implication requires a proof, and we will prove the
contrapositive. Assume $X$ is linearly ordered by $\prec$ and $F(n)$ is a
sequence of distinct nonempty finite subsets of $X$. By the pigeonhole
principle, $\bigcup_{n}F(n)$ is infinite. We can order $\bigcup_{n}F(n)$ by
(writing $n(x)=\min\\{n:x\in F(n)\\}$)

$x\prec_{1}y\text{ if and only if $n(x)<n(y)$ or $n(x)=n(y)$ and $x\prec y$}.$

Since each $F(n)$ is finite, this is a well-ordering of the infinite subset
$\bigcup_{n}F(n)$ of $X$. Therefore $X$ is not Dedekind-finite.

We have the following analog of 5.1.10:

##### 5.2.9

Proposition. (i) A finite Cartesian product of Cohen-finite sets is Cohen-
finite. A finite union of Cohen-finite sets is Cohen-finite.

(ii) A finite Cartesian product of power Dedekind-finite sets is power
Dedekind-finite. A finite union of power Dedekind-finite sets is power
Dedekind-finite.

Proof: The proof of (i), and (ii) for unions, is straightforward along the
lines of the proof of 5.1.10. For products of power Dedekind-finite sets, it
suffices to show the result for the Cartesian product of two power Dedekind-
finite sets $X$ and $Y$. Suppose $f$ is a surjective function from $X\times Y$
to $\mathbb{N}$. For each $x\in X$, the restriction of $f$ to $\\{x\\}\times
Y$ has finite range $A_{x}$ since $\\{x\\}\times Y$ is power Dedekind-finite.
Thus $x\mapsto A_{x}$ is a function from $X$ to
$\operatorname{Fin}(\mathbb{N})$, which must have infinite range since the
union of the $A_{x}$ is $\mathbb{N}$. But $\operatorname{Fin}(\mathbb{N})$ is
countable (5.1.3), contradicting that $X$ is power Dedekind-finite.

##### 5.2.10

An interesting consequence is that if $X$ and $Y$ are PF, then there is not a
sequence of distinct relations from $X$ to $Y$, and in particular there is no
sequence of distinct functions from $X$ to $Y$.

If $X$ is dually Dedekind-infinite, and $f:X\to X$ is surjective but not
injective, then $(f^{n})$ is a sequence of distinct functions from $X$ to $X$,
and hence $X$ cannot be power Dedekind-finite. Thus $X$ PF $\Rightarrow$ $X$
DDF.

##### 5.2.11

If $X$ is DF, then $\operatorname{Seq}(X)$ is also DF [a sequence of distinct
elements of $\operatorname{Seq}(X)$ can be concatenated and the pigeonhole
principle implies that it contains a subsequence of distinct elements of $X$].
However, if $X$ is infinite, $\operatorname{Seq}(X)$ is not DDF, since there
is a surjective and noninjective function from $\operatorname{Seq}(X)$ to
itself sending the empty sequence to itself and dropping the first term in
each nonempty sequence. Thus, if there exists a DF set, there exists a DF set
which is not DDF.

##### 5.2.12

The situation with $\operatorname{Fin}(X)$ is quite different, even though
there is a finite-to-one surjective function from $\operatorname{Seq}(X)$ to
$\operatorname{Fin}(X)$. If $X$ is DF, then $\operatorname{Fin}(X)$ can be DF,
even DDF, or not DF. See [KS20] for a study. We do have the following (for
$n\geq 1$ let $\operatorname{Fin}^{n}(X)$ be the $n$-th iterate of the
operation $\operatorname{Fin}$).

##### 5.2.13

Proposition. If $X$ is a CF set, then $\operatorname{Fin}^{n}(X)$ is CF for
all $n\geq 1$.

Proof: Suppose $X$ is CF. In order to prove that $\operatorname{Fin}(X)$ is
CF, towards contradiction let $F_{n}$ be an infinite sequence of disjoint
nonempty subsets of $\operatorname{Fin}(X)$. Let $G_{n}=\bigcup F_{n}$. Then
$G_{n}$ are finite subsets of $X$, and since $X$ is CF there is
$G\in\operatorname{Fin}(X)$ such that $G_{n}=G$ for infinitely many $n$.
However, both ${\mathcal{P}}(G)$ and ${\mathcal{P}}({\mathcal{P}}(G))$ are
finite, contradicting the assumption that $F_{n}$ were pairwise distinct.

This proves that $\operatorname{Fin}(X)$ is CF, and it also provides the
inductive step for the proof that $\operatorname{Fin}^{n}(X)$ is CF for all
$n$.

##### 5.2.14

If $X$ is infinite, then ${\mathcal{P}}(\operatorname{Fin}(X))$ (and a
fortiori ${\mathcal{P}}({\mathcal{P}}(X))$) is always Dedekind-infinite: the
sets $A_{n}$ of $n$-element subsets of $X$ form a sequence of distinct
elements in ${\mathcal{P}}(\operatorname{Fin}(X))$. However, if $X$ is a DF
subset of $\mathbb{R}$, then $X$ is CF (Proposition 5.2.8) and therefore
$\operatorname{Fin}(X)$ is DF and even CF (5.2.13), although $X$ is not PF
(5.2.5).

###  5.3. Amorphous Sets

We now describe some particularly strange DF sets whose existence cannot be
refuted in ZF set theory.222More precisely, if ZF has a model, then it has a
model with an amorphous set. This also applies to other flavors of DF sets
discussed in §5.1 and §5. The reason why it is necessary to assume that ZF has
a model is that By Gödel’s Incompleteness Theorem, in ZF there is no proof
that there is a model of ZF… unless ZF is inconsistent.

##### 5.3.1

Definition. An infinite set $X$ is amorphous if $X$ cannot be written as a
disjoint union of two infinite subsets, i.e. every subset of $X$ is either
finite or cofinite.

##### 5.3.2

Any infinite subset of an amorphous set is amorphous. If $X$ is an amorphous
set, $Y$ an infinite set, and there is a surjective function $f:X\to Y$, then
$Y$ is also amorphous [if $Y$ is the disjoint union of $Y_{1}$ and $Y_{2}$,
then $X$ is the disjoint union of $f^{-1}(Y_{1})$ and $f^{-1}(Y_{2})$]. In
particular, if $X$ is weakly Dedekind-infinite, since $\mathbb{N}$ is not
amorphous $X$ cannot be amorphous, so if $X$ is amorphous then $X$ is PF. Thus
amorphous sets are Dedekind-finite in a strong sense. In particular, amorphous
sets are incompatible with the CC.

As an example of how strange an amorphous set is, we have:

##### 5.3.3

Proposition. An amorphous set cannot be totally ordered.

See e.g. [Bla23, II.9.5.13] for a proof.

###  5.4. Permanence Properties of CF Sets

##### 5.4.1

For a relation $R\subseteq X\times Y$ consider its horizontal and vertical
sections, $R_{x}=\\{y\in Y:(x,y)\in R\\}$ and $R^{y}=\\{x\in X:(x,y)\in R\\}$.

##### 5.4.2

Lemma. Suppose that $X$ and $Y$ are sets and at least one of them is CF. Also
suppose that $R\subseteq X\times Y$, and all horizontal and vertical sections
of $R$ are finite. Then there are $Z$ and partitions $X=\bigsqcup_{z\in
Z}X_{z}$, $Y=\bigsqcup_{z\in Z}Y_{z}$ into finite sets such that
$\bigcup\\{R_{x}:x\in X_{z}\\}\subseteq Y_{z}$ and $\bigcup\\{R^{y}:y\in
Y_{z}\\}\subseteq X_{z}$ for all $z\in Z$. This implies that both sets are CF.

Proof: Note that all sections of $R$ are finite if and only if all sections of
the inverse relation $R^{-1}$ are finite. It therefore suffices to prove the
case when $X$ is CF, since otherwise we can exchange the roles of $X$ and $Y$
and consider $R^{-1}$ instead of $R$.

Fix for a moment $x\in X$. Recursively define $F(n)\subseteq X$ and
$G(n)\subseteq Y$ as follows. Let $F(0)=\\{x\\}$, and for all $n$ let

$\displaystyle G(n)$ $\displaystyle=\bigcup\\{R_{x^{\prime}}:x^{\prime}\in
F(n)\\},$ $\displaystyle F(n+1)$ $\displaystyle=\bigcup\\{R^{y}:y\in G(n)\\}.$

Clearly $F(n)\subseteq F(n+1)$ and $G(n)\subseteq G(n+1)$ for all $n$. By
induction on $n$ one proves that all $F(n)$ and all $G(n)$ are finite, being
unions of finitely many finite sets. Since $X$ is CF, there exists $m=m(x)$
such that $F(m)=F(n)$ for all $n\geq m$. This implies that $G(m)=G(n)$ for all
$n\geq m$. Let $F[x]=F(m)$ and $G[x]=G(m)$.

Fix $x^{\prime}\in F[x]$. By induction on the minimal $n$ such that
$x^{\prime}\in F(n)$, one proves that $F[x^{\prime}]=F[x]$, and therefore
$G[x^{\prime}]=G[x]$. Let $Z=\\{F[x]:x\in X\\}$. For every $z\in Z$ there is a
unique finite subset $G[z]$ of $Y$ which satisfies $z=G[x]$ for some (every)
$x$ such that $F[x]=z$. Then $X_{z}=z$ and $Y_{z}=G[z]$ define partitions of
$X$ and $Y$ into finite sets. By construction, $\bigcup\\{R_{x}:x\in
X_{z}\\}\subseteq Y_{z}$ and $\bigcup\\{R^{y}:y\in Y_{z}\\}\subseteq X_{z}$
for all $z\in Z$.

To see that $Y$ is CF, fix an increasing sequence of finite subsets of $Y$,
$G(n)$. Then $F(n)=\bigcup_{y\in G(n)}R^{y}$ is an increasing sequence of
finite subsets of $X$, hence for some $m$ and all $n\geq m$ we have
$F(m)=F(n)$. Then $G(n)\subseteq\bigcup_{x\in F(m)}R_{x}$ for all $n$, hence
all $G(n)$ are subsets of a fixed finite set. This implies that the sequence
$G(n)$ eventually stabilizes. Since this sequence was arbitrary, $Y$ is CF.

##### 5.4.3

A function is _finite-to-one_ if the preimage of every element of its range is
finite. The following is a notable consequence of Lemma 5.4.2.

Lemma. Suppose $f\colon X\to Y$ is finite-to-one and surjective. Then $X$ is
CF if and only if $Y$ is CF. If $Y$ is DF so is $X$, but the converse does not
hold in general.

Proof: Take $R\subseteq X\times Y$ to be the graph of $f$. Then all vertical
sections of $R$ are singletons and all horizontal sections are finite. Thus
Lemma 5.4.2 applies to imply that $X$ is CF if and only if $Y$ is CF.

If there is an injection from $\mathbb{N}$ into $X$, then the composition of
this function with $f$ has an infinite range, and from it one defines an
injection from $\mathbb{N}$ into $Y$.

Finally, the Russell set (Definition 5.7.1) is DF and it clearly has
$\mathbb{N}$ as a two-to-one image.

###  5.5. Examples and Constructions

##### 5.5.1

The first example of a DF set was due to A. Fraenkel [Fra22], who, assuming a
certain extension ZFA of ZF is consistent (ZFA was later proved to be
equiconsistent with ZF) constructed a model of ZF containing an amorphous set
$X$. This $X$ is not only DF, but even PF.

##### 5.5.2

The most famous Dedekind-finite set was constructed by P. Cohen in his work on
the independence of the Axiom of Choice and the Continuum Hypothesis [Coh66]:
a canonical infinite Dedekind-finite subset $X$ of $\mathbb{R}$ in a certain
model of ZF. This $X$ is weakly Dedekind-infinite (5.2.5) and of course can be
totally ordered, hence has quite different properties from Fraenkel’s example
(although Cohen’s construction was modeled on Fraenkel’s). Also note that
these sets exist in different universes: Fraenkel’s model is a model with
atoms while Cohen’s model is a model of ZF.

###  5.6. Rigid Sets

Recall that a permutation of a set $X$ is a bijection from $X$ to $X$.

##### 5.6.1

Definition. An infinite set $X$ is rigid if every permutation of $X$ moves
only finitely many elements (i.e. is the identity on the complement of a
finite subset).

##### 5.6.2

Any set equipotent with a rigid set is rigid. A subset of a rigid set is rigid
(since a permutation of a subset extends to a permutation of the whole set).
Since $\mathbb{N}$ is not rigid (it has many permutations moving infinitely
many elements, even all elements), a rigid set must be Dedekind-finite. But
“most” Dedekind-finite sets are not rigid (e.g. if $X$ is any infinite set,
then $X\times Y$ is not rigid for any $Y$ with more than one element).

Rigid sets are interesting in the context of associated Hilbert spaces, since
a permutation of $X$ naturally defines a unitary operator on $\ell^{2}(X)$;
this unitary is a finite-rank perturbation of a scalar if and only if the
permutation moves only finitely many elements of $X$.

