# On the cardinality of separable pseudoradial spaces

Alan Dow  Department of Mathematics, University of North Carolina at
Charlotte, Charlotte, NC 28223<EMAIL_ADDRESS>and  István Juhász  Alfréd Rényi
Institute of Mathematics, Hungarian Academy of Sciences<EMAIL_ADDRESS>
###### Abstract.

The aim of this paper is to consider questions concerning the possible maximum
cardinality of various separable pseudoradial (in short: SP) spaces. The most
intriguing question here is if there is in ZFC a regular (or just Hausdorff)
SP space of cardinality $>\mathfrak{c}$. While this question is left open, we
establish a number of non-trivial results that we list below.

  * •

It is consistent with $MA\,+\,\mathfrak{c}=\aleph_{2}$ that there is a
countably tight and compact SP space of cardinality $2^{\mathfrak{c}}$.

  * •

If $\kappa$ is a measurable cardinal then in the forcing extension obtained by
adding $\kappa$ many Cohen reals, every countably tight regular SP space has
cardinality at most $\mathfrak{c}$.

  * •

If $\kappa>\aleph_{1}$ Cohen reals are added to a model of GCH then in the
extension every pseudocompact SP space with a countable dense set of isolated
points has cardinality at most $\mathfrak{c}$.

  * •

If $\mathfrak{c}\leq\aleph_{2}$ then there is a 0-dimensional SP space with a
countable dense set of isolated points that has cardinality greater than
$\mathfrak{c}$.

###### Key words and phrases:

tightness, pseudoradial

###### 1991 Mathematics Subject Classification:

54A35

First author supported by NSF grant DMS-1501506 and
Second author supported by NKFIH grant number K 129211

##  1\. Introduction

The class of pseudoradial spaces is a natural and well-studied class that is a
generalization of the radial property in the same way that the class of
sequential spaces is a generalization of the class of Frechet-Urysohn spaces.
For a cardinal $\kappa$, a sequence $\\{x_{\alpha}:\alpha<\kappa\\}$ is said
to converge to $x$ in a space $X$, if for every neighborhood $U$ of $x$ there
is a $\beta<\kappa$ such that $\\{x_{\alpha}:\beta<\alpha<\kappa\\}$ is a
subset of $U$. A set $A$ is radially closed in $X$ if, for every cardinal
$\kappa\leq|X|$, no $\kappa$-sequence of points of $A$ converges to a point
not in $A$. A space $X$ is pseudoradial if every radially closed set is
closed. A set $A$ is sequentially closed if no $\omega$-sequence of points
from $A$ converges to a point not in $A$ and a space is sequential if every
sequentially closed set is closed.

As stated in the abstract, we are exploring the question of whether separable
pseudoradial spaces of cardinality greater than $\mathfrak{c}$ exist. We may
say that a space is large if it has cardinality greater than $\mathfrak{c}$.
We thank A. Bella for informing us that it was shown in [ArhBella93] that it
is consistent that the usual product space $2^{\omega_{1}}$ is SP.

One of the most interesting results about compact pseudoradial spaces arose
when Sapirovskii [SapPseudo] proved that the continuum hypothesis implied that
a compact space is pseudoradial so long as it is sequentially compact. A space
is sequentially compact if every infinite sequence has a limit point. This was
improved in [JSpseudo] where it was shown that it follows from
$\mathfrak{c}\leq\aleph_{2}$ that compact sequentially compact spaces are
pseudoradial and that this bound can not be improved in ZFC. In this paper we
are able to show that this same assumption of $\mathfrak{c}\leq\aleph_{2}$ is
sufficient to produce examples of regular Hausdorff separable pseudoradial
spaces of cardinality equal to $2^{\mathfrak{c}}$. In this paper we will
restrict our investigation to regular Hausdorff spaces and note that
$2^{\mathfrak{c}}$ is the upper bound on the cardinality of any separable
space.

Returning to the class of compact separable spaces, we recall that every
pseudoradial compact space is sequentially compact. More generally in a
pseudoradial space, a countable discrete set is closed if it contains no
converging sequence, and so clearly, a compact pseudoradial space is
sequentially compact. It is well-known that the sequential closure of a
countable set has cardinality at most $\mathfrak{c}$ and therefore every
sequential separable space has cardinality at most $\mathfrak{c}$. Balogh
[Balogh] proved that the proper forcing axiom implies that every compact space
of countable tightness is sequential, and therefore compact separable spaces
of countable tightness have cardinality at most $\mathfrak{c}$. Of course this
paper of Balogh’s was in answer to the celebrated Moore-Mrowka problem. Recall
that the proper forcing axiom also implies that $\mathfrak{c}=\aleph_{2}$
[Boban, Justin]. This background motivates one to ask more about separable
pseudoradial spaces of countable tightness both with and without the extra
assumption of compactness.

##  2\. Martin’s Axiom

This project began when we were made aware of a question in connection to the
Moore-Mrowka problem posed by S. Spadaro in MathOverFlow. We thank K.P. Hart
for bringing the question to our attention. We refer the reader to [Kunen] for
details about Martin’s Axiom and to [Balogh] for the statement of the proper
forcing axiom.

> Assume $MA_{\aleph_{1}}$. Is it true that every compact pseudoradial space
> of countable tightness is sequential?

We answer this question in the negative but will rely on quoting two related
results from the literature. It would be too ambitious to reproduce the proofs
from these publications. However we can connect the investigations there to
the current one.

######  Definition 2.1.

A space $X$ is initially $\aleph_{1}$-compact if every open cover of
cardinality at most $\aleph_{1}$ has a finite subcover.

An initially $\aleph_{1}$-compact space is countably compact and a first-
countable initially $\aleph_{1}$-compact space is sequentially compact. It is
shown in [DowSide]*7.1 that every compactification of an initially
$\aleph_{1}$-compact space of countable tightness has countable tightness. It
was proven in [JKL] that there is a model of $\mathfrak{c}=\aleph_{2}$ in
which there exists a separable first-countable locally compact initially
$\aleph_{1}$-compact space that is not compact. The one-point compactification
of this space is compact, separable, sequentially compact, countably tight and
not sequential. Since $\mathfrak{c}\leq\aleph_{2}$ holds in this model, this
space is also pseudoradial. Finally, it was shown explicitly in [DowFeng]*5.11
(and implicitly in [DowSide]*6.3) that one can perform a further ccc forcing
to make $\mbox{MA}_{\aleph_{1}}$ and $\mathfrak{c}=\aleph_{2}$ hold while
ensuring that the original space generates a space in the extension with all
the same properties. This answers Spadaro’s question in the negative.

Now we consider our generalization of Spadaro’s question and ask if Martin’s
Axiom is sufficient to ensure that compact separable pseudoradial spaces of
countable tightness have cardinality at most $\mathfrak{c}$. For this we use
the example constructed in [DowSide]*5.5,6.3 that had, in answer to a question
of Arhangelskii, an example of an initially $\aleph_{1}$-compact first-
countable space of cardinality greater than $\mathfrak{c}$. We cite the exact
statement that we will need. This was also shown to hold in a model of
Martin’s Axiom (MA) and $\mathfrak{c}=\aleph_{2}$. Clearly this space was not
compact and it also contained a separable closed subset that had a
compactification of cardinality $2^{\mathfrak{c}}$.

######  Proposition 2.2.

[DowSide]*5.5,6.3 There is a model of MA $+~{}\mathfrak{c}=\aleph_{2}$ in
which there is a separable first-countable initially $\aleph_{1}$-compact
space $X$ with the property that $\beta X\setminus X$ has cardinality
$2^{\aleph_{2}}$ and contains no infinite compact subsets.

######  Corollary 2.3.

It is consistent with $\mbox{MA}_{\aleph_{1}}$ that there is a compact
separable pseudoradial space of countable tightness and cardinality greater
than $\mathfrak{c}$.

###### Proof.

Let $X$ be the space and in the model as stated in Proposition 2.2. We show
that $\beta X$ is the desired example. The only property that needs to be
established is that $\beta X$ is pseudoradial and has countable tightness. We
remarked above that every compactification of a initially $\aleph_{1}$-compact
space has countable tightness. We are working in a model of
$\mathfrak{c}=\aleph_{2}$, so to prove that $\beta X$ is pseudoradial, it is
enough to prove that it is sequentially compact. The space $X$ is sequentially
compact so we consider an infinite countable subset $D$ of $\beta X\setminus
X$. The closure of $D$ is not contained in $\beta X\setminus X$ by the
statement in Proposition 2.2, hence there is a point $x\in X$ that is a limit
point of $D$. It is easily checked that if $\\{U_{n}:n\in\omega\\}$ is a local
base in $X$ for $x$ consisting of cozero subsets of $X$, then the family
$\\{\beta X\setminus\mbox{cl}_{\beta X}(X\setminus U_{n}):n\in\omega\\}$ is a
local base for $x$ in $\beta X$. It follows then that there is a sequence from
$D$ converging to $x$, and this completes the proof that $\beta X$ is
sequentially compact and pseudoradial. ∎

##  3\. cardinality of separable pseudoradial spaces

In this section we begin our investigation of separable pseudoradial spaces in
the absence of the assumption of compactness. We first consider the effect of
the proper forcing axiom in the context of a strengthening of countable
tightness and then, using large cardinals, we establish the consistency of
there being no large separable pseudoradial spaces with countable tightness.

We introduce a natural generalization of the property of a set being
sequentially closed in a space $X$. In particular, in this next definition, we
would say that a sequentially closed set is ${<}\aleph_{1}$-closed.

######  Definition 3.1.

For a cardinal $\kappa$ and a set $A$ in a space $X$, say that $A$ is
${<}\kappa$-radially closed if for every cardinal $\lambda<\kappa$, no
$\lambda$-sequence of points of $A$ converges to a point of $X$ outside of
$A$.

For a set $A$ in a space $X$, let $A^{(<\kappa)}$ denote the smallest
${<}\kappa$-radially closed set containing $A$.

Needless to say, $A^{(<\kappa)}$ exists in every space since the intersection
of ${<}\kappa$-radially closed sets is ${<}\kappa$-radially closed.
Alternatively, $A^{(<\kappa)}$ can be recursively constructed in a manner
analogous to the usual constructions of the sequential closure.

######  Proposition 3.2.

Let $A$ be a subset of a space $X$ and let $\kappa$ be a cardinal. By
induction on $\alpha\leq\kappa^{+}$, define $A_{\alpha}\subset X$ as follows:

  1. (1)

$A_{0}=A$,

  2. (2)

for limit $\alpha\leq\kappa^{+}$,
$A_{\alpha}=\bigcup\\{A_{\beta}:\beta<\alpha\\}$,

  3. (3)

$A_{\alpha+1}$ is the set of all points of $X$ such that there is a cardinal
$\lambda<\kappa$ and a $\lambda$-sequence of points of $A_{\alpha}$ that
converges to $x$.

Then $A_{\kappa^{+}}=A^{(<\kappa)}$.