##### 5.6.3

Being rigid seems to be similar to being amorphous. But it turns out that the
notions are distinct. Say an infinite set $X$ is strongly rigid if in any
partition of $X$ into nonempty subsets, all but finitely many cells are
singletons, and strongly amorphous if it is amorphous and strongly rigid. A
strongly rigid set is rigid. It can be shown that Fraenkel’s set is strongly
amorphous, hence rigid, and that Cohen’s generic subset of $\mathbb{R}$ is
rigid, but it is not amorphous. There are amorphous sets containing infinitely
many pairwise disjoint two-element sets (§6.2.2), and such a set is not rigid
(just interchange each pair of socks). And there are rigid sets which can be
written as a countable disjoint union of three-element sets in some models of
ZF.

###  5.7. Russell Sets and Russell Cardinals

##### 5.7.1

Definition. A set $X$ is a Russell set if it is Dedekind-finite but can be
written as a countable union of pairwise disjoint two-element sets (called
pairs of Russell socks). A cardinal $\kappa$ is a Russell cardinal if it is
the cardinal of a Russell set.

##### 5.7.2

Note that a Russell set $X$ is DF but not CF, and is not amorphous. There is a
permutation of $X$ interchanging each pair of socks, so $X$ is not rigid
either. If $\kappa$ is a Russell cardinal, then $\kappa\pm 1$ is not a Russell
cardinal, but is DF and not CF. In fact, $\kappa\pm n$ is a Russell cardinal
if and only if $n$ is even. Russell socks are a graphic example of a countable
union of countable, even finite, sets which is not countable.

##### 5.7.3

If $X$ is a Russell set, then $X$ is DDF. To prove this, let
$X=\bigsqcup_{n\in\mathbb{N}}X_{n}$ be the decomposition of $X$ into pairs.
Suppose that $f\colon X\to X$ is a surjection. In order to prove that it is
injective, let

$A=\\{m\mid(\exists k\geq 1)(\exists n)f^{k}[X_{n}]\text{ is a proper subset
of }X_{m}\\}.$

We claim that $A$ is finite. Otherwise, for every $m\in A$ choose the
lexicographically minimal pair $k(m),n(m)$ such that $f^{k(m)}[X_{n(m)}]$ is a
proper subset of $X_{m}$. Let $x_{m}$ be the unique element of $X_{m}\setminus
f^{k(m)}[X_{n(m)}]$. Then $\\{x_{m}\mid m\in A\\}$ is a countably infinite
subset of $X$; contradiction.

For $m\in\mathbb{N}\setminus A$ let $g(m)$ be the minimal $n$ such that
$f[X_{n}]=X_{m}$. Note that $g\colon(\mathbb{N}\setminus A)\to\mathbb{N}$ is
an injection. Suppose for a moment that for some $m_{0}\in\mathbb{N}\setminus
A$ we have $g(m_{0})\neq m_{0}$. Fix $x_{0}\in X_{m_{0}}$, and for $k\geq 1$
let $x_{k}$ be the unique element of $X_{g^{k}(m_{0})}$ such that
$f^{k}(x_{k})=x_{0}$. Since $g$ is an injection, the sequence $(m_{k})$ is
infinite and again we have a countably infinite subset $\\{x_{k}\mid
k\in\mathbb{N}\\}$ of $X$; contradiction.

We have proven that the set $B=\\{m\in\mathbb{N}\setminus A\mid g(m)=m\\}$ is
cofinite in $\mathbb{N}$, and $f[X_{m}]=X_{m}$ for all $m\in B$. Therefore $f$
is an injection on the cofinite subset $\bigcup_{m\in B}X_{m}$ of $X$. Since
it is surjection, $f$ is an injection as required.

##### 5.7.4

There are models of ZF containing Russell sets; for example, the second
Cohen’s model ([Jec08, §5.4]). For a thorough and entertaining description of
Russell sets, including motivation for the sock terminology, see [HT06]; some
results described there graphically show that the world of DF cardinals (when
they exist at all) is even more bizarre than we have indicated (for example,
there is a family of $2^{\aleph_{0}}$ pairwise incomparable DF cardinals). For
one interesting observation, if one has a set of Russell socks and adds
another pair of socks, the _same_ number ($\aleph_{0}$) of pairs of socks, but
more socks (a strictly larger cardinality). The latter fact is a consequence
of Corollary 5.1.7.

##### 5.7.5

There are various models of ZF in which all the types of DF sets described
above occur (assuming there is a model for ZF at all, i.e. that ZF is
consistent); see §10. There is even a single model of ZF in which all such
sets occur, including counterexamples for the false implications between the
properties.

##### 5.7.6

Figure (1) shows the flavors of DF sets considered in this paper.

strongly amorphousamorphousstrongly rigidPFRussellCFDDFrigidDF Figure 1.
Implications between different flavors of Dedekind–finite sets discussed in
this section. See also Fig. 2 in §9.

##  6\. $\ell^{2}(X)$ For DF Sets

If $X$ is a DF set, the Hilbert space $\ell^{2}(X)$ has some unusual
properties depending on the flavor of DF.

##### 6.0.1

Definition. Let $X$ be a set, $\eta\in\ell^{2}(X)$, and
$\eta(x)=\langle\eta,\xi_{x}\rangle$ the $x$’th coordinate of $\eta$ for each
$x\in X$. The support of $\eta$ is $supp(\eta)=\\{x\in X:\eta(x)\neq 0\\}$.

For the following one needs to keep in mind that the union of a sequence of
finite sets need not be countable (see Definition 5.7.1).

##### 6.0.2

Proposition. Let $X$ be a set, $\eta\in\ell^{2}(X)$, and
$\eta_{x}=\langle\eta,\xi_{x}\rangle$ the $x$’th coordinate of $\eta$ for each
$x\in X$. Then $\operatorname{supp}(\eta)$ is the union of a sequence of
finite subsets of $X$.

Proof: Let $\epsilon>0$, and set $A_{\epsilon}=\\{x:|\eta(x)|\geq\epsilon\\}$.
If $B=\\{x_{1},\dots,x_{n}\\}$ is an $n$-element subset of $A_{\epsilon}$, set

$\zeta=\sum_{k=1}^{n}\eta(x_{k})\xi_{x_{k}}\ .$

then

$\|\eta\|^{2}\geq\|\zeta\|^{2}=\sum_{k=1}^{n}|\eta(x_{k})|^{2}\geq
n\epsilon^{2}$

so $n\leq\left(\frac{\|\eta\|}{\epsilon}\right)^{2}$. Thus $A_{\epsilon}$ is a
finite set. We have $\operatorname{supp}(\eta)=\cup_{n}A_{1/n}$.

Suppose $X$ is CF. We then have severe restrictions on the support of vectors
(note that (ii) of Proposition 6.0.3 says that the canonical orthonormal basis
of $\ell^{2}(X)$ is a Hamel basis; we will return to this observation in
6.2.7).333Parts of Proposition 6.0.3 have been proved in [Bru86] in a slightly
more general context.

##### 6.0.3

Proposition. For every set $X$ the following are equivalent.

  1. (i)

$X$ is Cohen-finite.

  2. (ii)

Every vector in $\ell^{2}(X)$ has finite support.

  3. (iii)

Every sequence of disjointly supported unit vectors in $\ell^{2}(X)$ is
finite.

  4. (iv)

Every sequence of vectors in $\ell^{2}(X)$ has finite common support.

  5. (v)

The closed unit ball of $\ell^{2}(X)$ is sequentially compact.

  6. (vi)

The standard orthonormal basis for $\ell^{2}(X)$ is a Hamel basis.

Proof: (i) implies (ii) follows immediately from 6.0.2 and 5.2.6, and (vi) is
obviously equivalent to (ii).

(ii) implies (iii): If $\eta_{n}$ are disjointly supported unit vectors, then
$\eta=\sum_{n=0}^{\infty}2^{-n-1}\eta_{n}$ is a unit vector with infinite
support.

(iii) implies (i): Suppose $X$ is not CF. Then there is a sequence of pairwise
disjoint nonempty finite subsets of $X$, $F(n)$ for $n\in\mathbb{N}$. For each
$n$ let $\eta_{n}=|F(n)|^{-1/2}\sum_{x\in F(n)}\xi_{x}$. These are unit
vectors with disjoint supports.

It remains to prove (iv) and (v) are equivalent to the first three statements.
(iv) implies (ii) is trivial.

(i) and (ii) together imply (iv): If $(\eta_{n})$ is a sequence of vectors in
$\ell^{2}(X)$, then by (ii) $(\operatorname{supp}(\eta_{n}))$ is a sequence of
finite subsets of $X$. Since $X$ is CF, this sequence can contain only
finitely many distinct terms.

(iv) implies (v): If $(\eta_{n})$ is a sequence in $\ell^{2}(X)$, then the
$\eta_{n}$ all lie in a finite-dimensional subspace of $\ell^{2}(X)$. But the
closed unit ball of a finite-dimensional normed vector space is compact (see
4.0.1 and 4.0.2).

(v) implies (iii): Suppose that (iii) fails. An infinite sequence of unit
vectors with disjoint supports does not have a norm-convergent subsequence
(e.g. because it weakly converges to $0$).

##### 6.0.4

Thus if $X$ is CF, the closed unit ball of $\ell^{2}(X)$ is a
$\sigma$-complete metric space which is sequentially compact but not compact
(not totally bounded, cf. 4.0.2).

##### 6.0.5

In fact, if $X$ is CF, $\ell^{2}(X)$ is the union of the finite-dimensional
subspaces corresponding to finite subsets of $X$. There are two natural
topologies on $\ell^{2}(X)$: the norm topology and the weak topology.444This
is the pointwise convergence topology when vectors are identified with the
associated linear functionals via the Riesz Representation Theorem, Theorem
2.0.6. There is also the inductive limit topology from the finite-dimensional
subspaces. It may be that these three topologies coincide, at least on bounded
subsets of $\ell^{2}(X)$. (Note that Tikhonov’s Theorem is needed to prove
that the closed unit ball of a Hilbert space is weakly compact—an instance of
the Banach–Alaoglu Theorem—so in this setting we do not necessarily have weak
compactness of the ball.)

From this, we can obtain a nice example from [BSB96] of an inner product space
which is Cauchy-complete but not $\sigma$-complete (another claimed example
from [BSB96] is wrong, cf. 6.1.2):

##### 6.0.6

Example. Let $X$ be a CF set. Every vector in $\ell^{2}(X)$ is of the form
$\displaystyle{\sum_{x\in X}c_{x}\xi_{x}}$ with the $c_{x}\in\mathbb{C}$ and
only finitely many $c_{x}$ nonzero (Proposition 6.0.3). Let

${\mathcal{H}}_{0}=\left\\{\sum_{x\in X}c_{x}\xi_{x}\in\ell^{2}(X):\sum_{x\in
X}c_{x}=0\right\\}\ .$

Then ${\mathcal{H}}_{0}$ is a vector subspace of $\ell^{2}(X)$. If
$(\eta_{n})$ is a Cauchy sequence in ${\mathcal{H}}_{0}$, then there is a
finite subset $Y$ of $X$ such that all the $\eta_{n}$ are supported in the
span ${\mathcal{Y}}$ of $\\{\xi_{y}:y\in Y\\}$. Since ${\mathcal{Y}}$ is
finite-dimensional, $\eta_{n}\to\eta$ for some $\eta\in{\mathcal{Y}}$, and
$\eta\in{\mathcal{H}}_{0}$. Thus ${\mathcal{H}}_{0}$ is Cauchy-complete.
However, ${\mathcal{H}}_{0}$ is not closed in $\ell^{2}(X)$, and is in fact
dense. For if $\displaystyle{\zeta=\sum_{x\in X}c_{x}\xi_{x}}$ is in
${\mathcal{H}}_{0}^{\perp}$, all the $c_{x}$ must be equal (just compare with
vectors in ${\mathcal{H}}_{0}$ which are differences of two standard basis
vectors), hence all $c_{x}$ must be 0, so ${\mathcal{H}}_{0}^{\perp}=\\{0\\}$.
Alternatively, define a linear functional $\phi$ on $\ell^{2}(X)$ by

$\phi\left(\sum_{x\in X}c_{x}\xi_{x}\right)=\sum_{x\in X}c_{x}\ .$

It can be shown directly that $\phi$ is unbounded, and hence
${\mathcal{H}}_{0}={\mathcal{N}}(\phi)$ (the kernel of $\phi$) is dense in
$\ell^{2}(X)$. Thus ${\mathcal{H}}_{0}$ is not $\sigma$-complete.

###  6.1. $\ell^{2}(X)$ for Russell Sets

##### 6.1.1

We now let $X$ be a DF set which is not CF, i.e. $X$ contains a sequence
$(A_{n})$ of pairwise disjoint nonempty finite sets. For each $n$ set

$\eta_{n}=\frac{1}{\sqrt{|A_{n}|}}\sum_{x\in A_{n}}\xi_{x}\ .$

Then $(\eta_{n})$ is a sequence of unit vectors in $\ell^{2}(X)$ with disjoint
supports, hence orthonormal. Thus, although $\ell^{2}(X)$ has an orthonormal
basis whose cardinality is DF, $\ell^{2}(X)$ contains an orthonormal sequence
of vectors (whose cardinality is $\aleph_{0}$, hence not DF). In fact, see
6.3.1.

##### 6.1.2

Now let $X$ be a Russell set, with $|A_{n}|=2$ for all $n$. Then $\eta_{n}$ is
the (normalized) sum of the basis vectors in each pair of socks. Let
${\mathcal{Y}}$ be the closed subspace of $\ell^{2}(X)$ spanned by the
$\eta_{n}$. Then ${\mathcal{Y}}$ is isometrically isomorphic to
$\ell^{2}(\mathbb{N})$.

Let ${\mathcal{Z}}={\mathcal{Y}}^{\perp}$. Then ${\mathcal{Z}}$ is the closed
span of the differences of the vectors in each pair of socks. These
differences are only well defined up to sign, and there does not exist a
global choice of signs; however, the one-dimensional subspace spanned by the
difference is well defined for each pair. Thus ${\mathcal{Z}}$ is spanned by a
sequence of one-dimensional subspaces, although there is not a global choice
of unit vectors in these subspaces (since we are working with complex Hilbert
spaces, we have a circle of unit vectors in each subspace, but splitting the
circle into left and right halves would give a global choice of signs). But it
seems unlikely that ${\mathcal{Z}}$ has an orthonormal basis; in fact, it
seems likely that any orthonormal set in ${\mathcal{Z}}$ is finite (this is
claimed in [BSB96], with reference to a proof in [Bru90] which we are unable
to follow or see its relevance; it is also claimed that this proof shows that
${\mathcal{Z}}$ is not $\sigma$-complete, which is wrong since it is a closed
subspace). So ${\mathcal{Z}}$ may be a Hilbert space in which no infinite-
dimensional subspace has an orthonormal basis. In 6.2.3 we will prove that
with a stronger assumption on $X$ this is indeed the case. By Theorem 4.1.2,
neither ${\mathcal{Z}}$ nor $\ell^{2}(X)$ is separable, although $\ell^{2}(X)$
is the “direct sum” of countably many 2-dimensional Hilbert spaces (infinite
direct sums and products, and inductive limits, are not well defined without
some Choice, cf. 7.1.9).

It may be worth noting that if ${\mathcal{Z}}$ has a basis ${\mathcal{B}}$,
then $\ell^{2}(X)$ has two bases of different cardinalities, namely the
canonical basis and $\\{\eta_{n}:n\in\mathbb{N}\\}\cup{\mathcal{B}}$ (compare
with 6.3.2). Moreover, regardless of whether ${\mathcal{Z}}$ has a basis or
not, $\ell^{2}(X)$ has two bases of different cardinalities; see Example
6.3.1.

Note that there is a self-adjoint unitary operator on $\ell^{2}(X)$ which
interchanges the basis vectors in each pair of socks, for which
${\mathcal{Y}}$ and ${\mathcal{Z}}$ are the $+1$ and $-1$ eigenspaces
respectively.

###  6.2. Bases in Hilbert Spaces

##### 6.2.1

The Axiom of Choice implies that every Hilbert space has an orthonormal basis,
and that all of its bases have the same cardinality. We will see that the
first statement can fail, and give positive and negative results about the
second.

##### 6.2.2

For the following we need a strong assumption (stronger than the existence of
a Russell set, 5.7.1):

> There exists an amorphous set $X$ which can be presented as
> $X=\bigsqcup_{z\in Z}X_{z}$, where $Z$ is strongly amorphous and $X_{z}$ are
> pairwise disjoint two-element sets.

Relative consistency of the existence of such $X$ with ZF is given in [Tru95,
Theorem 6.3], but in Proposition 10.0.2 we show why this follows from general
theory as presented in [KS22]. If $X$ has this property, then there is no
choice function $f\colon Z\to X$, because the range of such function would be
an infinite, co-infinite subset of $X$. Moreover, $X$ is amorphous. [If $Y$ is
a subset of $X$, then for all but finitely many $z\in Z$, $Y$ either includes
$X_{z}$ or is disjoint from it. Since $Z$ is amorphous, the set $\\{z\mid
X_{z}\subseteq Y\\}$ is finite or cofinite, thus $Y$ is cofinite or cofinite.]
Since $X$ is amorphous, it is also CF. It is not rigid because the permutation
that interchanges the elements of $X_{z}$ for all $z$ has no fixed points.

##### 6.2.3

Theorem. Suppose $X$ is as in 6.2.2. Then the following holds.

  1. (1)

$\ell^{2}(X)$ has a closed subspace ${\mathcal{Y}}$ isomorphic to
$\ell^{2}(Z)$, and there is no injection from $Z$ into $X$.

  2. (2)

The orthogonal complement of ${\mathcal{Y}}$ has no basis.

  3. (3)

The orthogonal complement of ${\mathcal{Y}}$ contains no infinite orthonormal
set.

Proof: (1) Write $X=\bigsqcup_{z\in Z}X_{z}$. The canonical basis of
$\ell^{2}(X)$ is $\xi_{z,x}$, for $z\in Z$ and $x\in X_{z}$. (Clearly the
index $z$ is redundant, but it will come handy in the proof.) Consider the
subspace ${\mathcal{Y}}$ of $\ell^{2}(X)$ spanned by the vectors
$\eta_{z}=2^{-1/2}\sum_{x\in X_{z}}\xi_{z,x}$. Since each $X_{z}$ has exactly
two elements, each $\eta_{z}$ is a unit vector. These vectors have disjoint
supports and are therefore orthogonal. Clearly their closed linear span,
${\mathcal{Y}}$, is isomorphic to $\ell^{2}(Z)$.

Suppose that $f\colon Z\to X$ is an injection from $Z$ into $X$. Since $Z$ is
amorphous, the set $\\{z\mid f(z)\in X_{z}\\}$ is either finite or cofinite.
By the assumption that there is no choice function on $Z$, it is finite.
Therefore the function $g\colon Z\to Z$ uniquely defined by the requirement
$f(z)\in X_{g(z)}$ has at most finitely many fixed points. This function also
has the property that the preimage of every element has at most two points; we
will say that $g$ is _at most two-to-one_.

For $z\in Z$ consider the forward orbit $\\{g^{n}(z)\mid n\geq 1\\}$. It
cannot be infinite, since otherwise there would be an injection from
$\mathbb{N}$ into $Z$. Therefore for all but finitely many $z\in Z$ there are
$m(z)<n(z)$ such that $g^{m(z)}(z)=g^{n(z)}(z)$. Choose this pair so that
$(m(z),n(z))$ is the minimal possible (in the lexicographic order). Since $Z$
is amorphous, for every $S\subseteq\mathbb{N}^{2}$, the set $Z_{S}=\\{z\in
Z\mid(m(z),n(z))\in S\\}$ is finite or cofinite. Therefore

$\\{S\subseteq\mathbb{N}^{2}\mid Z_{S}\text{ is cofinite in
$Z$}\\}=\\{S\subseteq\mathbb{N}^{2}\mid Z_{S}\text{ is infinite}\\}$

and this set is a filter. Since $\mathbb{N}$ is not amorphous, it is a
principal filter, and there is a pair $(m,n)$ such that

$Z^{\prime}=\\{z\in Z\mid(m(z),n(z))=(m,n)\\}$

is cofinite in $Z$ (and therefore amorphous).

We claim that $m=1$. For $k\geq 1$ let

$Z[k]=\\{z\in Z’\mid(\exists y\in Z)g^{k}(y)=z\\}.$

If $1\leq k$ then $g[Z[k]]=Z[k+1]$. Since $g$ is at most two-to-one, if $m>1$
and $Z[m]$ is infinite then $Z[m-1]$ is infinite as well. By the minimality of
$m$ we conclude that $m=1$.

Therefore $g(z)\in X_{z}$ for all but finitely many $z\in Z$. This contradicts
the fact that $g$ has at most finitely many fixed points proven earlier.

(3) We claim that the orthogonal complement of ${\mathcal{Y}}$ in
$\ell^{2}(X)$ includes no infinite orthonormal set. Assume otherwise and fix
one, denoted ${\mathcal{B}}$. Let

$R=\\{(z,\eta)\in Z\times{\mathcal{B}}:\langle\eta,\xi_{z,x}\rangle\neq
0\text{ for some $x\in X_{z}$}\\}.$

We will use Lemma 5.4.2 and the notation $R_{z}$, $R^{\eta}$ for horizontal
and vertical sections of $R$ introduced there. Since $X$ is CF, every vector
in $\ell^{2}(X)$ has finite support by Proposition 6.0.3. This means that for
every $\eta$ in ${\mathcal{B}}$ the horizontal section $R^{\eta}$ of $R$ is
finite.

We claim that for every $z\in Z$, the vertical section $R_{z}$ of $R$ is
finite. To see this, fix $z$ and let $P_{z}$ denote the projection to the span
of $\xi_{z,x}$, for $x\in X_{z}$. For every $n$, the set
$\\{\eta\in{\mathcal{B}}:\|P_{z}(\eta)\|>1/n\\}$ is finite. Since $X$ is CF,
there is $n$ such that $P_{z}(\eta)$ is nonzero if and only if
$\|P_{z}(\eta)\|>1/n$. Since the set of $\eta$ such that $\|P_{z}(\eta)\|>1/n$
is finite, we conclude that $R_{z}$ is finite for all $z\in Z$.

Therefore the assumptions of Lemma 5.4.2 (i) are satisfied. This lemma implies
that there are a set $W$ and partitions $Z=\bigsqcup_{w\in W}Z_{w}$ and
${\mathcal{B}}=\bigsqcup_{w\in W}{\mathcal{B}}_{w}$ into finite sets such that
$\bigcup\\{R_{x}:x\in Z_{w}\\}\subseteq{\mathcal{B}}_{w}$ and
$\bigcup\\{R^{\eta}:\eta\in{\mathcal{B}}_{w}\\}\subseteq Z_{w}$ for all $w\in
W$.

Then $Z_{w}$, for $w\in W$, are equivalence classes of an equivalence relation
on $Z$. Since $Z$ is strongly amorphous, all but finitely many of the classes
are singletons. Let $Z^{\prime}$ be the union of the singleton classes. Since
$\ell^{2}(X_{z})$ is two-dimensional, for each $z\in Z^{\prime}$ there is a
unique $\eta(z)\in{\mathcal{B}}$ with $R^{\eta(z)}=\\{z\\}$. For each $z\in Z$
we have $\eta(z)=\sum_{x\in X_{z}}\lambda_{x}\xi_{z,x}$. Since $\eta(z)$ is
orthogonal to $2^{-1/2}\sum_{x\in X_{z}}\xi_{z,x}$ we have $\sum_{x\in
Z_{x}}\lambda_{x}=0$. We can therefore define a function $f\colon
Z^{\prime}\to X$ by

$f(z)=x\quad\text{if and only if}\quad 0\leq\arg(\lambda_{x})<\pi.$

Then $f$ is a choice function on $Z^{\prime}$. Since $Z\setminus Z^{\prime}$
is finite, it can be extended to a choice function on all of $Z$,
contradicting our assumption.

(2) clearly follows from (3).

##### 6.2.4

The space without a basis constructed in 6.2.3 is “equal” to $\bigoplus_{z\in
Z}{\mathcal{H}}_{z}$ (recall that infinite direct sums of Hilbert spaces are
not really well defined; we just mean the union of the finite direct sums is
dense), where (using the notation from the proof of 6.2.3) for every $z$,
${\mathcal{H}}_{z}=\ell^{2}(X_{z})\ominus\operatorname{span}(\sum_{x\in
X_{z}}\xi_{z,x})$ is a one-dimensional Hilbert space.

##### 6.2.5

Proposition. Suppose that ${\mathcal{H}}$ is a real Hilbert space and there is
a family ${\mathcal{H}}_{z}$, for $z\in Z$, of one-dimensional, pairwise
orthogonal, subspaces of ${\mathcal{H}}$ such that
$\bigcup_{F\in\operatorname{Fin}(Z)}\operatorname{span}\bigcup\\{{\mathcal{H}}_{z}\mid
z\in F\\}$ is dense in ${\mathcal{H}}$. Then ${\mathcal{H}}$ is a subspace of
a Hilbert space with a basis.

Proof: For each $z$, the unit ball $S(z)$ of ${\mathcal{H}}_{z}$ has two
elements. Let $Y=\sqcup_{z\in Z}S(z)$. Then
$v_{z}\colon{\mathcal{H}}_{z}\to\ell^{2}(S(H_{z}))$ defined by
$v_{z}(\xi)=2^{-1/2}\sum_{\eta\in S(z)}\langle\xi,\eta\rangle\eta$ is an
isometry for every $z$, and $\bigoplus_{z\in Z}v_{z}\colon H\to\ell^{2}(Y)$ is
an isometry.

##### 6.2.6

It is not clear how to extend the argument of 6.2.5 to any other space of the
form $\bigoplus_{z\in Z}{\mathcal{H}}_{z}$, even if
$\dim({\mathcal{H}}_{z})=2$ for all $z$ or if $\dim({\mathcal{H}}_{z})=1$ but
it is a complex vector space. (The latter is the case of 6.2.3, but the space
constructed in this proposition is rather specific.)

Compare the following with Proposition 6.0.3 of which it is a continuation.
Parts of it appear in [Bru86] in a slightly more general context.

##### 6.2.7

Proposition. For every infinite-dimensional Hilbert space ${\mathcal{H}}$ with
an orthonormal basis the following are equivalent.

  1. (i)

${\mathcal{H}}$ has a CF orthonormal basis.

  2. (ii)

Every infinite orthonormal set in ${\mathcal{H}}$ is CF.

  3. (iii)

Every orthonormal basis is a Hamel basis.

Proof: (ii) trivially implies (i).