###### Proof.

It follows easily, by induction on $\alpha\leq\kappa^{+}$ that $A_{\alpha}$ is
a subset of $A^{(<\kappa)}$. So it suffices to note that $A_{\kappa^{+}}$ is
itself ${<}\kappa$-radially closed. ∎

A $\kappa$-sequence $\\{x_{\alpha}:\alpha<\kappa\\}$ is a free sequence
[Arhangelskii] in a space $X$ if, for all $\delta<\kappa$, the initial segment
$\\{x_{\alpha}:\alpha<\delta\\}$ and the final segment
$\\{x_{\alpha}:\delta\leq\alpha<\kappa\\}$ have disjoint closures. The
tightness degree, $t(X)$, of a space $X$ is of course a well-known cardinal
invariant. Similarly, the invariant, $F(X)$, is the supremum of the lengths of
free sequences of $X$. Of course Arhangelskii showed that $t(X)\leq F(X)$
holds for any compact space $X$, and it was shown by Bella [Bella86] that this
inequality also holds for pseudoradial spaces. We record this for future
reference.

######  Proposition 3.3 ([Bella86]).

A pseudoradial space of uncountable tightness contains uncountable free
sequences.

###  3.1.  PFA and countable tightness

######  Proposition 3.4.

PFA implies that every separable regular pseudoradial space of cardinality
greater than $\mathfrak{c}$ contains uncountable free sequences.

###### Proof.

Let $X$ be a separable regular pseudoradial space of cardinality greater than
$\mathfrak{c}$. Since we simply wish to prove that $X$ contains an uncountable
free sequence, we may assume, by Corollary 3.3, that $X$ has countable
tightness. Let $\kappa$ be a large enough regular cardinal so that $X$ is an
element of $H(\kappa)$ (see [Kunen]*IV §6). Let $X$ be an element of an
elementary submodel $M$ of $H(\kappa)$ where $M$ is chosen so that
$|M|=2^{\aleph_{1}}=\mathfrak{c}$ and so that every subset of $M$ of
cardinality at most $\aleph_{1}$ is an element of $M$. We note that $Y=X\cap
M$ is ${<}\aleph_{2}$-radially closed. We may assume that $\omega$ is a dense
subset of $X$. Let
$\mathcal{W}_{z}=\\{W\subset\omega:z\in\mbox{int}_{X}\,\mbox{cl}_{X}(W)\\}$.
Define the family $\mathcal{H}_{z}$ to be all countable subsets of $Y$ that
have $z$ in their closure. Let ${\mathcal{F}}_{z}$ denote the family of all
closed subsets of $Y$ that contain a member of the family
$\\{\mbox{cl}_{Y}(H):H\in\mathcal{H}_{z}\\}$.

######  Claim 1.

$\mathcal{F}_{z}$ is a countably complete maximal filter of closed subsets of
$Y$.

###### Proof of Claim:.

Let $\\{H_{n}:n\in\omega\\}$ be a subset of $\mathcal{H}_{z}$. Let
$F_{\omega}=\bigcap\\{\mbox{cl}_{Y})(H_{n}):n\in\omega\\}$ and assume there is
$W\in\mathcal{W}_{z}$ such that $\mbox{cl}_{X}(W)$ is disjoint from
$F_{\omega}$. Since $W\in M$, we can replace each $H_{n}$ by
$\mbox{int}_{X}\,\mbox{cl}_{X}(W)\cap H_{n}$ since the latter is also in
$\mathcal{H}_{z}$. However, this is impossible since
$z\in\mbox{cl}_{X}(H_{n})$ for all $n\in\omega$, by elementarity, there are
points of $Y$ in $F_{\omega}$. Now it follows that $z$ is in the closure of
$F_{\omega}$, and by countable tightness, there is an
$H_{\omega}\in\mathcal{H}_{z}$ with $H_{\omega}\subset F_{\omega}$. ∎

It is evident from its definition that $\mathcal{F}_{z}$ has a base of
separable sets. Only minor modifications of any number of published proofs in
connection to the Moore-Mrowka problem (e.g. [PrahaIII]*Theorem 3.3) that
these hypotheses on the filter $\mathcal{F}_{z}$ are sufficient to conclude
that there is a proper poset $P$ that will force an uncountable free sequence
in the space $X$. We provide a sketch. Fix a neighborhood assignment
$\mathcal{U}_{Y}=\\{U_{y}:y\in Y\\}$ so that for each $y\in Y$, $z$ is not in
the closure of $U_{y}$. A member $p$ of the poset $P$ is a function into $Y$
with a finite domain $\mathcal{M}_{p}$. The domain $\mathcal{M}_{p}$ is a
finite $\in$-chain of countable elementary submodels of $H(\kappa)$ satisfying
that $\\{X,Y,\mathcal{F}_{z},\mathcal{U}_{Y}\\}$ are members of each. For each
$\\{M_{1},M_{2}\\}\subset\mathcal{M}_{p}$ with $M_{1}\in M_{2}$, the value of
$p(M_{1})$ is an element of $M_{2}\cap F$ for all $F\in\mathcal{F}_{z}\cap
M_{1}$. The poset is ordered by the conditions that $p<q$ providing $p\supset
q$ and for all $M_{1}\in\mathcal{M}_{p}\setminus\mathcal{M}_{q}$, if there is
a (minimal) $M_{2}\in\mathcal{M}_{q}$ such that $M_{1}\in M_{2}$, then
$p(M_{1})\in U_{p(M_{3})}$ for all $M_{3}\in\mathcal{M}_{q}$ satisfying that
$p(M_{2})\in U_{p(M_{3})}$. We omit the proof that $P$ is proper (see any of [
]). It is easily shown that for each $\delta\in\omega_{1}$, the set
$D_{\delta}=\\{p\in P:(\exists M_{p}\in\mathcal{M}_{p})~{}~{}\delta\in
M_{p}\\}$ is a dense subset of $P$. By PFA, there is a filter $G\subset P$
such that $G\cap D_{\delta}$ is not empty for all $\delta<\omega_{1}$. This
implies there is an uncountable set $C_{G}\subset\omega_{1}$ such that
$\gamma\in C_{G}$ if and only if there is an
$M_{\gamma}\in\bigcup\\{\mathcal{M}_{p}:p\in G\\}$ with
$M_{\gamma}\cap\omega_{1}=\gamma$. For each $\gamma\in C_{G}$, choose any
$p_{\gamma}\in G$ such that $M_{\gamma}\in\mathcal{M}_{p_{\gamma}}$. Since $G$
is a filter, we may assume that for all $\gamma\in C_{G}$,
$M_{\min(C_{G})}\in\mathcal{M}_{p_{\gamma}}$.

For each $\gamma\in C_{G}$, also let $F_{\gamma}=\bigcap\mathcal{F}_{z}\cap
M_{\gamma}$. The sequence $\\{x_{\gamma}=p_{\gamma}(M_{\gamma}):\gamma\in
C_{G}\\}$ satisfies that for each $\delta\in C_{G}$, $F_{\delta}$ contains the
set $\\{x_{\gamma}:\delta\leq\gamma\\}$. It also follows that, for
$\gamma<\delta$ both in $C_{G}$, the closure of $U_{x_{\gamma}}$ is disjoint
from $F_{\delta}$ because of the condition that $z$ is not in the closure of
$U_{x_{\gamma}}$. Finally, for each $\min(C_{G})<\delta\in C_{G}$, there is a
$\beta\in C_{G}\cap\delta$ such that $\\{x_{\xi}:\beta<\xi\in
C_{G}\cap\delta\\}$ is contained $U_{x_{\delta}}$. Indeed, then there is a
finite $S_{\delta}\subset\delta{+}1$ such that $\\{x_{\xi}:\xi\in
C_{G}\cap\delta\\}$ is contained in $\bigcup\\{U_{x_{\xi}}:\xi\in
S_{\delta}\\}$. This completes the proof that the closure of
$\\{x_{\xi}:\xi\in C_{G}\cap\delta\\}$ is disjoint from the closure of
$\\{x_{\xi}:\delta<\xi\in C_{G}\\}$. ∎

In compact spaces, the property of being countably tight is equivalent to the
property that every subset has countable $\pi$-character [Boris1]. A family of
non-empty open sets $\mathcal{U}$ is a local $\pi$-base in $X$ at $x$ if every
neighborhood of $x$ contains a member of $\mathcal{U}$. A point has countable
$\pi$-character if it has a countable local $\pi$-base. It was discovered by
Eisworth [Todd] that in the absence of compactness, the property of hereditary
countable $\pi$-character was an important strengthening of countable
tightness. In particular, it was shown in [DowTodd] and [picharacter] that it
was consistent that countably compact subsets of such spaces are closed. Using
this result we note the following application.

######  Corollary 3.5.

PFA implies that every separable countably compact pseudoradial regular space
has cardinality at most $\mathfrak{c}$ if the space also has hereditary
countable $\pi$-character.

###### Proof.

Assume that $X$ is a separable countably compact pseudoradial regular space.
The sequential closure of the countable dense set is countably compact and has
cardinality $\mathfrak{c}$. It was proven in [picharacter] that PFA implies
that countably compact subsets are closed in such spaces. ∎

######  Question 1.

Let $X$ be a countably compact regular pseudoradial space of countable
tightness. Does PFA imply any of the following?

  1. (1)

$X$ has cardinality at most $\mathfrak{c}$.

  2. (2)

Countably compact subsets of $X$ are closed.

  3. (3)

Every subset of $X$ has countable $\pi$-character.

###  3.2. Consistency of no large countably tight examples

In this section we prove that it is consistent, modulo large cardinals, that
every separable regular pseudoradial space of countable tightness has
cardinality at most $\mathfrak{c}$. We give the result using a measurable
cardinal for simplicity, but the same proof can be modified to work for a
weakly compact cardinal. A cardinal $\kappa$ is measurable if there is a
$\kappa$-complete free ultrafilter on $\kappa$.

Before proving the theorem we make note of a simple property shared by
pseudoradial spaces.

######  Proposition 3.6.

Assume that a point $z$ of a pseudoradial regular space $X$ is a limit point
of a countable subset $A$ of $X$. Let $A^{(1)}$ denote the set of points of
$X$ that are limit points of a converging sequence from $A$. Then $z$ is in
the closure of $A^{(1)}\setminus A$.

###### Proof.

Since $z$ is a limit point of $A$ we may assume that $z\notin A$ and, to be
clear, we note that $A\subset A^{(1)}$. If $z$ is an element of $A^{(1)}$ then
there is nothing to prove. Let $W$ be a neighborhood of $z$ and assume that
$A_{W}=\overline{W}\cap A^{(1)}$ is contained in $A$. Certainly $z$ is in the
closure of $A_{W}$ but this contradicts that $X$ is pseudoradial since $A_{W}$
is clearly radially closed. ∎

######  Theorem 3.7.

If $\kappa$ is a measurable cardinal then in the forcing extension obtained by
adding $\kappa$ many Cohen reals, every separable regular pseudoradial space
of countable tightness has cardinality at most $\mathfrak{c}$.

###### Proof.

Let $\kappa$ be a measurable cardinal. By [Jech]*(8.7),10.20 , there is a
normal $\kappa$-complete ultrafilter $\mathcal{U}$ on $\kappa$. This means
that for every $U\in\mathcal{U}$ and regressive function $f$ on $U$ (i.e.
$f(\alpha)<\alpha)$ for all $\alpha\in U$) there is a $U_{1}\in\mathcal{U}$
satisfying that $f(\lambda)=f(\mu)$ for all $\lambda,\mu\in U_{1}$. Note also
that it follows that each $U\in\mathcal{U}$ is a stationary subset of $\kappa$
[Jech]*10.19. More surprisingly, there is an element $U$ of $\mathcal{U}$
consisting only of regular limit cardinals. To review this, first we note that
the function sending $\mu^{+}$ to $\mu$ is regressive on the set of successor
cardinals and so this set can not be an element of $\mathcal{U}$. Now we prove
that there is a member of $\mathcal{U}$ consisting only of regular cardinals.
This is because if the function $f(\lambda)=\mbox{cf}(\lambda)$ was regressive
on a set $U$ of $\mathcal{U}$, i.e. every member of $U$ would have the same
cofinality $\delta<\kappa$, then we would have $\delta$-many regressive
functions (to elements of cofinal sequences) and one of these could not be
constant on a member of $\mathcal{U}$. Needless to say, it is also true that
$\kappa$ is a strongly inaccessible cardinal.