(i) implies (ii): If (i) holds, then by Proposition 3.0.6 we may assume
${\mathcal{H}}=\ell^{2}(X)$ for a CF set $X$. Fix an orthomormal set
${\mathcal{B}}$ in ${\mathcal{H}}$. Let

$R=\\{(x,\eta)\in X\times{\mathcal{B}}:\langle\xi_{x},\eta\rangle\neq 0\\}.$

For every $\eta\in{\mathcal{B}}$ the horizontal section $R^{\eta}$ is equal to
the support of $\eta$, and therefore finite by Proposition 6.0.3.

Fix $x\in X$. We will prove that $R_{x}$ is finite. Assume otherwise. Since
$\sum_{\eta\in R_{x}}|\langle\xi_{x},\eta\rangle|^{2}=1$, the set
$G(n)=\\{\eta\in{\mathcal{B}}:|\langle\xi_{x},\eta\rangle|\geq 1/n\\}$ is
finite for all $n$. Therefore $F(n)=\bigcup_{\eta\in G(n)}R^{\eta}$ is a
nondecreasing sequence of finite subsets of $X$. Since $X$ is CF, there exists
$m$ such that $F(n)=F(m)$ for all $n\geq m$.

Every $G\subseteq{\mathcal{B}}$ is included in the linear span of
$\\{\xi_{x}:x\in\bigcup_{\eta\in G}R^{\eta}\\}$. This implies that
$|G(n)|\leq|F(n)|$ (conveniently, this is the additional assumption of Lemma
5.4.2 (ii)) which is by the previous paragraph $\leq|F(m)|$ for all $n$. Since
the sequence $G(n)$ is increasing, it stabilizes. Hence its union is finite,
contradicting the assumption that $R_{x}$ was infinite.

Since $x\in X$ was arbitrary, all horizontal sections of $R$ are finite, and
Lemma 5.4.2 (ii) implies that ${\mathcal{B}}$ is CF.

(i) implies (iii): Assume (i) and fix an orthonormal basis of ${\mathcal{H}}$.
By (ii) we may assume ${\mathcal{H}}=\ell^{2}(X)$ for a CF set $X$.
Proposition 6.0.3 implies that every vector in $\ell^{2}(X)$ is a linear
combination of finitely many basis vectors.

(iii) implies (i): Assume (i) fails, hence ${\mathcal{H}}=\ell^{2}(X)$ and $X$
is not CF. By Proposition 6.0.3 (ii) implies (i), some vector in
${\mathcal{H}}$ has infinite support, thus $\xi_{x}$, for $x\in X$, is not a
Hamel basis.

###  6.3. Uniqueness of Orthogonal Dimension

Can we have a Hilbert space with bases of different cardinalities? The answer
is yes:

##### 6.3.1

Example. Let $X$ be a Dedekind-finite set which is not Cohen-finite, and let
${\mathcal{Y}}$ be the closed subspace of $\ell^{2}(X)$ spanned by an
orthonormal sequence (6.1.1) and ${\mathcal{Z}}={\mathcal{Y}}^{\perp}$. Then
${\mathcal{Y}}\cong\ell^{2}(\mathbb{N})$. If $X\sqcup\mathbb{N}$ is the
disjoint union of $X$ and $\mathbb{N}$, then

$\ell^{2}(X\sqcup\mathbb{N})\cong\ell^{2}(\mathbb{N})\oplus\ell^{2}(X)\cong\ell^{2}(\mathbb{N})\oplus{\mathcal{Y}}\oplus{\mathcal{Z}}\\\
\cong\ell^{2}(\mathbb{N})\oplus\ell^{2}(\mathbb{N})\oplus{\mathcal{Z}}\cong\ell^{2}(\mathbb{N})\oplus{\mathcal{Z}}\cong{\mathcal{Y}}\oplus{\mathcal{Z}}\cong\ell^{2}(X)$

since
$\ell^{2}(\mathbb{N})\oplus\ell^{2}(\mathbb{N})\cong\ell^{2}(\mathbb{N})$.
Thus, if $X$ is DF, $\ell^{2}(X)$ has an orthonormal basis indexed by the DF
set $X$, and an orthonormal basis indexed by the Dedekind-infinite set
$X\sqcup\mathbb{N}$.

It follows that if $Y$ is $X$ with any finite number of points added or
removed, i.e. $|Y|=|X|\pm n$ for some $n\in\mathbb{N}$, then
$\ell^{2}(Y)\cong\ell^{2}(X)$. For any $n\neq 0$, $|Y|\neq|X|$. If $X$ is a
Russell set, then $Y$ is a Russell set if and only if $n$ is even (§5.7.2).

The following shows that the cardinalities of orthonormal bases cannot be too
wild.

##### 6.3.2

Theorem. Suppose that $\ell^{2}(X)$ and $\ell^{2}(Y)$ are isomorphic,
$U\colon\ell^{2}(X)\to\ell^{2}(Y)$ is a unitary, and $X$ is CF.

  1. (i)

There are a set $Z$ and partitions $X=\bigsqcup_{z\in Z}X_{z}$ and
$Y=\bigsqcup_{z\in Z}Y_{z}$ into finite sets such that
$U[\ell^{2}(X_{z})]=\ell^{2}(Y_{z})$ (hence $|X_{z}|=|Y_{z}|$) for all $z\in
Z$.

  2. (ii)

If $X$ is in addition strongly amorphous, then $|X|=|Y|$ and there are
cofinite $X^{\prime}\subseteq X$, $Y^{\prime}\subseteq Y$, and a bijection
$f\colon X^{\prime}\to Y^{\prime}$ such that $U\xi_{x}=c\xi_{f(x)}$ for some
$c\in\mathbb{C}$ with $|c|=1$, for all $x\in X^{\prime}$.

###### Proof.

(i) By 6.2.7, $Y$ is CF, hence 6.0.3 implies that every vector in
$\ell^{2}(X)$ and $\ell^{2}(Y)$ has finite support. Therefore all vertical and
horizontal sections of $R\subseteq X\times Y$ defined by

$R=\\{(x,y)\in X\times Y:\langle U\xi_{x},\xi_{y}\rangle\neq 0\\}$

are finite. Since $X$ is CF, Lemma 5.4.2 implies that there are a set $Z$ and
partitions $X=\bigsqcup_{z\in Z}X_{z}$ and $Y=\bigsqcup_{z\in Z}Y_{z}$ into
finite sets such that $\bigcup\\{R_{x}:x\in X_{z}\\}\subseteq Y_{z}$ and
$\bigcup\\{R^{y}:y\in Y_{z}\\}\subseteq X_{z}$ for all $z\in Z$.

Then the restriction of $U$ to $\ell^{2}(X_{z})$ is an isometry onto
$\ell^{2}(Y_{z})$ for all $z\in Z$. Since these are finite-dimensional spaces,
$|X_{z}|=|Y_{z}|$.

To prove (ii), use the fact that since $X$ is strongly amorphous, $X_{z}$ is a
singleton for a cofinite $Z^{\prime}\subseteq Z$. Thus $Y_{z}$ is also a
singleton for all $z\in Z^{\prime}$, and we have a bijection between cofinite
subsets of $X$ and $Y$. Using $|X_{z}|=|Y_{z}|$ for all $z$ again, we extend
it to a bijection between $X$ and $Y$. Since $X$ is amorphous, for all but
finitely many $x\in X$ we have $U\xi_{x}=c\xi_{f(x)}$ for some fixed
$c\in\mathbb{C}$, $|c|=1$. ∎

By applying (ii) of Theorem 6.3.2 in the case when $X=Y$ is strongly
amorphous, we obtain the following.

##### 6.3.3

Corollary. If $X$ is strongly amorphous then every orthonormal basis of
$\ell^{2}(X)$ includes $\\{c\xi_{x}:x\in Y\\}$ for some cofinite subset $Y$ of
$X$ and some $c\in\mathbb{C}$, $|c|=1$.

###  6.4. Universality of Hilbert Spaces With a Basis?

##### 6.4.1

Can one prove in ZF that every Hilbert space is the quotient of a Hilbert
space with a basis, and/or that every Hilbert space is isomorphic to a
subspace of a Hilbert space with a basis? We will see that these two questions
are related (6.4.5). The only example of a Hilbert space without a basis known
to us, given in 6.2.3, is both a subspace and a quotient of a Hilbert space
with a basis (the latter follows by 6.4.5).

##### 6.4.2

A standard argument shows the following.

Lemma. Every Banach space is the quotient of a Banach space with a basis.

Proof: Suppose ${\mathcal{X}}$ is a Banach space and let $Y$ be the unit
sphere of ${\mathcal{X}}$. Denoting the canonical basis of $\ell^{1}(Y)$ by
$\delta_{y}$, for $y\in Y$ define $T\colon\ell^{1}(Y)\to{\mathcal{X}}$ by

$T(\sum_{y}\lambda_{y}\delta_{y})=\sum_{y}\lambda_{y}y.$

To see that $T$ is well-defined, note that every
$\sum_{y}\lambda_{y}\delta_{y}$ satisfies
$\|\sum_{y}\lambda_{y}\delta_{y}\|=\sum_{y}|\lambda_{y}|\geq\sum_{y}|\lambda_{y}|\|y\|\geq\|\sum_{y}y\|$.
Therefore , we have $\|T\|\leq K$.

By the choice of $Y$, $T$ is surjective.

##### 6.4.3

There is a lot of room for improvement in Lemma 6.4.2. If $X$ is a Hilbert
space, can we choose $Y$ to be a Hilbert space? If the set $Y$ in the proof of
Lemma 6.4.2 is a dense subset of the unit sphere of $X$, is the operator $T$
still surjective (the standard proof of this uses the axiom of Dependent
Choice)?

In order to prove 6.4.5 below, we need the following fact interesting in its
own right.

##### 6.4.4

Proposition. Every short exact sequence of Hilbert spaces in which the
connecting maps are isometries splits.

Proof: Suppose that $0\to{\mathcal{X}}\to{\mathcal{H}}\to{\mathcal{Y}}\to 0$
is a short exact sequence of Hilbert spaces with isometries as connecting
maps. Thus the image of ${\mathcal{X}}$ is a closed subspace of
${\mathcal{H}}$. By Corollary 2.0.5, this subspace has an orthogonal
complement ${\mathcal{X}}^{\perp}$ and there is an orthogonal projection
$P_{{\mathcal{X}}}$ from ${\mathcal{H}}$ onto ${\mathcal{X}}^{\perp}$. The
restriction of the quotient map $q\colon{\mathcal{H}}\to{\mathcal{Y}}$ to
${\mathcal{X}}^{\perp}$ is a surjective isometry, and therefore its inverse is
continuous.

It would be of interest to generalize 6.4.4 to the case when the connecting
maps are arbitrary bounded linear operators. The Polar Decomposition Theorem
(7.0.6) may be relevant.

The following is an immediate consequence of 6.4.4.

##### 6.4.5

Corollary. Every Hilbert space which is the quotient of a Hilbert space with a
basis is a subspace of a Hilbert space with a basis. ∎

##  7\. Bounded Operators

If ${\mathcal{H}}$ is a Hilbert space, denote by
${\mathcal{B}}({\mathcal{H}})$ the set of bounded operators on
${\mathcal{H}}$.

##### 7.0.1

Proposition. If ${\mathcal{H}}$ is a Hilbert space, then
${\mathcal{B}}({\mathcal{H}})$ is $\sigma$-complete (i.e. a Banach algebra)
under the operator norm.

Proof: Let $(A_{n})$ be a decreasing sequence of closed bounded sets in
${\mathcal{B}}({\mathcal{H}})$ whose diameters go to 0. Fix a vector
$\xi\in{\mathcal{H}}$, and set $A_{n}^{\xi}=\\{T\xi:T\in A_{n}\\}$. Then the
sets $(\overline{A_{n}^{\xi}})$ form a decreasing sequence of closed bounded
subsets of ${\mathcal{H}}$ whose diameters go to 0, so
$\cap_{n}\overline{A_{n}^{\xi}}$ is a singleton we call $S\xi$. This defines a
function $S:{\mathcal{H}}\to{\mathcal{H}}$, which is clearly linear. If
$\|T\|\leq M$ for all $T\in A_{1}$, then for any $\xi$ $\|T\xi\|\leq M\|\xi\|$
for all $T\in A_{1}$, and hence $\|S\xi\|\leq M\|\xi\|$, so $S$ is bounded and
$\|S\|\leq M$. To show $S\in\cap_{n}A_{n}$, let $\epsilon>0$, and fix $m$ with
$\operatorname{diam}(A_{m})\leq\epsilon$. Let $n\geq m$. If $\xi$ is any unit
vector in ${\mathcal{H}}$, the set $\overline{A_{n}^{\xi}}$ has diameter
$\leq\epsilon$, hence $\|T\xi-S\xi\|\leq\epsilon$ for all $T\in A_{n}$. Thus
$\|T-S\|\leq\epsilon$ for every $T\in A_{n}$. Since the $A_{n}$ are decreasing
and closed, $S\in\cap_{n}A_{n}$.

This result holds more generally if ${\mathcal{H}}$ is just a
($\sigma$-complete) Banach space; in fact, if ${\mathcal{X}}$ and
${\mathcal{Y}}$ are normed vector spaces with ${\mathcal{Y}}$
$\sigma$-complete, then ${\mathcal{B}}({\mathcal{X}},{\mathcal{Y}})$ is
$\sigma$-complete.