######  Claim 2.

If $\\{S_{\xi}:\xi\in\kappa\\}$ is any sequence of countable subsets of
$\kappa$, then there is a set $\bar{S}$ and a set $U\in\mathcal{U}$ such that
for all $\lambda,\mu\in U$ with $\lambda<\mu$,
$\bar{S}=S_{\lambda}\cap\lambda$ and $S_{\lambda}\subset\mu$.

###### Proof of Claim:.

For each $\xi\in\kappa$, let $\delta_{\xi}<\omega_{1}$ be the order-type of
$S_{\xi}\cap\xi$ and let $\\{\beta^{\xi}_{\eta}:\eta<\delta_{\xi}\\}$ be an
enumeration of $S_{\xi}\cap\xi$. Choose $U_{0}\in\mathcal{U}$ and
$\delta<\omega_{1}$ so that $\delta_{\lambda}=\delta$ for all $\lambda\in
U_{0}$. Similarly, for each $\eta<\delta$, choose $\beta_{\eta}<\kappa$ and
$U_{\eta+1}\in\mathcal{U}$ so that $\beta^{\lambda}_{\eta}=\beta_{\eta}$ for
all $\lambda\in U_{\eta+1}$. Let $\bar{U}\in\mathcal{U}$ equal
$U_{0}\cap\bigcap\\{U_{\eta+1}:\eta<\delta\\}$. Choose any cub
$C\subset\kappa$ so that for all $\lambda\in C$ and $\gamma<\lambda$,
$S_{\gamma}\subset\lambda$. Now it follows that $C\cap\bar{U}\in\mathcal{U}$
has the required properties. ∎

We say that an indexed subset $\\{p_{\xi}:\xi<\delta\\}$ of the poset
$\operatorname{\operatorname{Fn}}(\kappa,2)$ is a $\Delta$-system if there is
a $\bar{p}\in\operatorname{\operatorname{Fn}}(\kappa,2)$ such that for all
$\xi\neq\eta<\delta$,
$p_{\xi}\restriction\operatorname{\mathop{dom}}(\bar{p})=\bar{p}$ and the
members of the family
$\\{\operatorname{\mathop{dom}}(p_{\xi})\setminus\operatorname{\mathop{dom}}(\bar{p}):\xi<\delta\\}$
are pairwise disjoint. We say that $\bar{p}$ is the root of the
$\Delta$-system. We record two well-know facts about Cohen forcing.

######  Claim 3.

If $\\{p_{\xi}:\xi<\delta\\}$ is a $\Delta$-system in
$\operatorname{\operatorname{Fn}}(\kappa,2)$ with root $\bar{p}$, then for
every generic filter $G$ with $\bar{p}\in G$ and every infinite subset
$J\subset\delta$ the set $\\{\xi\in J:p_{\xi}\in G\\}$ has cardinality $|J|$.

######  Claim 4.

For any uncountable regular cardinal $\mu\leq\kappa$ and sequence
$\\{p_{\xi}:\xi<\mu\\}$, there is a set $J$ cofinal in $\mu$ so that
$\\{p_{\xi}:\xi\in J\\}$ is a $\Delta$-system.

Now suppose that $\theta>\kappa$ is a cardinal and that in the extension by
the poset $\operatorname{\operatorname{Fn}}(\kappa,2)$ there is a regular
pseudoradial topology on $\theta$ in which $\omega$ is dense and which has
countable tightness.

If $G$ is a $\operatorname{\operatorname{Fn}}(\kappa,2)$-generic filter, then
in $V[G]$, by [Kunen]*VII), $\mathfrak{c}=\kappa$ and, for all infinite
cardinals $\lambda<\kappa$, $2^{\lambda}=\kappa$. Similarly, the cardinals of
$V$ and the cardinals of $V[G]$ coincide. It therefore follows that, in
$V[G]$, the cardinality of $\omega^{(<\kappa)}$ is at most $\kappa$. We will
use the convention that a nice name of a subset of $\theta$ is a set $\dot{A}$
of the form $\bigcup\\{\\{\alpha\\}\times A_{\alpha}:\alpha\in I\\}$ where $I$
is any subset of $\theta$ and for each $\alpha\in I$, $A_{\alpha}$ is a
countable subset of $\operatorname{\operatorname{Fn}}(\kappa,2)$. We will say
that $\dot{A}$ is a nice name of a subset of $I$ to mean that
$\\{\alpha\in\theta:A_{\alpha}\neq\emptyset\\}\subset I$. When we say that
$\dot{A}$ is an $I$-name we will mean in this sense. Usually each $A_{\alpha}$
is taken to be an antichain but it will be convenient to use our
generalization. If $G$ is a
$\operatorname{\operatorname{Fn}}(\kappa,2)$-generic filter, then
$\operatorname{\mathop{val}}_{G}(\dot{A})$ is defined as $\\{\alpha:G\cap
A_{\alpha}\neq\emptyset\\}$. It is shown in [Kunen] that, in $V[G]$, every
subset of $\theta$ is equal to $\operatorname{\mathop{val}}_{G}(\dot{A})$ for
some nice name $\dot{A}$ (in the ground model). Moreover, for every infinite
subset $J$ of $\theta$ in $V[G]$ there is a ground model set $I$ and a nice
name of a subset of $I$ for $J$, where $I$ and $J$ have the same cardinality
in $V[G]$. Similarly, it can be shown that for every subset $\mathcal{A}$ of
the power set of $\theta$ in $V[G]$, there is a family $\mathcal{N}$ of nice
names of subsets of $\theta$ such that $\mathcal{A}$ is equal to
$\\{\operatorname{\mathop{val}}_{G}(\dot{A}):\dot{A}\in\mathcal{N}\\}$.
Therefore, we may suppose that we have families
$\\{\mathcal{N}_{\alpha}:\alpha\in\theta\\}$ where, for each $\alpha<\theta$,
$\mathcal{N}_{\alpha}$ is the family of nice names of the open subsets of
$\theta$ that contain $\alpha$.

Another subset of $\theta$ that is of interest is the set that will evaluate
to $\omega^{(<\kappa)}$ in $V[G]$. There is a set $I\subset\theta$ of
cardinality at most $\kappa$ so that there is a nice name $\dot{Y}$ of a
subset of $I$ satisfying that, for any generic filter
$G\subset\operatorname{\operatorname{Fn}}(\kappa,2)$,
$\operatorname{\mathop{val}}_{G}(\dot{Y})$ contains $\omega^{(<\kappa)}$.
Next, we may choose a condition $\bar{p}\in G$ and an ordinal
$z\in\theta\setminus I$ such that $\bar{p}$ forces that there is a
$\kappa$-sequence from $\omega^{(<\kappa)}$ that converges to $z$. Again we
may suppose that there is a sequence $\\{\dot{y}_{\xi}:\xi<\kappa\\}$ where
for each $\xi<\kappa$, $\dot{y}_{\xi}$ is a nice name of a single ordinal that
is forced by $\bar{p}$ to be in $\dot{Y}$ and so that the sequence is forced
to converge to $z$. It follows that each $\dot{y}_{\xi}$ is an $I_{\xi}$-name
for some countable (or finite) set $I_{\xi}$.

For each $\xi<\kappa$, we may choose a pair $(\alpha_{\xi},p_{\xi})$ that is
in the name $\dot{y}_{\xi}$ and such that $\bar{p}$ is compatible with
$p_{\xi}$. That is, if $\bar{p}\cup p_{\xi}\in G$, then
$\operatorname{\mathop{val}}_{G}(\dot{y}_{\xi})$ is equal to $\alpha_{\xi}$
and is in $\omega^{(<\kappa)}$. A minor modification of Claim 2 show that
there is a $\bar{q}\leq\bar{p}$ and a set $U_{0}\in\mathcal{U}$ such that, for
all $\lambda<\mu$ with both from $U_{0}$, $\bar{q}=(\bar{p}\cup
p_{\lambda})\restriction\lambda$ and
$\operatorname{\mathop{dom}}(p_{\lambda})\subset\mu$.

We define a new nice name $\dot{Z}=\\{(\alpha_{\lambda},\bar{p}\cup
p_{\lambda}):\lambda\in U_{0}\\}$ and observe that, by Claim 3, if $\bar{p}\in
G$ (for a $\operatorname{\operatorname{Fn}}(\kappa,2)$-generic filter $G$) the
sequence $\operatorname{\mathop{val}}_{G}(\dot{Z})$ is a $\kappa$-sequence
that converges to $z$. Let $C_{0}$ be a cub in $\kappa$ satisfying that, for
each $\mu\in C_{0}$, $U_{0}\cap\mu$ is cofinal in $\mu$. Since $z$ is forced
to not be an element of $\omega^{(<\kappa)}$, we may choose, for each $\mu\in
C_{0}$, a nice name $\dot{A}_{\mu}\in\mathcal{N}_{z}$, so that there is a
$I_{\mu}\subset U_{0}\cap\mu$ and a nice name $\dot{J}_{\mu}$ for a subset of
$I_{\mu}$ satisfying that $\dot{J}_{\mu}$ is forced to be cofinal in $\mu$ and
$\alpha_{\xi}$ is not $\dot{A}_{\mu}$ for all $\xi\in\dot{J}_{\mu}$. We
emphasize that, in $V[G]$, $z$ is not in the closure of
$\\{\alpha_{\lambda}:\lambda\in\dot{J}_{\mu}\\}$. Recall that a condition
$q\in\operatorname{\operatorname{Fn}}(\kappa,2)$ forces that an ordinal
$\xi\in\dot{J}_{\mu}$ if there is some pair $(\xi,p)\in\dot{J}_{\mu}$ with
$q\leq p$. Similarly, $q$ forces that $\xi\in\dot{Z}$ if $\xi\in U_{0}$ and
$q\leq\bar{p}\cup p_{\xi}$.

For each $\lambda\in U_{0}$ and $\lambda<\mu\in C_{0}$, choose a pair
$(\xi(\lambda,\mu),q(\lambda,\mu))$ so that

  1. (1)

$\lambda\leq\xi(\lambda,\mu)\in U_{0}\cap\mu$,

  2. (2)

$q(\lambda,\mu)\in\operatorname{\operatorname{Fn}}(\kappa,2)$ and
$q(\lambda,\mu)$ forces $\xi(\lambda,\mu)\in\dot{Z}$,

  3. (3)

$q(\lambda,\mu)$ forces that $\xi(\lambda,\mu)\in\dot{J}_{\mu}$.

There is a cub $C_{1}$ satisfying that, for all $\zeta\in C_{1}$,
$\\{q(\lambda,\mu):\mu<\zeta\ \mbox{and}\ \lambda\in U_{0}\cap\mu\\}$ is a
subset of $\operatorname{\operatorname{Fn}}(\zeta,2)$. Choose any
$U_{1}\in\mathcal{U}$ so that $U_{1}\subset C_{1}$ and every element of
$U_{1}$ is a regular cardinal.