The next important fact is a corollary of the Riesz Representation Theorem
(2.0.6); the standard proof (cf. [Bla23, XVI.9.11.2]) works verbatim.

##### 7.0.2

Corollary. Let ${\mathcal{H}}$ be a Hilbert space, and
$T\in{\mathcal{B}}({\mathcal{H}})$. Then there is a unique
$S\in{\mathcal{B}}({\mathcal{H}})$ with

$\langle Tx,y\rangle=\langle x,Sy\rangle$

for each $x,y\in{\mathcal{H}}$. Write $T^{*}$ for this $S$.

##### 7.0.3

The adjoint operation $T\mapsto T^{*}$ has the usual properties ([Bla06,
I.2.3.1]), which show that ${\mathcal{B}}({\mathcal{H}})$ is a C*-algebra.
7.0.2 is what allows reasonable operator theory to be done on ${\mathcal{H}}$.

##### 7.0.4

In the example of a Cauchy-complete, not $\sigma$-complete space
${\mathcal{H}}_{0}$ obtained from a CF set $X$ in 6.0.6, all the conclusions
2.0.4, 2.0.5, and 2.0.6 fail. Operators on ${\mathcal{H}}_{0}$ do not have
adjoints in general, or more precisely the adjoint of a bounded operator on
${\mathcal{H}}_{0}$ is a (bounded) operator on ${\mathcal{H}}$, the
$\sigma$-completion of ${\mathcal{H}}_{0}$. Here is a simple example of an
operator on ${\mathcal{H}}_{0}$ with no adjoint. Let $\zeta$ be a unit vector
in $\ell^{2}(X)\setminus{\mathcal{H}}_{0}$, $\eta$ a unit vector in
${\mathcal{H}}_{0}$, and let $T(\xi)=\langle\xi,\zeta\rangle\eta$ for
$\xi\in{\mathcal{H}}_{0}$. This is a rank one operator on ${\mathcal{H}}_{0}$,
but its adjoint (in ${\mathcal{B}}(\ell^{2}(X))$) is a rank-one operator whose
range is the span of $\zeta$, so it is not an operator on ${\mathcal{H}}_{0}$.
Note that this example also gives failure of the Riesz Representation theorem
for ${\mathcal{H}}_{0}$ and a bounded sesquilinear form on ${\mathcal{H}}_{0}$
that does not correspond to a bounded linear operator on ${\mathcal{H}}_{0}$.

##### 7.0.5

As usual, we define a positive operator on ${\mathcal{H}}$ to be a
$T\in{\mathcal{B}}({\mathcal{H}})$ with $T=T^{*}$ and $\langle
T\xi,\xi\rangle\geq 0$ for all $\xi\in{\mathcal{H}}$. If
$T\in{\mathcal{B}}({\mathcal{H}})$, then $T^{*}T\geq 0$. The converse also
holds: in fact, every $T\geq 0$ has a unique positive square root, an $S\geq
0$ with $S^{2}=T$ (write $S=T^{1/2}$). This is a special case of continuous
functional calculus of self-adjoint elements. There is a direct proof not
requiring any Choice, cf. [RSN90] or [Hal82].

##### 7.0.6

Theorem.[Polar Decomposition Theorem] Let ${\mathcal{H}}$ be a Hilbert space,
and $T$ a bounded linear operator on ${\mathcal{H}}$. Then there are a unique
positive operator $|T|$ and a partial isometry $V$ such that $T=V|T|$. If we
require $V$ to have the property that the range of $V^{*}V$ is
${\mathcal{N}}(T)^{\perp}$, then $V$ is unique.

Proof: Set $|T|=(T^{*}T)^{1/2}$. For every vector $\xi$, by a standard
argument we have

$\||T|\xi\|^{2}=\langle|T|\xi,|T|\xi\rangle=\langle
T^{*}T\xi,\xi\rangle=\langle T\xi,T\xi\rangle=\|T\xi\|^{2}.$

Thus $|T|\xi\mapsto T\xi$ is an isometry from the range of $|T|$ to the range
of $T$, and hence by 2.0.7 extends to an isometry between the closures. Extend
this to $V\in{\mathcal{B}}({\mathcal{H}})$ by setting $V=0$ on
$\operatorname{Ran}(|T|)^{\perp}={\mathcal{N}}(T)$ (the null space of $T$).
Then ${\mathcal{N}}(V)={\mathcal{N}}(|T|)={\mathcal{N}}(T)$, and
$T\xi=V|T|\xi$ for all $\xi\in{\mathcal{H}}$, as required.

##### 7.0.7

As a consequence, many other standard properties of
${\mathcal{B}}({\mathcal{H}})$ persist in this setting, such as:

  1. (i)

Every $T\in{\mathcal{B}}({\mathcal{H}})$ has left and right support
projections.

  2. (ii)

The projections in ${\mathcal{B}}({\mathcal{H}})$ form a complete lattice.

  3. (iii)

The right annihilator of any subset of ${\mathcal{B}}({\mathcal{H}})$ is
generated by a projection (i.e. ${\mathcal{B}}({\mathcal{H}})$ is an
$\mathrm{AW}^{*}$-algebra).

  4. (iv)

${\mathcal{B}}({\mathcal{H}})$ admits not only continuous functional calculus
of self-adjoint elements [BF23], but even Borel functional calculus; in
particular, if $T=T^{*}\in{\mathcal{B}}({\mathcal{H}})$ and $B$ is a Borel
subset of $\sigma(T)$, the spectral projection $E_{B}(T)$ is well defined with
the usual properties.

  5. (v)

A bounded increasing net of positive operators converges in the strong
operator topology.

These results will appear in [BF23].

###  7.1. Compact Operators

##### 7.1.1

Under usual versions of Choice, there are several equivalent conditions which
characterize compact operators. These conditions may not be equivalent in the
absence of CC, however, so the proper definition must be nailed down. The
following definition seems to be the best and most useful one (whether or not
the AC is assumed):

##### 7.1.2

Definition. Let ${\mathcal{H}}$ be a Hilbert space, and
$T:{\mathcal{H}}\to{\mathcal{H}}$ a linear operator. Then $T$ is compact if,
whenever $A$ is a bounded subset of ${\mathcal{H}}$, $T(A)$ is totally
bounded. Denote by ${\mathcal{K}}({\mathcal{H}})$ the set of compact operators
on ${\mathcal{H}}$.

##### 7.1.3

If $T$ is a compact operator, the image of the unit ball of ${\mathcal{H}}$ is
totally bounded, hence bounded, so $T$ is bounded, i.e.
${\mathcal{K}}({\mathcal{H}})\subseteq{\mathcal{B}}({\mathcal{H}})$. Since
bounded sets in a finite-dimensional normed vector space are totally bounded
(Choice is not needed to prove this), every finite-rank bounded operator is
compact. Using our definition, we have the next standard result (it is unclear
whether it holds using other definitions):

##### 7.1.4

Proposition. Let ${\mathcal{H}}$ be a Hilbert space. Then
${\mathcal{K}}({\mathcal{H}})$ is a two-sided ideal in
${\mathcal{B}}({\mathcal{H}})$, and is the norm-closure of the set of finite-
rank operators on ${\mathcal{H}}$.

Proof: Since bounded operators send bounded sets to bounded sets and totally
bounded sets to totally bounded sets, ${\mathcal{K}}({\mathcal{H}})$ is closed
under left and right multiplication by bounded operators.
${\mathcal{K}}({\mathcal{H}})$ is obviously closed under scalar
multiplication. If $S$ and $T$ are compact operators, $A$ is a bounded subset
of ${\mathcal{H}}$, and $\epsilon>0$, cover $S(A)$ by open balls of radius
$\frac{\epsilon}{2}$ centered at $\\{x_{1},\dots,x_{n}\\}$ and cover $T(A)$ by
open balls of radius $\frac{\epsilon}{2}$ centered at
$\\{y_{1},\dots,y_{m}\\}$; then the open balls of radius $\epsilon$ centered
at the $x_{j}+y_{k}$ cover $S(A)+T(A)\supseteq(S+T)(A)$. So $S+T$ is compact.
Thus ${\mathcal{K}}({\mathcal{H}})$ is a two-sided ideal in
${\mathcal{B}}({\mathcal{H}})$.

Let $T$ be in the norm-closure of ${\mathcal{K}}({\mathcal{H}})$, $A$ a
bounded subset of ${\mathcal{H}}$, and $\epsilon>0$. Choose
$S\in{\mathcal{K}}({\mathcal{H}})$ with $\|S-T\|<\frac{\epsilon}{2}$. Cover
$S(A)$ with open balls of radius $\frac{\epsilon}{2}$ centered at
$\\{x_{1},\dots,x_{n}\\}$. Then the balls of radius $\epsilon$ centered at
$\\{x_{1},\dots,x_{n}\\}$ cover $T(A)$. Thus
$T\in{\mathcal{K}}({\mathcal{H}})$. So ${\mathcal{K}}({\mathcal{H}})$ is norm-
closed.

Finally, let $T\in{\mathcal{K}}({\mathcal{H}})$ and $\epsilon>0$. Cover the
image of the closed unit ball of ${\mathcal{H}}$ under $T$ by open balls of
radius $\frac{\epsilon}{2}$ centered at $\\{x_{1},\dots,x_{n}\\}$. Let $P$ be
the orthogonal projection of ${\mathcal{H}}$ onto the finite-dimensional
subspace spanned by $\\{x_{1},\dots,x_{n}\\}$. [Existence of $P$ follows from
2.0.5, but there is a more elementary argument: any finite-dimensional
subspace of an inner product space has an orthogonal complement by applying
the Gram-Schmidt process to an orthonormal basis of the subspace and a given
vector in the space.] Since $P$ has norm 1, it follows that
$\|PT-T\|<\epsilon$, and $PT$ has finite rank. Thus $T$ is in the closure of
the finite-rank operators.

##### 7.1.5

Corollary. Let ${\mathcal{H}}$ be a Hilbert space, and
$T\in{\mathcal{K}}({\mathcal{H}})$. Then $T^{*}$ is also in
${\mathcal{K}}({\mathcal{H}})$.

##### 7.1.6

Thus ${\mathcal{K}}({\mathcal{H}})$ is a C*-subalgebra of the C*-algebra
${\mathcal{B}}({\mathcal{H}})$. (Actually any closed two-sided ideal in a
C*-algebra is a C*-subalgebra, but this is an elementary observation in the
case of ${\mathcal{K}}({\mathcal{H}})$.)

A $\mathrm{C}^{*}$-algebra is AF (_approximately finite_) if it is an
“inductive limit” of a directed system of finite-dimensional
$\mathrm{C}^{*}$-algebras (since inductive limits are tricky and not well
defined by Bratteli diagrams without the AC, we will take this to mean there
is a system of finite-dimensional C*-subalgebras, directed by inclusion, whose
union is dense).

##### 7.1.7

Proposition. If ${\mathcal{H}}$ is a Hilbert space, then
${\mathcal{K}}({\mathcal{H}})$ is a simple C*-algebra, which is nonunital if
${\mathcal{H}}$ is infinite-dimensional. For every set $X$,
${\mathcal{K}}(\ell^{2}(X))$ is an AF algebra.

##### 7.1.8

We may more generally define adjoints and compactness for operators between
two Hilbert spaces in the same way, with analogous properties.

##### 7.1.9

If $X$ is a set of Russell socks, then $X$ is the union of an increasing
sequence of finite sets, and hence there is an increasing sequence of finite-
dimensional matrix algebras, each embedded as a corner in the next, whose
union is dense in ${\mathcal{K}}(\ell^{2}(X))$.
${\mathcal{K}}(\ell^{2}(\mathbb{N}))$ has a seemingly identical such sequence.
But ${\mathcal{K}}(\ell^{2}(X))$ is not isomorphic to
${\mathcal{K}}(\ell^{2}(\mathbb{N}))$; in fact, ${\mathcal{K}}(\ell^{2}(X))$
is not separable since no countable set can approximate the rank one
projections onto the spans of the standard basis elements (cf. the proof of
4.1.2). Thus inductive limits of sequences of C*-algebras, even finite-
dimensional ones, are not well defined in the usual sense without CC.

##### 7.1.10

There is a notable analogy of 7.1.9 to a result of Katsura ([Kat06, Theorem
6]) showing that nonisomorphic non-separable AF-algebras may have the same
Bratteli diagram.

##  8\. ${\mathcal{B}}(\ell^{2}(X))$ For DF Sets

##### 8.0.1

If $X$ is a set, then the Hilbert space $\ell^{2}(X)$ and the C*-algebras
${\mathcal{B}}(\ell^{2}(X))$, ${\mathcal{K}}(\ell^{2}(X))$, and the Calkin
algebra
${\mathcal{Q}}(\ell^{2}(X))={\mathcal{B}}(\ell^{2}(X))/{\mathcal{K}}(\ell^{2}(X))$
are invariants of the cardinal $|X|$, along with associated objects like the
lattice of ideals of ${\mathcal{B}}(\ell^{2}(X))$. Properties of these objects
reflect properties of $|X|$, in a more interesting and varied way than in the
AC case.

We now examine properties of ${\mathcal{B}}({\mathcal{H}})$,
${\mathcal{K}}({\mathcal{H}})$, and the Calkin algebra
${\mathcal{Q}}({\mathcal{H}})$ for ${\mathcal{H}}=\ell^{2}(X)$, $X$ DF.