Applying Claim 4, we may choose, for each $\mu\in U_{1}$, a strictly
increasing function $h_{\mu}:\mu\mapsto U_{0}\cap\mu$ so that the sequence
$\\{q(h_{\mu}(\alpha),\mu):\alpha\in\mu\\}$ is a $\Delta$-system. For each
$\mu\in U_{1}$, let $\tilde{q}_{\mu}$ denote the root of this $\Delta$-system,
and, by possibly shrinking $U_{1}$, we can also assume that the sequence
$\\{\tilde{q}_{\mu}:\mu\in U_{1}\\}$ is a $\Delta$-system.

Now we come to a stronger version of Claim 3.

######  Claim 5.

There is a strictly increasing function $h$ from $\kappa$ into $U_{0}$ and a
sequence $\\{(\bar{\xi}_{\alpha},\bar{q}_{\alpha}):\alpha<\kappa\\}$ such that
for all $\omega<\delta<\kappa$, there is a set $U_{\delta}\in\mathcal{U}$ so
that, for all $\mu\in U_{\delta}$,

$\\{(h(\alpha),\bar{\xi}_{\alpha},\bar{q}_{\alpha}):\alpha<\delta\\}=\\{(h_{\mu}(\alpha),\xi(h_{\mu}(\alpha),\mu),q(h_{\mu}(\alpha),\mu)\restriction\mu):\alpha<\delta\\}~{}.$

###### Proof of Claim:.

Choose any enumeration
$e:\kappa\mapsto\kappa\times\kappa\times\operatorname{\operatorname{Fn}}(\kappa,2)$.
Choose a cub $C_{e}\subset\kappa$ so that for all $\gamma<\delta\in C_{e}$,
$e(\gamma)\in\delta\times\delta\times\operatorname{\operatorname{Fn}}(\delta,2)$.
For each $\alpha\in\kappa$, the function sending $\mu$ to the ordinal
$\gamma_{\mu}$ satisfying that
$(h_{\mu}(\alpha),\xi(h_{\mu}(\alpha)),q(h_{\mu}(\alpha),\mu)\restriction\mu)=e(\gamma_{\mu})$
is regressive on a set in $\mathcal{U}$. Therefore there is a
$W_{\alpha}\in\mathcal{U}$ and a value $g(\alpha)\in\kappa$ so that

$(h_{\mu}(\alpha),\xi(h_{\mu}(\alpha),\mu),q(h_{\mu}(\alpha),\mu)\restriction\mu)=e(g(\alpha))\
\ \mbox{ for all}\ \ \mu\in W_{\alpha}~{}.$

Let $(h(\alpha),\bar{\xi}_{\alpha},\bar{q}_{\alpha})$ be defined so as to
equal $e(g(\alpha))$. Now, for each $\delta<\kappa$, the set
$U_{\delta}=\bigcap_{\alpha<\delta}W_{\alpha}$ is an element of $\mathcal{U}$
and, after some straightforward unravelling, we have that for all $\mu\in
U_{\delta}$,

$\\{(h(\alpha),\bar{\xi}_{\alpha},\bar{q}_{\alpha}):\alpha<\delta\\}=\\{(h_{\mu}(\alpha),\xi(h_{\mu}(\alpha),\mu),q(h_{\mu}(\alpha),\mu)\restriction\mu):\alpha<\delta\\}~{}~{}.$

∎

The family $\\{\bar{q}_{\alpha}:\alpha\in\kappa\\}$ is a $\Delta$-system of
$\operatorname{\operatorname{Fn}}(\kappa,2)$ since for each $\delta<\kappa$,
there is a $\mu\in U_{1}$ such that the initial segment
$\\{\bar{q}_{\alpha}:\alpha<\delta\\}$ is a subset of the $\Delta$-system
$\\{q(h_{\mu}(\alpha),\mu)\restriction\mu:\alpha<\delta\\}~{}~{}.$ In
addition, each $\bar{q}_{\alpha}$ forces that $\bar{\xi}_{\alpha}$ is an
element of the set $\\{\dot{y}_{\zeta}:\alpha\leq\zeta<\kappa\\}$.

We briefly pass to the extension $V[G]$ and note that, by Claim 3, the set
$J=\\{\alpha:\bar{q}_{\alpha}\in G\\}$ is cofinal in $\kappa$. Since the
sequence $\\{\operatorname{\mathop{val}}_{G}(\dot{y}_{\zeta}):\zeta<\kappa\\}$
converges to $z$, and by countable tightness, there is a countable subset
$J_{1}$ of $J$ such that $\\{\bar{\xi}_{\alpha}:\alpha\in J_{1}\\}$ has $z$ in
its closure. Now we apply Proposition 3.6 to the point $z$ and the countable
set $A=\\{\bar{\xi}_{\alpha}:\alpha\in J_{1}\\}$ and we choose a countable set
$J_{2}$ of $\theta$ (i.e. the space $X$) so that $J_{2}\subset
A^{(1)}\setminus A$ and $z$ is in the closure of $J_{2}$.

Now we return to $V$ and we choose countable $I_{1}\subset\kappa$ and
countable $I_{2}\subset\theta$ so that, in $V[G]$, $J_{1}\subset I_{1}$ and
$J_{2}\subset I_{2}$. Fix a nice name, $\dot{J}_{2}$, for the subset $J_{2}$
of $I_{2}$ and notice that the name $\dot{J}_{1}$ for $J_{1}$ is simply the
set $\\{(\alpha,\bar{q}_{\alpha}):\alpha\in I_{1}\\}$. Next we want to choose
sufficiently many nice names for sequences contained in
$\\{\bar{\xi}_{\alpha}:\alpha\in J_{1}\\}$ that converge to points of $J_{2}$.
We can do this by a simple examination of the name $\dot{J}_{2}$. For each
ordinal $x\in I_{2}$ and condition
$p\in\operatorname{\operatorname{Fn}}(\kappa,2)$ such that $(x,p)$ is an
element of the name $\dot{J}_{2}$ (thus $p\Vdash x\in J_{2}$) choose a nice
name $\dot{S}(x,p)$ of a subset of $I_{1}$ such that $p$ forces that
$\dot{S}(x,p)$ is a subset of $J_{1}$ and that
$\\{\bar{\xi}_{\alpha}:\alpha\in\dot{S}(x,p)\\}$ converges to $x$. Certainly
we have only chosen countably many nice names of the form $\dot{S}(x,p)$. Fix
any $\delta<\kappa$ so that

  1. (1)

$I_{1}\cup I_{2}\subset\delta$,

  2. (2)

$\dot{J}_{2}$ is a nice $\operatorname{\operatorname{Fn}}(\delta,2)$-name,

  3. (3)

for all $x\in I_{2}$ and $(x,p)\in\dot{J}_{2}$, $\dot{S}(x,p)$ is a nice
$\operatorname{\operatorname{Fn}}(\delta,2)$-name,

and choose any $\mu\in U_{\delta}$ such that $\tilde{q}_{\mu}\in G$.

We are ready for our final contradiction (to the assumption that $X$ has
cardinality greater than $\kappa$). The contradiction will be that $z$ is not
in the closure of the set $A_{\mu}=\\{\xi(h_{\mu}(\alpha),\mu):\alpha\in
I_{1}\ \mbox{and}\
h_{\mu}(\alpha)\in\operatorname{\mathop{val}}_{G}(\dot{J}_{\mu})\\}$, but $z$
is in the closure of $J_{2}$. We will show that $J_{2}$ is contained in the
closure of $A_{\mu}$. To show this, it suffices to prove that for each $x\in
I_{2}$ and $p\in G$ such that $(x,p)\in\dot{J}_{2}$, the sequence
$\\{\bar{\xi}_{\alpha}:\alpha\in\operatorname{\mathop{val}}_{G}(\dot{S}(x,p))\\}$
meets $A_{\mu}$ in an infinite set. Fix any $x\in I_{2}$ and $p$ so that
$(x,p)\in\dot{J}_{2}$. Let $H$ be any finite subset of $I_{1}$ and we will
show that there is an
$\alpha\in\operatorname{\mathop{val}}_{G}(\dot{S}(x,p))\setminus H$ such that
$\bar{\xi}_{\alpha}\in A_{\mu}$. To do so, we will work in $V$ and have to use
the genericity of $G$. Following [Kunen]*VII, it suffices to show that if $r$
is any element of $G$, there is an $\alpha\in I_{1}\setminus H$ and an
extension $r_{\alpha}\in\operatorname{\operatorname{Fn}}(\kappa,2)$ so that,
with $\bar{\xi}_{\alpha}$ $r_{\alpha}$ forces that $\alpha\in\dot{S}(x,p)$ and
$r_{\alpha}$ forces that $h_{\mu}(\alpha)\in\dot{J}_{\mu}$. This is the same
as proving that the set of conditions of the form $r_{\alpha}$ is dense below
the condition $p$. This ensures that $G$ will include one such $r_{\alpha}$
and that $\bar{\xi}_{\alpha}$ will be the desired member of $A_{\mu}$ with
$\alpha\in\dot{S}(x,p)$.

Now we use that $\\{q(h_{\mu}(\alpha),\mu):\alpha\in I_{1}\setminus H\\}$ is a
$\Delta$-system. By possibly increasing the size of $H$ we can assume that
$\operatorname{\mathop{dom}}(r)\cup\operatorname{\mathop{dom}}(\tilde{q}_{\mu})\cup\operatorname{\mathop{dom}}(p)$
is disjoint from
$\operatorname{\mathop{dom}}(q(h_{\mu}(\alpha),\mu))\setminus\operatorname{\mathop{dom}}(\tilde{q}_{\mu})$
for all $\alpha\in I_{1}\setminus H$. Next we simply note that, since
$\\{\bar{\xi}_{\alpha}:\alpha\in\operatorname{\mathop{val}}_{G}(\dot{S}(x,p))\\}$
converges to $x$, there an $\alpha\in I_{1}\setminus H$ and a condition
$\bar{r}_{\alpha}\in G$ so that $(\alpha,\bar{r}_{\alpha})$ is an element of
the name $\dot{S}(x,p)$. Recalling that $p$ forces that $\dot{S}(x,p)$ is a
subset of $J_{1}$, it follows also that
$\bar{r}_{\alpha}\leq\bar{q}_{\alpha}$. By the choice of $\delta$,
$\bar{r}_{\alpha}\in\operatorname{\operatorname{Fn}}(\delta,2)$ and
$\bar{r}_{\alpha}\leq q(h_{\mu}(\alpha),\mu)\restriction\mu$. Finally, it
follows that with $r_{\alpha}=\bar{r}_{\alpha}\cup q(h_{\mu}(\alpha),\mu)$ we
have completed the proof. ∎

##  4\. Pseudoradial spaces in models of $\mathfrak{c}\leq\aleph_{2}$

In this section we prove that if $\mathfrak{c}$ is most $\aleph_{2}$, then
there are separable regular pseudoradial spaces of cardinality
$2^{\mathfrak{c}}$. These examples are also 0-dimensional with a countable
dense set of isolated points. We first prove that CH implies the stronger
result that there are such space that are compact.

######  Theorem 4.1 (CH).

There is a compactification of $\omega$ that is pseudoradial and has
cardinality $2^{\aleph_{1}}$.

###### Proof.

Let $X$ be an $\eta_{1}$-set of cardinality $\aleph_{1}$ and let $\prec$
denote the corresponding (strict) linear ordering on $X$. Recall that an
$\eta_{1}$-set has the property that if $A$ and $B$ are countable subsets of
$X$ and that $a\prec b$ for all $a\in A$ and $b\in B$, then there is an $x\in
X$ with $a\prec x\prec b$ for all $a\in A$ and $b\in B$. Next, let
$\mathbb{D}$ denote the standard Dedekind completion of $X$. Since $X$ is
dense in $\mathbb{D}$ it follows that $\mathbb{D}$ has weight $\aleph_{1}$ and
no points of countable character. It also follows then that $\mathbb{D}$ has
cardinality $2^{\aleph_{1}}$. Next, let $K$ be the space obtained by doubling
every point of $\mathbb{D}$ that has countable one-sided character. Again,
using that $X$ is dense, it follows that $K$ has weight $\aleph_{1}$. Since
$K$ is a linearly ordered space, it follows that $K$ is radial. We observe
that $K$ has the property that every countably infinite set has subsequence
converging to a point of countable character.

Finally, it follows from CH that there is a compactification $\gamma\omega$
that has remainder $K$. This is the requisite example. It remains only to
prove that $\gamma\omega$ is sequentially compact. Since $K$ is sequentially
compact, it suffices to prove that every infinite subset $A$ of $\omega$ has a
converging subsequence. If $A$ has only finitely many accumulation points then
this is clear. On the other hand, if $A$ has infinitely many accumulation
points, one of them will be a point of countable character. ∎

Now we turn to the proof in the case that $\mathfrak{c}=\aleph_{2}$. Before
constructing the example we establish an instructive constraint on the nature
of possible examples. This is based on the paper [BaumWeese] concerning
partitioner algebras. A similar application in [Dowlarge] established that it
was consistent that compact separable spaces of cardinality greater than
$\mathfrak{c}$ were not sequentially compact. The proof is similar to the
proof of Theorem 3.7.

######  Theorem 4.2.

It is consistent with $\mathfrak{c}=\aleph_{2}$ that every pseudocompact
pseudoradial regular space that contains a countable dense discrete space has
cardinality at most $\mathfrak{c}$. Moreover, if $V$ is a model of GCH and
$\kappa>\aleph_{1}$ is a regular cardinal, then in the forcing extension by
$\operatorname{\operatorname{Fn}}(\kappa,2)$ every pseudocompact pseudoradial
regular space with a countable dense discrete subset has cardinality at most
$\kappa$.

###### Proof.

Clearly it suffices to prove the second statement of the Theorem. Suppose that
GCH holds and that $\kappa>\aleph_{1}$ is a regular cardinal. Let $G$ be a
$\operatorname{\operatorname{Fn}}(\kappa,2)$-generic filter. We assume that
$\theta>\kappa$ is a cardinal and that we have a
$\operatorname{\operatorname{Fn}}(\kappa,2)$-name for a topology on $X=\theta$
such that, in $V[G]$, $\omega$ is open, dense, and discrete. Similar to
Theorem 3.7, let, for each $\alpha<\lambda$ and
$p\in\operatorname{\operatorname{Fn}}(\kappa,2)$, $\mathcal{N}_{\alpha}(p)$ be
the set of all nice names of subsets of $\omega$ that are forced by $p$ to
have $\alpha$ in the interior of their closure. Also let
$\mathcal{S}_{\alpha}(p)$ be the set of all nice names of subsets of $\omega$
that are forced by $p$ to simply have $\alpha$ in their closure. To avoid
confusion, we will let $1_{P}$ denote the maximal element $\emptyset$ of
$\operatorname{\operatorname{Fn}}(\kappa,2)$.

Since $\kappa$ is a regular cardinal, and we are assuming GCH, we may fix an
elementary submodel $M$ of $H(\theta^{+})$ of cardinality $\kappa$ such that
$\\{\\{\mathcal{N}_{\alpha}(1_{P}),\mathcal{S}_{\alpha}(1_{P})\\}:\alpha\in\theta\\}$
is an element of $M$, and every subset of $M$ of cardinality less than
$\kappa$ is an element of $M$. Let $z$ be any element of $\theta\setminus M$.
In this proof we will choose a $\kappa$-sequence
$\\{y_{\xi}:\xi<\kappa\\}\subset M\cap\theta$ satisfying that, in $V[G]$, the
sequence $\\{y_{\xi}:\xi<\kappa\\}$ converges to $z$. Fix enumerations
$\\{\dot{S}_{\beta}:\beta<\kappa\\}$ of $\mathcal{S}_{z}(1_{P})$ and
$\\{\dot{W}_{\beta}:\beta<\kappa\\}$ of $\mathcal{N}_{z}(1_{P})$. Fix any
$\lambda<\kappa$. By assumption, the sequence
$\\{\dot{S}_{\beta}:\beta<\lambda\\}$ and
$\\{\dot{W}_{\beta}:\beta<\lambda\\}$ are elements of $M$. Since

$H(\theta^{+})\models(\exists
z\in\theta)(\\{\dot{S}_{\beta}:\beta<\lambda\\}\subset\mathcal{S}_{z}(1_{P})\
\mbox{and}\ \\{\dot{W}_{\beta}:\beta<\lambda\\}\subset\mathcal{N}_{z}(1_{P}))$

there is a $y_{\lambda}\in M\cap\theta$ satisfying that
$\\{\dot{S}_{\beta}:\beta<\lambda\\}\subset\mathcal{S}_{y_{\lambda}}(1_{P})$
and
$\\{\dot{W}_{\beta}:\beta<\lambda\\}\subset\mathcal{N}_{y_{\lambda}}(1_{P})$.

It is easily verified that $\\{y_{\lambda}:\lambda<\kappa\\}$ converges to $z$
in $V[G]$ but this is not actually needed in the remainder of the proof.
Rather, we will prove that $\omega$ is forced to have an infinite subset $S$
that contains no converging sequence. If $X$ is pseudoradial, $S$ would be
closed and, since $\omega$ is open in $X$, this would imply that $X$ is not
pseudocompact.

For limit $\lambda<\kappa$ of uncountable cofinality, no cofinal subsequence
of $\\{y_{\alpha}:\alpha<\lambda\\}$ converges to $z$ since $z$ is not an
element of $M$. We translate this into the forcing language.

######  Claim 6.

For each limit $\lambda<\kappa$ of uncountable cofinality, there is a value
$\zeta_{\lambda}<\kappa$ and
$\dot{W}_{\zeta_{\lambda}}\in\mathcal{N}_{z}(1_{P})$ satisfying that, for all
$\beta<\lambda$ and $p\in\operatorname{\operatorname{Fn}}(\kappa,2)$, there
are $\beta<\alpha<\lambda$ and $p_{\alpha}<p$ such that the name
$\omega\setminus\dot{W}_{\zeta_{\lambda}}$ is an element of
$\mathcal{N}_{y_{\alpha}}(p_{\alpha})$.

The last statement in Claim 6 is simply asserting that $p_{\alpha}$ forces
that $y_{\alpha}$ is not in the closure of $\dot{W}_{\zeta_{\lambda}}$. We
also note that the statement of Claim 6 asserts that, for each
$\beta<\lambda$, the set of $p_{\alpha}$ as in the statement is a dense subset
of $\operatorname{\operatorname{Fn}}(\kappa,2)$. For each $\lambda$, choose
another ordinal $\rho_{\lambda}$ so that $1_{P}$ forces that the closure of
$\dot{W}_{\rho_{\lambda}}\in\mathcal{N}_{z}(1_{P})$ is contained in the
interior of the closure of $\dot{W}_{\zeta_{\lambda}}$.

Let $\Lambda$ denote the set of $\lambda<\kappa$ of uncountable cofinality.
For all $\lambda\in\Lambda$, let $\dot{W}_{\zeta_{\lambda}}$ be the name
$\\{(n,p(\lambda,n,k)):n\in B_{\lambda},k\in\omega\\}$ and let
$\dot{W}_{\rho_{\lambda}}$ be the name $\\{(n,q(\lambda,n,k)):n\in
D_{\lambda},k\in\omega\\}$. Note that $D_{\lambda}\subset B_{\lambda}$ and we
may assume that, for all $n\in D_{\lambda}$ and $k\in\omega$, there is a
$j\in\omega$ so that $q(\lambda,n,k)<p(\lambda,n,j)$ (since
$q(\lambda,n,k)\Vdash n\in\dot{W}_{\zeta_{\lambda}}$). Let $I_{\lambda}$
denote the union of the set
$\bigcup\\{\operatorname{\mathop{dom}}(q(\lambda,n,k)):n\in
B_{\lambda},k\in\omega\\}$. For any
$p\in\operatorname{\operatorname{Fn}}(\kappa,2)$ and partial injection
$\sigma:\kappa\mapsto\kappa$ such that
$\operatorname{\mathop{dom}}(p)\subset\operatorname{\mathop{dom}}(\sigma)$, we
let $\widehat{\sigma}(p)$ be the condition with domain
$\sigma(\operatorname{\mathop{dom}}(p))$ and satisfying that
$\widehat{\sigma}(p)(\sigma(\alpha))=p(\alpha)$ for all
$\alpha\in\operatorname{\mathop{dom}}(p)$.

Fix a cub $C\subset\kappa$ satisfying that for all $\mu\in C$,
$\\{\zeta_{\lambda}\\}\cup I_{\lambda}\subset\mu$ for all
$\lambda\in\Lambda\cap\mu$. By the usual pressing down lemma, and some
straightforward enumeration techniques as used in the proof of Theorem 3.7,
there is a stationary set $E\subset\Lambda\cap C$ satisfying that, for all
$\lambda,\mu\in E$ with $\lambda<\mu$, there is an order preserving
isomorphism $\sigma_{\lambda,\mu}$ from $I_{\lambda}$ to $I_{\mu}$ so that:

  1. (1)

$B_{\lambda}=B_{\mu}$ and $I_{\lambda}\cap\lambda=I_{\mu}\cap\mu$,

  2. (2)

$\sigma_{\lambda,\mu}(I_{\lambda}\setminus\lambda)=I_{\mu}\setminus\mu$,

  3. (3)

for all $n\in B_{\lambda}$ and $k\in\omega$,
$p(\mu,n,k)=\widehat{\sigma}_{\lambda,\mu}(p(\lambda,n,k))$,

  4. (4)

for all $n\in D_{\lambda}$ and $k\in\omega$,
$q(\mu,n,k)=\widehat{\sigma}_{\lambda,\mu}(q(\lambda,n,k))$ .

Let $\mu_{0}$ be any element of $E$. Next, using that $E$ is stationary it
follows that we can find $\mu_{0}<\mu_{1}\in E$ so that every
$\operatorname{\operatorname{Fn}}(\mu_{1},2)$-name in $\mathcal{S}_{z}(1_{P})$
is in the list $\\{\dot{S}_{\beta}:\beta<\mu_{1}\\}$. Additionally we can
ensure that, for every nice
$\operatorname{\operatorname{Fn}}(\mu_{0},2)$-name, $\dot{S}$, for a subset of
$\omega$, and every $p\in\operatorname{\operatorname{Fn}}(\kappa,2)$ such that
$\dot{S}\in\mathcal{S}_{z}(p)$, then
$\dot{S}\in\mathcal{S}_{z}(p\restriction\mu_{1})$. This latter condition is a
simple consequence of the fact that
$\operatorname{\operatorname{Fn}}(\kappa,2)$ is ccc.