A prominent open problem in theory of $\mathrm{C}^{*}$-algebras is whether
every stably finite $\mathrm{C}^{*}$-algebra has a tracial state; we recall
the definitions for the benefit of the readers with stronger background in set
theory than in operator algebras; note the analogy with the definition of
Dedekind—finiteness.

##### 8.0.2

Definition. Suppose that $A$ is a unital $\mathrm{C}^{*}$-algebra. It is
called _finite_ if there is no $v\in A$ such that $1_{A}=v^{*}v$ and
$vv^{*}\neq v^{*}v$,555Such $v$, if it exists, is called a _proper isometry_
since its image under any GNS representation is a proper isometry. _infinite_
if it is not finite, and _stably finite_ if $M_{n}(A)$ is finite for all $n$.
It is _properly infinite_ if there are no $v$ and $w$ in $A$ such that
$v^{*}v=w^{*}w=1_{A}$ and $v^{*}w=0$.

See [Bla06, V.2.1] for more information.

One might hope (and we did hope) that if ${\mathcal{H}}=\ell^{2}(X)$ for a DF
set $X$ then ${\mathcal{B}}({\mathcal{H}})$ would be finite, i.e. have no
nonunitary isometries (even stably finite since
${\mathcal{H}}^{n}\cong\ell^{2}(Y)$ with $Y=X\times\\{0,\dots,n-1\\}$ which is
also DF). But things are not so simple:

##### 8.0.3

Example. Let $X$ be a set which is DF but not CF. Then
$\ell^{2}(X)\cong{\mathcal{Y}}\oplus{\mathcal{Y}}^{\perp}$ (6.1.1), where
${\mathcal{Y}}\cong\ell^{2}(\mathbb{N})$. The unilateral shift $S$ is a
nonunitary isometry on $\ell^{2}(\mathbb{N})$, so $S\oplus I$ is a nonunitary
isometry in
${\mathcal{B}}({\mathcal{Y}}\oplus{\mathcal{Y}}^{\perp})={\mathcal{B}}(\ell^{2}(X))$.

By replacing $S$ by an isometry with infinite codimension, we obtain a
nonunitary isometry in ${\mathcal{Q}}({\mathcal{H}})$. It is unclear whether
${\mathcal{B}}({\mathcal{H}})$ or ${\mathcal{Q}}({\mathcal{H}})$ is properly
infinite, or whether ${\mathcal{Q}}({\mathcal{H}})$ is simple: it could be
that every bounded operator from ${\mathcal{Y}}$ to ${\mathcal{Y}}^{\perp}$ is
compact, and if true then
${\mathcal{Q}}({\mathcal{H}})\cong{\mathcal{Q}}({\mathcal{Y}})\oplus{\mathcal{Q}}({\mathcal{Y}}^{\perp})$.
See however Corollary 8.1.3 and §8.4.

###  8.1. ${\mathcal{B}}(\ell^{2}(X))$ for CF Sets

Now suppose $X$ is CF. Then every operator on $\ell^{2}(X)$ has a lot of
finite-dimensional invariant subspaces:

##### 8.1.1

Lemma. Let $X$ be a CF set, $m\geq 1$, and
$T_{1},\dots,T_{m}\in{\mathcal{B}}(\ell^{2}(X))$. Then every finite subset $E$
of $X$ is contained in a finite subset $F$ such that
$\operatorname{span}\\{\xi_{x}:x\in F\\}$ is invariant under
$T_{1},\dots,T_{m}$ (by abuse of terminology, we say $F$ is invariant under
$T_{1},\dots,T_{m}$).

Proof: Let $E$ be a nonempty finite subset of $X$ (e.g. a singleton). Set
$F_{0}=E$, and inductively let $F_{n+1}$ be the union of $F_{n}$ and the
supports of $T_{k}\xi_{x}$ for all $x\in F_{n}$ and all $k$ (these supports
are finite by 6.0.3). The $F_{n}$ form an increasing sequence of finite
subsets of $X$, and hence must stabilize since $X$ is CF. If $F$ is the final
$F_{n}$, then $F$ is invariant under $T_{1},\dots,T_{k}$.

##### 8.1.2

Proposition. Let $X$ be a CF set, and
$T_{1},\dots,T_{m}\in{\mathcal{B}}(\ell^{2}(X))$. Then $X$ partitions into the
disjoint union of finite subsets, each invariant under $T_{1},\dots,T_{m}$.

Proof: Replacing each $T_{j}$ by its real and imaginary parts, we may assume
the $T_{j}$ are self-adjoint. Define a relation $\sim$ on $X$ by $x\sim y$ if
there are $n$ and $x=x_{1},x_{2},\dots,x_{n}=y$ such that for each $k$, $1\leq
k\leq n-1$, there is a $j$ with $\langle
T_{j}\xi_{x_{k}},\xi_{x_{k+1}}\rangle\neq 0$ ($j$ can vary with $k$). This
relation is symmetric since $T_{j}=T_{j}^{*}$ for all $j$, transitive, and
reflexive (if $T_{j}\neq 0$ for some $j$). Thus $X$ partitions into
$\sim$-equivalence classes. If $X$ is CF, each equivalence class is finite by
8.1.1. Each equivalence class is invariant under $T_{1},\dots,T_{m}$.

##### 8.1.3

Corollary. Let $X$ be a CF set. Then ${\mathcal{B}}(\ell^{2}(X))$ is stably
finite.

Proof: Let $T$ be an isometry in ${\mathcal{B}}(\ell^{2}(X))$. Suppose $T$ is
not unitary, i.e. not surjective, and let $\eta$ be a vector not in the range
of $T$. Let $E$ be the support of $\eta$, and let $F$ be a finite set
containing $E$ invariant under $T$. Then the restriction of $T$ to
${\mathcal{Y}}=\operatorname{span}\\{\xi_{x}:x\in F\\}$ is an isometry in
${\mathcal{B}}({\mathcal{Y}})$. But ${\mathcal{Y}}$ is finite-dimensional, so
$T|_{{\mathcal{Y}}}$ is unitary, i.e. surjective, contradicting that
$\eta\in{\mathcal{Y}}$ is not in the range of $T$. Thus $T$ is unitary. We
conclude that ${\mathcal{B}}(\ell^{2}(X))$ is finite. Since
$X\times\\{0,\dots,n-1\\}$ is also CF for any $n$,
${\mathcal{B}}(\ell^{2}(X))$ is stably finite.

Although a stably finite $\mathrm{C}^{*}$-algebra can have an infinite
quotient (e.g., the cone over the Toeplitz algebra), we can even conclude that
the Calkin algebra ${\mathcal{Q}}(\ell^{2}(X))$ is stably finite (cf. 9.1.8).

###  8.2. ${\mathcal{K}}(\ell^{2}(X))$ for CF Sets

##### 8.2.1

Proposition. Let $X$ be a CF set. Then every compact operator on $\ell^{2}(X)$
has finite rank.

Proof: Let $T\in{\mathcal{K}}(\ell^{2}(X))$. For $\epsilon>0$ set

$A_{\epsilon}=\\{x\in X:\|T\xi_{x}\|\geq\epsilon\\}\ .$

Then $A_{\epsilon}$ is a finite set: if $F$ is a finite subset of $X$ with
$\|T-TP_{F}\|<\epsilon$, and $x\notin F$, then

$\|T\xi_{x}\|=\|T\xi_{x}-TP_{F}\xi_{x}\|\leq\|T-TP_{F}\|\|\xi_{x}\|<\epsilon$

so $A_{\epsilon}\subseteq F$. Then $(A_{1/n})$ is an increasing sequence of
finite subsets of $X$, which must stabilize at some finite $A$ since $X$ is
CF. The range of $T$ is contained in $\operatorname{span}\\{\xi_{x}:x\in
A\\}$.

###  8.3. How Big Is ${\mathcal{Q}}(\ell^{2}(X))$?

##### 8.3.1

One might expect that if $X$ is sufficiently DF, then
${\mathcal{Q}}({\mathcal{H}})$ (${\mathcal{H}}=\ell^{2}(X)$) might be “small.”
We could even potentially have ${\mathcal{Q}}({\mathcal{H}})\cong\mathbb{C}$,
i.e. ${\mathcal{B}}({\mathcal{H}})={\mathcal{K}}({\mathcal{H}})+\mathbb{C}1$.
(Note that if ${\mathcal{H}}$ is any infinite-dimensional Hilbert space, the
identity operator on ${\mathcal{H}}$ is not compact since the closed unit ball
of ${\mathcal{H}}$ is not totally bounded, cf. 4.0.2). This could potentially
happen, though, only if $X$ is rigid:

##### 8.3.2

Any permutation $\pi$ of a set $X$ defines a unitary operator $U_{\pi}$ on
$\ell^{2}(X)$ by

$U_{\pi}\left(\sum_{x\in X}c_{x}\xi_{x}\right)=\sum_{x\in X}c_{x}\xi_{\pi(x)}\
.$

If $\pi$ moves infinitely many elements of $X$, then $U_{\pi}$ is not a
compact perturbation of a scalar. Thus permutations of $X$ moving infinitely
many elements define nontrivial unitaries in ${\mathcal{Q}}(\ell^{2}(X))$.

##### 8.3.3

Proposition. If $X$ is strongly amorphous, then ${\mathcal{B}}(\ell^{2}(X))$
is equal to ${\mathcal{K}}(\ell^{2}(X))+\mathbb{C}1$, hence
${\mathcal{Q}}(\ell^{2}(X))$ is one-dimensional and
${\mathcal{B}}(\ell^{2}(X))$ is an AF algebra.

Proof: Let $T\in{\mathcal{B}}(\ell^{2}(X))$. By 8.1.2, $X$ partitions into
finite subsets invariant under $T$. All but finitely many equivalence classes
must be singletons, i.e. all but finitely many $\xi_{x}$ are eigenvectors, so
$T$ is a compact perturbation of a diagonal operator. All but finitely many of
the diagonal eigenvalues must be the same since $X$ is amorphous. Thus $T$ is
a compact (even finite-rank) perturbation of a scalar operator.

The last sentence follows by Proposition 7.1.7.

There is an interesting reinterpretation of this result:

##### 8.3.4

Corollary. Let $X$ be a strongly amorphous set, and
${\mathcal{H}}=\ell^{2}(X)$. Then ${\mathcal{H}}$ is infinite-dimensional, and
every closed subspace of ${\mathcal{H}}$ has either finite dimension or finite
codimension.

##### 8.3.5

This is a “hereditarily indecomposable Hilbert space,” a Hilbert space version
of [GM93], where (under AC) a hereditarily indecomposable Banach space is
constructed. In fact, it is “better,” since under AC every infinite-
dimensional Banach space has an infinite-dimensional closed subspace of
infinite codimension (an indecomposable Banach space is just one where no
infinite-dimensional closed subspace has a closed infinite-dimensional
complement). Our ${\mathcal{H}}$ also has the property that every compact
operator on ${\mathcal{H}}$ has finite rank; under CC every infinite-
dimensional Banach space has an infinite-rank compact operator. (However, by
[AH11] there is a separable Banach space such that every operator on it is of
the form scalar + compact.) Of course, ${\mathcal{H}}$ only exists under set-
theoretic assumptions analysts might find objectionable.

It is possible that “closed” can be removed from the statement of 8.3.4 (note
that ${\mathcal{H}}$ has nonclosed subspaces by 6.0.6). This is just a vector
space problem, since ${\mathcal{H}}$ is a complex vector space with a Hamel
basis indexed by $X$.

##### 8.3.6

In the situation of Proposition 8.3.3, ${\mathcal{B}}(\ell^{2}(X))$ is also
peculiar. By a classical result of Szankowski ([Sza81]),
${\mathcal{B}}(\ell^{2}(\mathbb{N}))$ does not have the Approximation Property
for Banach spaces. (Recall that a Banach space $X$ has the _Approximation
Property_ if the identity operator on $X$ can be approximated by finite rank
operators uniformly on compact sets.) This implies that if ${\mathcal{H}}$ has
a closed subspace isomorphic to $\ell^{2}(\mathbb{N})$, then
${\mathcal{B}}({\mathcal{H}})$ does not have the Approximation Property.
However, if $X$ is strongly amorphous then Propositon 8.3.3 implies that
${\mathcal{B}}(\ell^{2}(X))$ is AF. In this case conditional expectations to
finite-dimensional $\mathrm{C}^{*}$-subalgebras witness the Approximation
Property of ${\mathcal{B}}(\ell^{2}(X))$.

##### 8.3.7

If $X$ is a strongly amorphous set, and $Y=X\times\\{0,\dots,n-1\\}$, then
${\mathcal{Q}}(\ell^{2}(Y))\cong M_{n}(\mathbb{C})$. The most interesting case
may be if $X$ is CF but far from rigid, i.e. has many nontrivial permutations;
then ${\mathcal{Q}}(\ell^{2}(X))$ is infinite-dimensional and stably finite.
It is hard to see how such a ${\mathcal{Q}}$ could have any natural traces.
(Our original motivation for this work was to try to find an example of a
stably finite unital C*-algebra without trace, although it now does not seem
promising that such an example can be constructed by these methods unless we
give up the Hahn-Banach Theorem, without which functional analysis is
essentially impossible.)