Next, extend $\\{\mu_{0},\mu_{1}\\}$ to any strictly increasing sequence
$\\{\mu_{\ell}:\ell\in\omega\\}\subset E_{1}$. Our next step is to prove that
the family $\\{\dot{W}_{\zeta_{\mu_{\ell}}}:\ell\in\omega\\}$ is forced to be
an independent family, or rather that
$\\{\\{\dot{W}_{\rho_{\mu_{\ell}}},\omega\setminus\dot{W}_{\zeta_{\mu_{\ell}}}\\}:\ell\in\omega\\}$
is forced to be an independent family of pairs. We will use the fact
$\operatorname{\operatorname{Fn}}(\kappa,2)$ is isomorphic to
$\operatorname{\operatorname{Fn}}(\mu_{0},2)\times\operatorname{\operatorname{Fn}}(\kappa\setminus\mu_{0},2)$
and so when we pass to the extension $V[G_{\mu_{0}}]$ (where
$G_{\mu_{0}}=G\cap\operatorname{\operatorname{Fn}}(\mu_{0},2)$), we can use
the fact that $V[G]$ is equal to
$V[G_{\mu_{0}}][G\cap\operatorname{\operatorname{Fn}}(\kappa\setminus\mu_{0},2)]$.
We are trying to minimize our appeal to advanced forcing techniques but we
need a minor one here.

######  Claim 7.

For each $S\in[\omega]^{\aleph_{0}}\cap V[G_{\mu_{0}}]$, if $V[G]\models
z\in\mbox{cl}_{X}(S)$, then there is an $\beta<\mu_{1}$ such that
$\dot{S}_{\beta}\in\mathcal{S}_{z}(1_{P})$ is a
$\operatorname{\operatorname{Fn}}(\mu_{1},2)$-name and
$\operatorname{\mathop{val}}_{G}(\dot{S}_{\beta})=S$.

###### Proof of Claim:.

By definition, if $S\in V[G_{\mu_{0}}]\cap[\omega]^{\aleph_{0}}$ there is a
$\operatorname{\operatorname{Fn}}(\mu_{0},2)$-name $\dot{S}$ such that
$S=\operatorname{\mathop{val}}_{G_{\mu_{0}}}(\dot{S})$. Now assume that, in
$V[G]$, that $z$ is in the closure of $\dot{S}$. By the forcing theorem, there
is a condition $p\in G$ that forces $z$ is in the closure of $\dot{S}$. Of
course this means that $\dot{S}\in\mathcal{S}_{z}(p)$. By the assumption on
$\mu_{1}$, we can assume that
$p\in\operatorname{\operatorname{Fn}}(\mu_{1},2)$. It is a routine exercise to
prove that there is a nice $\operatorname{\operatorname{Fn}}(\mu_{1},2)$-name
$\dot{T}$ such that $p\Vdash\dot{T}=\dot{S}$ and, all
$q\in\operatorname{\operatorname{Fn}}(\mu_{1},2)$ that are incomparable with
$p$ force that $\dot{T}=\omega$. It follows now, from the assumption on
$\mu_{1}$, that $\dot{T}$ is in the list
$\\{\dot{S}_{\alpha}:\alpha<\mu_{1}\\}$. ∎

Let $B=\\{n\in B_{\mu_{0}}:(\exists k)p(\mu_{0},n,k)\restriction\mu_{0}\in
G_{\mu_{0}}\\}$ and $D=\\{n\in D_{\mu_{0}}:(\exists
k)q(\mu_{0},n,k)\restriction\mu_{0}\in G_{\mu_{0}}\\}$. Let $I$ denote the
countable set $I_{\mu_{0}}\setminus\mu_{0}$ and let, for each $\ell<\omega$,
$\dot{U}_{\ell}$ denote the canonical
$\operatorname{\operatorname{Fn}}(I_{\mu_{\ell}}\setminus\mu_{\ell},2)$-name
in $V[G_{\mu_{0}}]$ for the set $\dot{W}_{\zeta_{\mu_{\ell}}}$ and let
$\dot{V}_{\ell}$ denote the canonical
$\operatorname{\operatorname{Fn}}(I_{\mu_{\ell}}\setminus\mu_{\ell},2)$-name
in $V[G_{\mu_{0}}]$ for the set $\dot{W}_{\rho_{\mu_{\ell}}}$. More precisely,
for each $n\in B$ and $k\in\omega$,

$(n,p(\mu_{\ell},n,k)\restriction I)\in\dot{U}_{0}\ \ \mbox{if and only if}\ \
p(\mu_{\ell},n,k)\restriction\mu_{0}\in G_{\mu_{0}}.$

For all $\ell>0$,
$\dot{U}_{\ell}=\\{(n,\widehat{\sigma}_{\mu_{0},\mu_{\ell}}(p)):(n,p)\in\dot{U}_{0}\\}$.
For convenience, let $\sigma_{\mu_{0},\mu_{0}}$ denote the identy function on
$I$ and recall that the sets
$\\{I_{\mu_{\ell}}\setminus\mu_{\ell}=\sigma_{\mu_{0},\mu_{\ell}}(I):\ell\in\omega\\}$
are pairwise disjoint. The definition of $\dot{V}_{\ell}$ is defined
similarly.

For each $p\in\operatorname{\operatorname{Fn}}(I,2)$, let $T_{p}=\\{n\in
B:p\Vdash n\in\dot{U}_{0}\\}$ and let $V_{p}=\\{n\in D:(\exists q<p)~{}q\Vdash
n\in\dot{V}_{0}\\}$. We note that $T_{p}$ and $V_{p}$ are elements of
$V[G_{\mu_{0}}]$.

######  Claim 8.

For each $p\in\operatorname{\operatorname{Fn}}(I,2)$, $z$ is not in the
closure, in $V[G]$, of $T_{p}$. Similarly, $z$ is in the interior of the
closure of $V_{p}$.

###### Proof of Claim:.

Assume that the first statement fails for some
$p\in\operatorname{\operatorname{Fn}}(I,2)$. By Claim 7, there is an
$\beta<\mu_{1}$ such that
$T_{p}=\operatorname{\mathop{val}}_{G}(\dot{S}_{\beta})$. The family
$\\{\sigma_{\mu_{0},\mu_{\ell}}(p):1\leq\ell\in\omega\\}$ is a $\Delta$-system
(with empty root) and so there is an $\ell>0$ such that
$\sigma_{\mu_{0},\mu_{\ell}}(p)\in G$. By Claim 6, there is an $\alpha$ and a
condition $p_{\alpha}<\sigma_{\mu_{0},\mu_{\ell}}(p)$ such that
$p_{\alpha}\restriction\mu_{0}\in G_{\mu_{0}}$, $\beta<\alpha<\mu_{\ell}$ and
$p_{\alpha}\Vdash\omega\setminus\dot{W}_{\ell}\in\mathcal{N}_{y_{\alpha}}(p_{\alpha})$.
This is equivalent to the assertion that $p_{\alpha}$ forces that $y_{\alpha}$
is not in the closure of $\dot{W}_{\ell}$. On the other hand, we have that
$\dot{S}_{\beta}\in\mathcal{S}_{y_{\alpha}}(1_{p})$. This implies that $1_{P}$
forces that $y_{\beta}$ is in the closure of $T_{p}$ and yet, in
$V[G_{\mu_{0}}]$, $p_{\alpha}$ forces $T_{p}\subset\dot{U}_{\ell}$ is disjoint
from a neighborhood of $y_{\alpha}$.

The second statement is proven similarly by simply noting that
$\sigma_{\mu_{0},\mu_{\ell}}(p)$ forces that $\dot{V}_{\ell}$ is a subset of
$V_{p}$. ∎

Now we prove that
$\\{\\{\dot{V}_{\ell},\omega\setminus\dot{U}_{\ell}\\}:\ell\in\omega\\}$ is
forced to be an independent family. Fix any pair of disjoint finite sets
$L_{0},L_{1}$ of $\omega$. Let
$q\in\operatorname{\operatorname{Fn}}(\kappa\setminus\mu_{0},2)$ be any
condition. We produce a condition $\tilde{q}<q$ and an integer $n$ so that
$\tilde{q}$ forces that $n\in\dot{V}_{\ell}$ for all $\ell\in L_{0}$ and
$\tilde{q}$ forces that $n\notin\dot{U}_{\ell}$ for all $\ell\in L_{1}$. For
each $\ell\in L_{0}\cup L_{1}$, let
$p_{\ell}\in\operatorname{\operatorname{Fn}}(I,2)$ be chosen so that
$\sigma_{\mu_{0},\mu_{\ell}}(p_{\ell})=q\restriction
I_{\mu_{\ell}}\setminus\mu_{\ell}$. Let $T=\bigcup\\{T_{p_{\ell}}:\ell\in
L_{1}\\}$ and $U=\bigcap\\{V_{p_{\ell}}:\ell\in L_{0}\\}$. Choose any integer
$n\in U\setminus T$. For each $\ell\in L_{0}$, $n\in U_{p_{0}}$ and so we may
choose $q_{\ell}<p_{\ell}$ in $\operatorname{\operatorname{Fn}}(I,2)$, such
that $q_{\ell}\Vdash n\in\dot{V}_{0}$. For each $\ell\in L_{1}$, since $n\in
T_{p_{\ell}}$, $p_{\ell}\not\Vdash n\in\dot{U}_{0}$, and so we may choose
$q_{\ell}<p_{\ell}$ in $\operatorname{\operatorname{Fn}}(I,2)$ such that
$q_{\ell}\Vdash n\notin\dot{U}_{0}$. It now follows that
$\tilde{q}=q\cup\bigcup\\{\sigma_{\mu_{0},\mu_{\ell}}(q_{\ell}):\ell\in
L_{0}\cup L_{1}\\}$ is an element of
$\operatorname{\operatorname{Fn}}(\kappa\setminus\mu_{0},2)$ and that
$\tilde{q}\Vdash n\in\dot{V}_{\ell}$ for all $\ell\in L_{0}$ and
$\tilde{q}\Vdash n\notin\dot{U}_{\ell}$ for all $\ell\in L_{1}$ as required.

We reassure the reader that we are nearly at the end of the proof. Fix any
$\lambda<\kappa$ such that $\mu_{\ell}<\lambda$ for all $\ell\in\omega$ and
let $\dot{L}_{\lambda}$ be the canonical subset of $\omega$ given by
$n\in\dot{L}_{\lambda}$ if and only if $\bigcup G(\lambda+n)=1$. That is
$\dot{L}_{\lambda}$ is the name
$\\{(n,\\{\langle\lambda+n,1\rangle\\}):n\in\omega\\}$. We have proven that
the family
$\\{\dot{V}_{\ell}:\ell\in\dot{L}_{\lambda}\\}\cup\\{\omega\setminus\dot{U}_{\ell}:\ell\in\omega\setminus\dot{L}_{\lambda}\\}$
has the finite intersection property. We can define $\dot{S}$ to be the
sequence $\\{\dot{n}_{\ell}:\ell\in\omega\\}$ where $\dot{n}_{\ell}$ is the
minimum element above $\dot{n}_{\ell-1}$ (for $\ell>0$) of the set

$\bigcap\\{\dot{V}_{j}:j\leq\ell\ \mbox{and}\
j\in\dot{L}_{\lambda}\\}\setminus\bigcup\\{\dot{U}_{j}:j\leq\ell\ \mbox{and}\
j\notin\dot{L}_{\lambda}\\}.$

Each $\dot{n}_{\ell}$ exists was proven in the previous paragraph. Now suppose
that $\dot{J}$ is a nice
$\operatorname{\operatorname{Fn}}(\kappa\setminus\mu_{0},2)$-name of an
infinite subset of $\omega$. Choose any $\mu\in E$ satisfying that
$\lambda+\omega<\mu$ and $\dot{J}$ is a
$\operatorname{\operatorname{Fn}}(\mu\setminus\mu_{0},2)$-name. We complete
the proof by showing that each of $\dot{W}_{\rho_{\mu}}$ and $\omega\setminus
W_{\zeta_{\mu}}$ hit the sequence $\\{\dot{n}_{\ell}:\ell\in\dot{J}\\}$ in an
infinite set. Since $\dot{W}_{\rho_{\mu}}$ and $\omega\setminus
W_{\zeta_{\mu}}$ have disjoint closures, and $\dot{J}$ was arbitrary, this
shows that $\dot{S}$ does not have any limit points. The proof is a density
argument. Analogous to the definitions of $\dot{V}_{\ell},\dot{W}_{\ell}$ from
$\dot{W}_{\rho_{\mu_{\ell}}},\dot{W}_{\zeta_{\mu_{\ell}}}$, let us define
$\dot{V}_{\mu}=\\{(n,\widehat{\sigma}_{\mu_{0},\mu}(p)):(n,p)\in\dot{V}_{0}\\}$
and
$\dot{U}_{\mu}=\\{(n,\widehat{\sigma}_{\mu_{0},\mu}(p)):(n,p)\in\dot{U}_{0}\\}$.
It follows that $\dot{V}_{\mu}$ is the
$\operatorname{\operatorname{Fn}}(\kappa\setminus\mu_{0},2)$-name for
$\operatorname{\mathop{val}}_{G}(\dot{W}_{\rho_{\mu}})$ and $\dot{U}_{\mu}$ is
the $\operatorname{\operatorname{Fn}}(\kappa\setminus\mu_{0},2)$-name for
$\operatorname{\mathop{val}}_{G}(\dot{W}_{\zeta_{\mu}})$. Let $p$ be an
arbitrary element of
$\operatorname{\operatorname{Fn}}(\kappa\setminus\mu_{0},2)$ and let $m$ be
any integer. We will produce $q<p$ and a pair $m<\ell_{1}<\ell_{2}$ so that
$q$ forces that $\\{\ell_{1},\ell_{2}\\}\subset\dot{J}$,
$\dot{n}_{\ell_{1}}\in\dot{V}_{\mu}$ and
$\dot{n}_{\ell_{2}}\notin\dot{U}_{\mu}$. By possibly increasing $m$, we can
assume that $\operatorname{\mathop{dom}}(p)\cap[\lambda,\lambda+\omega)$ is
contained in $[\lambda,\lambda+m]$. Similarly, we can assume that
$\operatorname{\mathop{dom}}(p)\cap I_{\mu_{\ell}}\setminus\mu_{\ell}$ is
empty for all $\ell>m$. Choose
$\bar{p}\in\operatorname{\operatorname{Fn}}(I,2)$ so that
$\sigma_{\mu_{0},\mu}(\bar{p})$ is equal to $p\restriction
I_{\mu}\setminus\mu$. Let
$q_{1}\in\operatorname{\operatorname{Fn}}(\mu\setminus\mu_{0},2)$ be any
extension of $(p\restriction\mu)\cup\sigma_{\mu_{0},\mu_{m+1}}(\bar{p})$ such
that $q_{1}(\lambda{+}m{+}1)=1$, and there is an $\ell_{1}>m$ and an
$n_{\ell_{1}}\in\omega$ such that $q_{1}\Vdash\ell_{1}\in\dot{J}$ and
$q_{1}\Vdash\dot{n}_{\ell_{1}}=n_{\ell_{1}}$. Notice that $q_{1}\Vdash
n_{\ell_{1}}\in\dot{V}_{m+1}$. Now let
$\bar{p}_{1}\in\operatorname{\operatorname{Fn}}(I,2)$ be chosen so that
$\sigma_{\mu_{0},\mu_{m+1}}(\bar{p}_{1})=q_{1}\restriction
I_{\mu_{m+1}}\setminus\mu_{m+1}$ and note that $\bar{p}_{1}<\bar{p}$. Let us
note that $\sigma_{\mu_{0},\mu}(\bar{p}_{1})$ forces that
$n_{\ell_{1}}\in\dot{V}_{\mu}$. Choose next a sufficiently large
$m_{2}>\ell_{1}$ so that
$\operatorname{\mathop{dom}}(q_{1})\cap[\lambda,\lambda+\omega)\subset[\lambda,\lambda+m_{2})$
and $\operatorname{\mathop{dom}}(q_{1})\cap I_{\mu_{\ell}}\setminus\mu_{\ell}$
is empty for all $\ell>m_{2}$. Choose
$q_{2}\in\operatorname{\operatorname{Fn}}(\mu\setminus\mu_{0},2)$ so that
$q_{2}<\sigma_{\mu_{0},\mu_{m_{2}}}(\bar{p}_{1})$, $q_{2}(\lambda+m_{2})=0$,
and, such that there are an $\ell_{2}>m_{2}$ and $n_{\ell_{2}}\in\omega$ such
that $q_{2}\Vdash\ell_{2}\in\dot{J}$ and
$q_{2}\Vdash\dot{n}_{\ell_{2}}=n_{\ell_{2}}$. Let
$\bar{p}_{2}\in\operatorname{\operatorname{Fn}}(I,2)$ such that
$\sigma_{\mu_{0},\mu_{m_{2}}}(\bar{p}_{2})=q_{2}\restriction
I_{\mu_{m_{2}}}\setminus\mu_{m_{2}}$. Since $q_{2}\Vdash
m_{2}\notin\dot{L}_{\lambda}$, we have that $q_{2}\Vdash
n_{\ell_{2}}\in\omega\setminus\dot{U}_{{m_{2}}}$. Therefore $\bar{p}_{2}\Vdash
n_{\ell_{2}}\notin\dot{U}_{0}$, which implies that
$\sigma_{\mu_{0},\mu}(\bar{p}_{2})<\sigma_{\mu_{0},\mu}(\bar{p}_{1})$ forces
that $n_{\ell_{2}}\notin\dot{U}_{\mu}$ and $n_{\ell_{1}}\in\dot{V}_{\mu}$. The
required condition $q$ is $q_{2}\cup\sigma_{\mu_{0},\mu}(\bar{p}_{2})$. We
have that $\sigma_{\mu_{0},\mu}(\bar{p}_{2})$ forces that
$n_{\ell_{1}}\in\dot{V}_{\mu}$ and that $n_{\ell_{2}}\notin\dot{U}_{\mu}$. ∎

We do not know if we had to assume that $\omega$ was a discrete subset in
Theorem 4.2 so was raise this question.

######  Question 2.

Is it consistent that every separable pseudocompact pseudoradial regular space
has cardinality at most $\mathfrak{c}$?

Now we prove what may be the main result of the paper.

######  Theorem 4.3.

If $\mathfrak{c}\leq\aleph_{2}$, then there is a separable 0-dimensional
pseudoradial space of cardinality greater than $\mathfrak{c}$.

###### Proof.

As we learned in our previous results the space will necessarily contain a
large independent family of clopen sets, and in fact the family must have
special properties that will support the existence of many converging
sequences. We will use the ideas introduced by P. Simon [Petr] to facilitate
the construction of such families by using a countable set with special
structure. For each $n\in\omega$, let $D_{n}=\left(2^{n}\right)^{2^{n}}$ Our
countable dense set will be $D=\bigcup\\{D_{n}:n\in\omega\\}$.

Our construction will be to define a Boolean subalgebra $\mathcal{B}$ of
$\mathcal{P}(D)$ and to then choose a subspace of the space of ultrafilters,
i.e. Stone space, of $\mathcal{B}$. As usual, the finite subsets of $D$ will
be members of $\mathcal{B}$ and we identify the fixed ultrafilters of
$\mathcal{B}$ with the points of $D$.

We will define a subtree of $2^{<\omega_{2}}$ that has cardinality
$\mathfrak{c}$ and has more than $\mathfrak{c}$ many cofinal branches. If
$2^{\aleph_{1}}=\aleph_{2}$ then this tree will necessarily have height
$\omega_{2}$. We first present the proof in the case that
$2^{\aleph_{1}}=\aleph_{2}$ and indicate at the end of the proof the minor
modification needed for the case that $2^{\aleph_{1}}>\aleph_{2}$.

We will let $T$ denote the chosen subtree of $2^{\leq\omega_{2}}$ (including
its cofinal branches) and use the elements of $T$ to enumerate the
ultrafilters of $\mathcal{B}$ that are chosen to be members of our space $X$.
This approach ensures that our space is zero-dimensional and Hausdorff. Thus
$X$ will, technically have as base set $D\cup\\{\mathcal{F}_{t}:t\in T\\}$ but
when proving that $X$ is pseudoradial it will be convenient to identify $t\in
T$ with the point $\mathcal{F}_{t}$ in $S(\mathcal{B})$. The structure of $T$
will be very helpful in understanding the convergence structure of $X$. In
particular, for many, but not all, elements $t$ of $T$ that lie on limit
levels of uncountable cofinality, the set of predecessors of $t$ will be a
well-ordered sequence that converges to $t$. This gives some of the insight
into the proof that $X$ is pseudoradial. As mentioned in the first paragraph,
the structure of $D$ is chosen to allow us to more strategically define the
sequences from $D$ that converge to points of $X$.

For $t\in 2^{\leq\omega_{2}}$, let $o(t)$ denote the domain or level of $t$.
Let $S^{2}_{0}$ and $S^{2}_{1}$ denote the stationary subsets of $\omega_{2}$
consisting of the cofinality $\omega$ and, respectively, $\omega_{1}$ limits.
Let $(S^{2}_{1})^{\prime}$ denote the set of all limits of $S^{2}_{1}$ and let
$S_{0}=S^{2}_{0}\cap(S^{2}_{1})^{\prime}$ and
$S_{1}=S^{2}_{1}\cap(S^{2}_{1})^{\prime}$. Let $T_{0}$ be the subtree of
$2^{\leq\omega_{2}}$ where $t\in T_{0}$ if and only if $t^{-1}(1)$ is a subset
of $S_{1}$ (i.e. we are only branching at levels in $S_{1}$). It should be
clear that $T_{0}\cap 2^{\omega_{2}}$ has cardinality $2^{\aleph_{2}}$. We
will need to add some more nodes to our tree because we want to have a large
set of auxillary successors to associate with each $t\in T_{0}$ where $o(t)\in
S_{0}$. Therefore $T$ will consist of $T_{0}$ together with all
$t^{\frown}\sigma$ (concatentation) where $t\in T_{0}$, $\sigma\in
2^{\leq\omega}$, and $o(t)\in S_{0}$. For each $t\in T\setminus T_{0}$, let
$\delta_{t}\in S_{0}$ and $\sigma_{t}\in 2^{\leq\omega}$ denote the values
such that $t=t\restriction\delta_{t}{}\frown\sigma_{t}$. For $s,t\in T$, let
$s\wedge t$ be the maximum element of $T$ satisfying that $s\wedge t\leq s$
and $s\wedge t\leq s$.