However, there is no reason to believe that ${\mathcal{Q}}({\mathcal{H}})$ is
necessarily simple (in the case where $X$ is DF of some kind). Even in the
presence of AC, the Calkin algebra associated to a non-separable Hilbert space
is not simple: the lattice of closed ideals of
${\mathcal{Q}}(\ell^{2}(\kappa))$ is isomorphic to the poset of infinite
cardinals $\leq\kappa$ (see e.g., [Far19, §12.3.1]). Something similar, but
more complicated, may be true for ${\mathcal{Q}}(\ell^{2}(\kappa))$ for
$\kappa$ DF (if $\kappa$ is DF, $\ell^{2}(\kappa)$ is non-separable). In the
absence of CC, there may even be Hilbert spaces which are so different that
every operator between them is compact. For such pair of spaces, the Calkin
algebra of their direct sum is the direct sum of their Calkin algebras, and
therefore non-simple and even with nontrivial center. Therefore direct sums of
such Hilbert spaces would have nonsimple Calkin algebras ($\ell^{2}(X)$, when
$X$ is a Russell set, may have this property; see 6.1.1).

Actually, this happens. The following uses a property of Cohen’s original CF
set ([Coh66]) that sits between being rigid and strongly rigid and whose proof
is included in Proposition 10.0.3.

##### 8.3.8

Proposition. If $X$ is the CF set that has the property that for every
partition of $X$ into nonempty finite sets, all but finitely many cells are
singletons, then the Calkin algebra ${\mathcal{Q}}(\ell^{2}(X))$ is a non-
separable abelian $\mathrm{C}^{*}$-algebra.

Proof: We first prove that ${\mathcal{Q}}(\ell^{2}(X))$ is equal to its
commutative subalgebra, $\ell^{\infty}(X)/c_{0}(X)$. Fix
$T\in{\mathcal{B}}(\ell^{2}(X))$. Then Proposition 8.1.2 implies that $X$ can
be partitioned into finite sets $(X_{i})$ such that $\ell^{2}(X_{i})$ is a
reducing subspace for $T$ for all $i$. By the assumption, all but finitely
many of the $X_{i}$ are singletons. This implies that $T$ is a compact
perturbation of a diagonal operator. Since $T$ was arbitrary,
${\mathcal{Q}}(\ell^{2}(X))$ is included in $\ell^{\infty}(X)/c_{0}(X)$, and
therefore equal to it.

In order to prove that this algebra is non-separable, we recall that $X$ is a
set of reals and consider it with the subspace topology. We first note that
$X$ has at most finitely many isolated points,666As a matter of fact, by
construction $X$ has no isolated points. and we can therefore assume that $X$
has no isolated points. We may also assume $X$ is bounded, and therefore the
closure $\overline{X}$ of $X$ is a perfect subset of $\mathbb{R}$. For every
function $f\colon X\to\mathbb{C}$, the operator with eigenvalue $f(x)$
corresponding to the eigenvector $\delta_{x}$ for $x\in X$ belongs to
$\ell^{\infty}(X)/c_{0}(X)$. For $c\in\overline{X}$ let $f_{c}$ be the
characteristic function of the interval $(-\infty,c)$. If $c<d$, then
$X\cap(c,d)$ is infinite, hence (with
$\pi\colon{\mathcal{B}}(\ell^{2}(X))\to{\mathcal{Q}}(\ell^{2}(X))$ we have
that $\|\pi(f_{c}-f_{d})\|=1$. Therefore $\\{\pi(f_{c})\mid c\in(a,b)\\}$ is
an uncountable discrete subset of $\ell^{\infty}(X)/c_{0}(X)$, as required.

The following complements Proposition 8.3.8.

##### 8.3.9

Proposition. If there exists a CF set, then there exists a set $X$ such that
${\mathcal{B}}(\ell^{2}(X))$ is stably finite and ${\mathcal{Q}}(\ell^{2}(X))$
is noncommutative.

Proof: If $Y$ is CF, then so is $X=Y\times Y$ (Proposition 5.2.9), and
${\mathcal{B}}(\ell^{2}(X))$ is stably finite (Corollary 8.1.3). We will prove
that ${\mathcal{Q}}(\ell^{2}(X))$ contains an isomorphic copy of
$M_{n}(\mathbb{C})$ for every $n\in\mathbb{N}$. Fix $F\subseteq Y$ of
cardinality $n$. Every permutation $\pi$ of $F$ defines a permutation
$\tilde{\pi}$ of $X$ that agrees with $\pi\times\operatorname{id}_{Y}$ on
$F\times Y$ and fixes all points in $(Y\setminus F)\times X$. This defines a
unitary $U_{\pi}$ on ${\mathcal{B}}(\ell^{2}(X))$. Similarly, every
$g\in\ell^{\infty}(F)$ defines an element of $\ell^{\infty}(X)$ by
$T_{g}((a,b))=g(a)$ is $a\in F$ and $T_{g}((a,b))=0$ otherwise. The algebra
generated by all $U_{\pi}$ and all $T_{g}$ is isomorphic to the subalgebra of
$M_{n}(\mathbb{C})$ generated by all permutation matrices and all diagonal
matrices, which is $M_{n}(\mathbb{C})$. Since $Y$ is infinite, the quotient
map acts on this algebra as an isometry, and this concludes the proof.

###  8.4. Stably Finite $\mathrm{C}^{*}$-algebras With no Quasitraces

The original impetus for this work came from one of the most notorious open
problems on $\mathrm{C}^{*}$-algebras (see [Haa14]): Does every stably finite
$\mathrm{C}^{*}$-algebra have a tracial state? In the absence of the Axiom of
Choice we provide two examples of stably finite $\mathrm{C}^{*}$-algebras
without tracial states (and even without quasitraces) of varying levels of
satisfactoriness.

#####  8.4.1

Known results provide an easy example of a $\mathrm{C}^{*}$-algebra which is
stably finite but has no quasitraces in every model from the most studied
family of models of ZF in which the Axiom of Choice fails. In the original
Solovay’s model in which all sets of reals are Lebesgue measurable ([Sol70]),
and in other models of the Axiom of Determinacy (see e.g.[Kan08]), all sets of
reals have the Property of Baire. (Notably, unlike the statement that all sets
of reals are Lebesgue measurable, the assertion that all sets of reals, and
all subsets of any Polish space, have the Property of Baire does not require
an inaccessible cardinal. By [She84, §7], the consistency of ZFC implies the
consistency of ZF+‘all sets of reals have the property of Baire’.)

#####  8.4.2

Lemma. The $\mathrm{C}^{*}$-algebra $\ell^{\infty}(X)/c_{0}(X)$ has a state if
and only if there is a finitely additive probability measure on
${\mathcal{P}}(X)$ that vanishes on singletons.

Proof: Identify ${\mathcal{P}}(X)$ with the lattice of projections of
$\ell^{\infty}(X)$. We claim that finitely additive probability measures on
${\mathcal{P}}(X)$ are in a bijective correspondence with states on
$\ell^{\infty}(X)$. The restriction of a state of the latter gives a finitely
additive probability measure. Conversely, a finitely additive probability
measure on ${\mathcal{P}}(X)$ can be linearly extended to a state on the
$*$-algebra of all finite linear combinations of projections in
$\ell^{\infty}(X)$. This state has norm 1, and the set of linear combinations
of projections is dense in $\ell^{\infty}(X)$, and therefore its continuous
extension is a state on $\ell^{\infty}(X)$.

Finally, a finitely additive measure on ${\mathcal{P}}(X)$ vanishes on the
singletons if and only if the associated state vanishes on $c_{0}(X)$, and
this completes the proof.

#####  8.4.3

Corollary. If all sets of reals have the Property of Baire then the
commutative $\mathrm{C}^{*}$-algebra
$\ell^{\infty}(\mathbb{N})/c_{0}(\mathbb{N})$ has no nontrivial bounded linear
functionals.

In particular, it does not have a nontrivial representation on a Hilbert space
and it is not isomorphic to $C(X)$ for a compact Hausdorff space $X$.

Proof: As pointed out without a proof in [Sol70, p. 3], if all sets of reals
have the Property of Baire then there is no finitely additive probability
measure on ${\mathcal{P}}(\mathbb{N})$ that vanishes on singletons (for a
proof see e.g., [Pin06, (8) on p. 206]). This implies that the commutative
$\mathrm{C}^{*}$-algebra $A=\ell^{\infty}(\mathbb{N})/c_{0}(\mathbb{N})$ has
no nontrivial bounded linear functionals.

Suppose that $\pi\colon A\to{\mathcal{B}}({\mathcal{H}})$ is a nontrivial
representation on some Hilbert space ${\mathcal{H}}$. Then there are
$\xi\in{\mathcal{H}}$ and $T$ in the image such that $\langle
T\xi,\xi\rangle\neq 0$, and therefore $\varphi(a)=\langle\pi(a)\xi,\xi\rangle$
is a nontrivial state on $A$; contradiction.

Finally, suppose that $A$ is isomorphic to $C(X)$ for a compact Hausdorff
space $X$, or even that $\Phi\colon A\to C(X)$ is a nontrivial *-homomorphism.
Then for some $x\in X$ the composition of $\Phi$ with the evaluation
functional on $C(X)$ is a nontrivial state on $A$; contradiction.

Corollary 8.4.3 is not satisfactory, since the reason for the absence of
tracial states in $\ell^{\infty}(\mathbb{N})/c_{0}(\mathbb{N})$ is the failure
of the Hahn–Banach theorem. The assumption that all sets of reals have the
Property of Baire implies that ${\mathcal{B}}(\ell^{2}(\mathbb{N}))$ has a
unique pure state, up to the unitary equivalence ([Far19, Exercise 12.6.6]),
making it a counterexample to Naimark’s Problem (see [AW04] and [Far19,
§11.2]).

We don’t know whether the assumption of the following proposition is
relatively consistent with ZF (see 11.0.2).

##### 8.4.4

Proposition. Suppose that there is a CF set $X$ such that ${\mathcal{P}}(X)$
has no finitely additive probability measure that vanishes on singletons. Then
${\mathcal{B}}(\ell^{2}(X))$ is stably finite, it has no tracial states, but
its states separate the points.

Proof: Since $X$ is CF, ${\mathcal{B}}(\ell^{2}(X))$ is stably finite by
Corollary 8.1.3. Suppose $\tau$ is a tracial state of
${\mathcal{B}}(\ell^{2}(X))$. We claim that $\tau$ vanishes on compact
operators. Otherwise, since projections of finite rank form an approximate
unit in ${\mathcal{K}}(\ell^{2}(X))$, we can fix a projection $p$ of rank 1
and $n\geq 1$ such that $\tau(p)>1/n$. Since $X$ is infinite, we can find $n$
orthogonal projections of rank 1 in ${\mathcal{B}}(\ell^{2}(X))$,
$p_{1},\dots,p_{n}$. All of these projections are Murray–von Neumann
equivalent, and therefore $\tau(p_{j})=\tau(p)>1/n$ for all $j$. This implies
$\tau(\sum_{j}p_{j})>1$, contradicting the assumption that $\tau$ is a state.

Therefore the restriction of $\tau$ to $\ell^{\infty}(X)$ defines a state on
$\ell^{\infty}(X)/c_{0}(X)$. Since no finitely additive measure on
${\mathcal{P}}(X)$ vanishes on singletons, we have contradiction by Lemma
8.4.2.

Finally, any two distinct operators in ${\mathcal{B}}(\ell^{2}(X))$ are
separated by a vector state associated with some $\xi\in\ell^{2}(X)$, and this
completes the proof.

###  8.5. Spectrum of Operators

##### 8.5.1

If $X$ is sufficiently DF, one might expect operators on
${\mathcal{H}}=\ell^{2}(X)$ to have small spectrum. For example, if $X$ is CF,
8.2.1 implies that the spectrum of every compact operator on ${\mathcal{H}}$
is finite.

##### 8.5.2

Theorem. Let $X$ be a power Dedekind-finite set, and
$T\in{\mathcal{B}}(\ell^{2}(X))$, Then $\sigma(T)$ is finite, and each
$\lambda\in\sigma(T)$ is an eigenvalue.

Proof: We may assume $X$ is infinite (PF) since the finite case is well known.
Thus $X$ is CF, so by 8.1.2 $X$ partitions into finite subsets $\\{X_{i}:i\in
I\\}$ whose spans ${\mathcal{H}}_{i}$ are invariant under $T$. There is a
finite-to-one function $f:X\to I$. Let $Y\subseteq\mathbb{C}$ be $\cup_{i\in
I}S_{i}$, where $S_{i}$ is the (finite) set of eigenvalues of the restriction
of $T$ to ${\mathcal{H}}_{i}$.

We claim $Y$ is finite. If not, then there is a partition of $Y$ into a
disjoint sequence of nonempty subsets $Y_{n}$ (5.2.5). For each $n$ let
$I_{n}$ be the set of all $i\in I$ such that $S_{i}\cap Y_{n}\neq\emptyset$.
Since $S_{i}$ is finite, each $i\in I$ is in only finitely many $I_{n}$. But
$\cup_{n}I_{n}=I$, an infinite set, so there are an infinite number of
distinct $I_{n}$. If $A_{n}=f^{-1}(I_{n})$, then each $A_{n}$ is a subset of
$X$, and infinitely many are distinct, contradicting that $X$ is PF.

Now we claim that $Y$ is the entire spectrum of $T$. Let
$\lambda\in\mathbb{C}\setminus Y$. Then $T-\lambda 1$ is bounded below on
${\mathcal{H}}_{i}$ for each $i\in I$. For each $n$, let $J_{n}$ be the set of
$i\in I$ such that $T-\lambda 1$ is bounded below by $\frac{1}{n}$ on
${\mathcal{H}}_{i}$, and let $B_{n}=f^{-1}(J_{n})\subseteq X$. We have
$J_{n}\subseteq J_{n+1}$ and hence $B_{n}\subseteq B_{n+1}$ for all $n$;
$\cup_{n}J_{n}=I$ so $\cup_{n}B_{n}=X$. Since $X$ is PF, we must have
$B_{n}=X$ for some $n$, i.e. $T-\lambda 1$ is bounded below by $\frac{1}{n}$
on all of $\ell^{2}(X)$ and hence invertible.

The conclusion of Theorem 8.5.2 does not imply that every operator is the sum
of a scalar and a compact operator. For example, if $X=\bigsqcup_{z\in
Z}X_{z}$, where $Z$ is strongly amorphous, is a set as in §6.2.2, then the
permutation that swaps each of the pairs $X_{z}$ moves all points of $X$, and
therefore defines a unitary which is not a compact perturbation of a scalar
(§8.3.2).

##### 8.5.3

If $X$ is weakly Dedekind-infinite, there are operators on
${\mathcal{H}}=\ell^{2}(X)$ with large spectrum. Write $X$ as a countable
union of pairwise disjoint nonempty sets $A_{n}$. If $K$ is a separable
compact subset of $\mathbb{C}$ (separability of subsets of $\mathbb{R}$ or
$\mathbb{C}$ is not automatic without CC, but there are many separable compact
subsets, e.g. the closure of any countable bounded subset), then there is a
diagonal operator $T$ on ${\mathcal{H}}$ whose spectrum is $K$, defined by
taking a dense sequence $(a_{n})$ in $K$, for $x\in X$ setting $n(x)$ the $n$
for which $x\in A_{n}$, and setting

$T\left(\sum_{x\in X}c_{x}\xi_{x}\right)=\sum_{x\in X}a_{n(x)}c_{x}\xi_{x}\ .$

If all the $A_{n}$ are infinite (which can be arranged by combining them), the
spectrum of the image of $T$ in ${\mathcal{Q}}({\mathcal{H}})$ is also $K$.

##### 8.5.4

Let $X$ be a CF set which is not PF, and write $X$ as a disjoint union of a
sequence $(A_{n})$ of nonempty subsets. Let $T$ be the operator on
$\ell^{2}(X)$ which is $\frac{1}{n}$ times the identity on the span of
$A_{n}$. Then $T$ is bounded ($\|T\|=1$) and injective, and is also surjective
since every vector in $\ell^{2}(X)$ has finite support (Proposition 6.0.3).
But the inverse $T^{-1}$ is not bounded. Thus the Bounded Inverse Theorem,
Open Mapping Theorem, and Closed Graph Theorem all fail in this
situation.777It is known that the Open Mapping Theorem is roughly equivalent
to the countable Axiom of Choice, [Fel17]. By considering the finite
truncations of $T^{-1}$, a sequence of pointwise bounded but not uniformly
bounded operators is obtained, showing that the Uniform Boundedness Theorem
also fails. The usual proofs of these theorems use some version of the Baire
Category Theorem, equivalent to the Axiom of Dependent Choice, which implies
CC and thus is incompatible with the existence of DF sets.

There is another similar example: the Banach spaces $\ell^{p}(X)$, $1\leq
p<\infty$, are all the same set of functions, and the norms are comparable but
pairwise nonequivalent.

##### 8.5.5

It would also be interesting to study the properties of
${\mathcal{B}}({\mathcal{H}})$, etc., for Hilbert spaces ${\mathcal{H}}$ which
do not have orthonormal bases. It is difficult to find tools to analyze these
cases, however. Even a Hilbert space ${\mathcal{H}}$ with an orthonormal basis
may have closed subspaces without one (6.2.3 (2)), hence
${\mathcal{B}}({\mathcal{H}})$ can have corners without sufficiently many rank
1 projections.

##  9\. New Size and Comparison Relations on Hilbert Spaces and Sets

###  9.1. Hilbert Dedekind-finiteness

There are Hilbert space analogs of the principal flavors of Dedekind-
finiteness, which lead to potentially new flavors on the set-theoretic level:

##### 9.1.1

Definition. Let ${\mathcal{H}}$ be a Hilbert space. Then

  1. (i)

${\mathcal{H}}$ is Hilbert Dedekind-infinite if ${\mathcal{H}}$ is
isometrically isomorphic to a proper subspace of itself. Otherwise
${\mathcal{H}}$ is Hilbert Dedekind-finite.

  2. (ii)

${\mathcal{H}}$ is Hilbert Cohen-infinite if ${\mathcal{H}}$ has a sequence of
mutually orthogonal nonzero finite-dimensional subspaces. Otherwise
${\mathcal{H}}$ is Hilbert Cohen-finite.

  3. (iii)

${\mathcal{H}}$ is Hilbert dually Dedekind-infinite if
${\mathcal{B}}({\mathcal{H}})$ contains an operator which is surjective but
not injective. Otherwise ${\mathcal{H}}$ is Hilbert dually Dedekind-finite.

  4. (iv)

${\mathcal{H}}$ is Hilbert weakly Dedekind-infinite if ${\mathcal{H}}$ has a
sequence of mutually orthogonal nonzero subspaces. Otherwise ${\mathcal{H}}$
is Hilbert Power Dedekind-finite.

  5. (v)

${\mathcal{H}}$ is Hilbert-amorphous if ${\mathcal{H}}$ is infinite-
dimensional and does not contain two orthogonal infinite-dimensional
subspaces, i.e. every closed subspace either has finite dimension or finite
codimension.

If $X$ is a set, say $X$ is Hilbert Dedekind-finite, etc., if $\ell^{2}(X)$ is
Hilbert Dedekind-finite, etc.

#### Notation

For a Hilbert space, HDF (resp. HCF, HDDF, HPF) means “infinite-dimensional
and Hilbert Dedekind-finite (resp. Hilbert Cohen-finite, Hilbert dually
Dedekind-finite, Hilbert power Dedekind-finite), and similarly for a set.

##### 9.1.2

It is clear that Hilbert-amorphous $\Rightarrow$ HPF $\Rightarrow$ HCF
$\Rightarrow$ HDF (cf. 9.1.3), and that for sets HDF $\Rightarrow$ DF, HCF
$\Rightarrow$ CF, HPF $\Rightarrow$ PF, and Hilbert-amorphous $\Rightarrow$
amorphous (the usual notions are the special cases of the Hilbert versions for
subspaces spanned by sets of standard basis vectors). Also, ${\mathcal{H}}$ is
Hilbert Dedekind-finite if and only if every isometry in
${\mathcal{B}}({\mathcal{H}})$ is unitary, i.e. if and only if
${\mathcal{B}}({\mathcal{H}})$ is finite.

There is considerable collapsing among these notions, particularly the set
ones:

##### 9.1.3

Proposition. Let ${\mathcal{H}}$ be a Hilbert space. The following are
equivalent:

  1. (i)

${\mathcal{H}}$ contains an infinite orthonormal sequence.

  2. (ii)

${\mathcal{H}}$ is Hilbert Dedekind-infinite.

  3. (iii)

${\mathcal{H}}$ is Hilbert dually Dedekind-infinite.

Proof: (i) $\Rightarrow$ (ii): If ${\mathcal{H}}$ contains an orthonormal
sequence, then ${\mathcal{H}}$ contains a closed subspace isometrically
isomorphic to $\ell^{2}(\mathbb{N})$, and there is a nonunitary isometry on
this subspace, which can be expanded to a nonunitary isometry on
${\mathcal{H}}$ by making it the identity on the orthogonal complement.

(ii) $\Rightarrow$ (i): if $V$ is a nonunitary isometry in
${\mathcal{B}}({\mathcal{H}})$, and ${\mathcal{Y}}$ is the range of $V$, let
$\xi$ be a unit vector in ${\mathcal{Y}}^{\perp}$. Then $\xi$, $V\xi$,
$V^{2}\xi$, $\dots$ is an orthonormal sequence in ${\mathcal{H}}$.

(ii) $\Rightarrow$ (iii): If $V$ is a nonunitary isometry in
${\mathcal{B}}({\mathcal{H}})$, then $V^{*}$ is surjective but not injective.

(iii) $\Rightarrow$ (ii): If $T$ is surjective but not injective, let
${\mathcal{Y}}={\mathcal{N}}(T)^{\perp}$. Then the restriction of $T$ to
${\mathcal{Y}}$ is an injective map from ${\mathcal{Y}}$ to ${\mathcal{H}}$,
so if $T=V|T|$ is the polar decomposition of $T$, then $V$ is a coisometry,
and $V^{*}$ is an isometry from ${\mathcal{H}}$ onto ${\mathcal{Y}}$.

##### 9.1.4

Corollary. Let $X$ be an infinite set. The following are equivalent:

  1. (i)

$X$ is CF.

  2. (ii)

$X$ is HDF.

  3. (iii)

$X$ is HDDF.

Proof: By 9.1.3, (ii) $\Longleftrightarrow$ (iii).

(ii) $\Longleftrightarrow$ (i): By 9.1.3, $X$ is not HDF if and only if
$\ell^{2}(X)$ contains a countably infinite orthonormal sequence. By 6.2.7,
this is equivalent to $X$ not being CF.

##### 9.1.5

It is unclear whether HCF is equivalent to CF (and hence to HDF and HDDF) for
sets (but see 6.0.3 and 6.2.7). HPF is distinct from the conditions of 9.1.4
for sets since HPF $\Rightarrow$ PF and there are sets which are CF but not
PF. It is unclear whether HPF is the same as PF since closed subspaces of
$\ell^{2}(X)$, $X$ PF, are potentially skew from the standard basis. Thus HPF
is potentially a Dedekind-finiteness condition on sets strictly more
restrictive than PF, and HCF is also a potentially new condition.

##### 9.1.6

Similarly, Hilbert-amorphous is potentially strictly stronger than amorphous
for sets. However, by Corollary 8.3.4 a strongly amorphous set is Hilbert-
amorphous. The converse is quite unclear. Thus Hilbert-amorphous is
potentially a Dedekind-finite condition strictly between being amorphous and
strictly amorphous.

Here is an extended version of Fig. 1 from Section 5.7.

strongly amorphousHilbert-amorphousHPFamorphousstrongly
rigidHCFPFRussellCF=HDFDDFrigidDF Figure 2. Implications between different
flavors of Dedekind–finite sets discussed in this section. Hilbert-amorphous,
HPF, HCF, and HDF will be defined in §9.

A simple but interesting observation is:

##### 9.1.7

Proposition. A finite direct sum of HDF Hilbert spaces is HDF.

Proof: Let ${\mathcal{H}}_{1},\dots,{\mathcal{H}}_{n}$ be Hilbert spaces.
Suppose $(\xi_{j})$ is an orthonormal sequence in
${\mathcal{H}}={\mathcal{H}}_{1}\oplus\cdots\oplus{\mathcal{H}}_{n}$. Then the
subspace of ${\mathcal{H}}$ spanned by the $\xi_{j}$ is infinite-dimensional.
For each $k\leq n$, let $(\xi_{j,k})$ be the sequence of $k$’th coordinates of
$\xi_{j}$, and let ${\mathcal{X}}_{k}$ the closed subspace of
${\mathcal{H}}_{k}$ spanned by the $\xi_{j,k}$. The ${\mathcal{X}}_{k}$ cannot
all be finite-dimensional, so there is a $k$ such that the sequence
$(\xi_{j,k})$ contains a linearly independent subsequence, from which an
orthonormal sequence can be made by Gram–Schmidt. Thus ${\mathcal{H}}_{k}$ is
not HDF.

##### 9.1.8

Proposition.

Let ${\mathcal{H}}$ be a Hilbert space. Then the following are equivalent:

  1. (i)

${\mathcal{B}}({\mathcal{H}})$ is finite.

  2. (ii)

${\mathcal{B}}({\mathcal{H}})$ is stably finite.

  3. (iii)

${\mathcal{Q}}({\mathcal{H}})$ is finite.

  4. (iv)

${\mathcal{Q}}({\mathcal{H}})$ is stably finite.

Proof: Since $M_{n}({\mathcal{B}}(H))$ is isomorphic to
${\mathcal{B}}(H^{n})$, (i) $\Longleftrightarrow$ (ii) follows from 9.1.7.

(i) $\Longleftrightarrow$ (iii): If ${\mathcal{B}}({\mathcal{H}})$ is
infinite, then by 9.1.3 ${\mathcal{H}}$ contains an orthonormal sequence,
hence an isometry of infinite codimension, whose image in
${\mathcal{Q}}({\mathcal{H}})$ is a nonunitary isometry, so
${\mathcal{Q}}({\mathcal{H}})$ is also infinite. Conversely, if $v$ is a
nonunitary isometry in ${\mathcal{Q}}({\mathcal{H}})$, let
$T\in{\mathcal{B}}({\mathcal{H}})$ have image $v$. If $T=V|T|$ is the polar
decomposition, then the image of $V$ is also $v$. $V$ is a partial isometry
whose source projection $V^{*}V$ has finite codimension and range projection