Now we choose our special independent subfamily of $\mathcal{P}(D)$ and a
special family of converging sequences. It will make the proof more readable
to use the set $T\cap 2^{<\omega_{2}}$ to enumerate the family. Let
$\\{r_{t}:t\in T_{0}\cap 2^{<\omega_{2}}\\}$ be any one-to-one enumeration of
$2^{\omega}\setminus\\{\vec{0}\\}$ (where $\vec{0}$ is the constant zero
function).

$\mbox{For each}~{}t\in T_{0}\cap 2^{<\omega_{2}},\ \ A_{t}=\bigcup_{n}\\{d\in
D_{n}:d(r_{t}\restriction n)=r_{t}\restriction n\\}~{}.$

Let $\mathcal{L}$ denote the family of finite non-empty chains of $T_{0}\cap
2^{<\omega_{2}}$. For each $L\in\mathcal{L}$, let $B_{L}$ be the set of
$d\in\bigcap\\{A_{t}:t\in L\\}$ satisfying that $d(\sigma)=\vec{0}\restriction
n$ providing $d\in D_{n}$ and $\sigma\in 2^{n}\setminus\\{r_{t}\restriction
n:t\in L\\}$ (i.e. $d\in B_{L}\cap D_{n}$ if $d$ is the identity on
$\\{r_{t}\restriction n:t\in L\\}$ and otherwise takes on value
$\vec{0}\restriction n$).

######  Claim 9.

For each $L\in\mathcal{L}$, $B_{L}$ is an infinite subset of
$\bigcap\\{A_{t}:t\in L\\}$ and for $L\neq L^{\prime}\in\mathcal{L}$,
$B_{L}\cap B_{L^{\prime}}$ is finite.

###### Proof of Claim:.

By symmetry we may assume that $L\setminus L^{\prime}\neq\emptyset$ and let
$t_{L}\in L\setminus L^{\prime}$. Choose $n_{0}$ large enough so that the
elements of $\\{r_{t}\restriction n_{0}:t\in L\cup L^{\prime}\\}$ are pairwise
distinct and not equal to $\vec{0}\restriction{n_{0}}$. For all $n>n_{0}$ and
$d\in B_{L^{\prime}}$, $d(r_{t_{L}}\restriction n)=\vec{0}\restriction n$ and,
for all $d\in B_{L}$, $d(r_{t_{L}}\restriction n)=r_{t_{L}}\restriction
n\neq\vec{0}\restriction n$. ∎

It follows from Claim 9 that the family $\\{A_{t}:t\in T_{0}\cap
2^{<\omega_{2}}\\}$ is an independent family. Let $\mathcal{I}$ be the ideal
of subsets of $D$ that are almost disjoint from every such $B_{L}$, i.e.
$\mathcal{I}$ is often denoted as
$\left(\\{B_{L}:L\in\mathcal{L}\\}\right)^{\perp}$.

Let $\mathcal{B}$ denote the Boolean subalgebra of $\mathcal{P}(D)$ generated
by the family

$[D]^{<\aleph_{0}}\cup\\{A_{t}:t\in T_{0}\cap
2^{<\omega_{2}}\\}\cup\\{B_{L}:L\in\mathcal{L}\\}\cup\mathcal{I}$

and we will choose, for each $t\in T$, a (free) ultrafilter $\mathcal{F}_{t}$
on $\mathcal{B}$.

For each $t\in T$, we will first make an assignment $H_{t}$ consisting of a
non-empty set of nodes in $T$ that are less or equal to $t$, and this
assignment will determine the ultrafilter $\mathcal{F}_{t}$ as described
below.

We begin with the assignment $\langle H_{t}:t\in T\rangle$.

  1. (1)

For $t\in T_{0}$ with $o(t)\in S_{1}\cup\\{\omega_{2}\\}$,
$H_{t}=\\{t\restriction\alpha:\alpha\in o(t)\\}$.

  2. (2)

For each $t\in T_{0}$ such that $o(t)=\mu+\omega_{1}$ for some $\mu\in S_{1}$,
ensure that $\\{H_{t\restriction\xi}:\mu<\xi<\mu+\omega_{1}\\}$ is a one-to-
one listing of all the finite subsets of
$\\{t\restriction\alpha:\alpha<\mu+\omega_{1}\\}$ that have at least one
element strictly extending $t\restriction\mu$ (and satisfy the requirement
that $t\restriction\xi\in H_{t\restriction\alpha}$ implies $\xi\leq\alpha$).

  3. (3)

For each $t\in T_{0}$ such that $o(t)=\mu+\omega_{1}$ for some
$\mu\in\\{0\\}\cup\overline{S^{2}_{1}}\setminus S_{1}$, ensure that
$\\{H_{t\restriction\xi}:\mu\leq\xi<\mu+\omega_{1}\\}$ is a one-to-one listing
of all the finite subsets of $\\{t\restriction\alpha:\alpha<\mu+\omega_{1}\\}$
that have at least one element above $t\restriction\mu$ (and satisfy the
requirement that $t\restriction\xi\in H_{t\restriction\alpha}$ implies
$\xi\leq\alpha$).

  4. (4)

For each $t\in T_{0}$ with $o(t)\in S_{0}$, ensure that
$\\{H_{t^{\frown}\sigma}:\sigma\in 2^{\leq\omega}\ \mbox{and}\
t^{\frown}\sigma\notin T_{0}\\}$ is a one-to-one listing of all countable
cofinal subsets of $\\{t\restriction\alpha:\alpha<o(t)\\}$. Note that, in the
previous item, $H_{t^{\frown}\sigma}$ has already been defined as a finite set
if $t^{\frown}\sigma\in T_{0}$.

######  Claim 10.

If $t\neq s$ and $t,s\in T$, then $H_{t}\neq H_{s}$.

###### Proof of Claim:.

It is clear that $H_{t}\neq H_{s}$ if either of them is uncountable.
Similarly, if they are countably infinite, then it follows from clause (4)
that $\delta_{t}=\delta_{s}$, and thus that $H_{t}\neq H_{s}$ since the
assignment in (4) was chosen to be one-to-one. Finally we suppose that $H_{t}$
and $H_{s}$ are finite. We may choose minimal $\mu_{t}$ so that $\mu_{t}\leq
o(t)\leq\mu_{t}+\omega_{1}$. Similarly choose minimal $\mu_{s}$ so that
$\mu_{s}\leq o(s)\leq\mu_{s}+\omega_{1}$. It follows by the minimality, that
$\mu_{t}$ and $\mu_{s}$ are limit points of $S^{2}_{1}$. It follows easily
from clauses (2) and (3) that if $H_{t}=H_{s}$, then $\mu_{t}=\mu_{s}$.

If $\mu_{t}\in S_{1}$ clause (2) ensures that if
$t\restriction\mu_{t}+1=s\restriction\mu_{t}+1$ then $H_{s},H_{t}$ are distint
elements of $\\{H_{\bar{t}\restriction\xi}:\mu_{t}<\xi<\mu_{t}+\omega_{1}\\}$
for some $\bar{t}\in 2^{\mu_{t}+\omega_{1}}$. Otherwise
$t\restriction\mu_{t}+1\neq s\restriction\mu_{t}+1$ and clause (2) ensures
that $H_{t}$ and $H_{s}$ contain incomparable elements. Finally, if
$\mu_{t}\notin S_{1}$, then $s$ and $t$ are distinct but comparable and again
we have that $H_{t}$ and $H_{s}$ are distinct elements of the list
$\\{H_{\bar{t}\restriction\xi}:\mu_{t}\leq\xi<\mu_{t}+\omega_{1}\\}$ for some
$\bar{t}\in 2^{\mu_{t}+\omega_{1}}$. ∎

Now we make the simple assignment of the family $\langle\mathcal{F}_{t}:t\in
T\rangle$:

  1. (5)

If $H_{t}$ is finite, then define $\mathcal{F}_{t}$ to be the unique free
ultrafilter of $\mathcal{B}$ with $B_{H_{t}}\in\mathcal{F}_{t}$.

  2. (6)

If $H_{t}$ is infinite, then $\mathcal{F}_{t}$ is the ultrafilter satisfying

    1. (a)

$A_{s}\in\mathcal{F}_{t}$ if and only if $s\in H_{t}$,

    2. (b)

$B_{L}\notin\mathcal{F}_{t}$ for all $L\in\mathcal{L}$,

    3. (c)

$I\notin\mathcal{F}_{t}$ for all $I\in\mathcal{I}$,

    4. (d)

$[D]^{<\aleph_{0}}\cap\mathcal{F}_{t}$ is empty.

It should be clear that the subspace $D\cup X_{1}$ is simply equal to the
traditional Mrowka-Isbell type space constructed from the almost disjoint
family $\\{B_{L}:L\in\mathcal{L}\\}$. Let us also note that for each
$I\in\mathcal{I}$, $I$ is closed in $X$, and this implies this next claim.

######  Claim 11.

If $Y\subset D$ and $\overline{Y}\cap X_{1}$ is empty, then $Y$ is closed.

###### Proof of Claim:.

If no point of $X_{1}$ is in the closure of $Y$, then $Y\cap B_{L}$ is finite
for all $L\in\mathcal{L}$. Thus $Y\in\mathcal{I}$. ∎

It will be convenient to let $K_{t}=\\{A_{s}:s\in(T_{0}\cap
2^{<\omega_{2}})\setminus H_{t}\\}$, i.e.
$K_{t}=\\{A_{s}:A_{s}\notin\mathcal{F}_{t}\\}$. For disjoint finite subsets
$H,K$ of $T_{0}\cap 2^{<\omega_{2}}$, let $[H;K]$ denote the clopen subset of
$X$ corresponding to the closure of this element of $\mathcal{B}$:

$[H;K]=\bigcap\\{A_{t}:t\in H\\}\setminus\left(\bigcup\\{A_{s}:s\in
K\\}\cup\bigcup\\{B_{L}:\emptyset\neq L\subset H\\}\right)~{}.$

For any $t\in X_{2}\cup X_{3}$, the family
$\\{[H;K]:H\in[H_{t}]^{<\aleph_{0}},K\in[K_{t}]^{<\aleph_{0}}\\}$ is easily
seen to be a local filter base in the subspace $X\setminus D$. Let $X_{0}=D$,
$X_{1}=\\{t\in T_{0}:|H_{t}|<\aleph_{0}\\}$, $X_{2}=\\{t\in
T:|H_{t}|=\aleph_{0}\\}$ and let $X_{3}=\\{t\in T:|H_{t}|>\aleph_{0}\\}$.

######  Claim 12.

$X_{3}$ is a closed radial subspace of $X$.

###### Proof of Claim:.

We first show that $X_{3}$ is closed. Clearly $X_{0}\cup X_{1}$ is open so
consider any $t\in X_{2}$. Since $H_{t}$ is a countable cofinal subset of the
uncountable ordinal $\delta_{t}$, there are $\beta<\alpha<\delta_{t}$ such
that $t\restriction\beta\in K_{t}$ and $t\restriction\alpha\in H_{t}$. Note
that $[\\{t\restriction\alpha\\};\\{t\restriction\beta\\}]$ is disjoint from
$X_{3}$ since $H_{s}$ is a downward closed subset of $T_{0}$.

Now consider any $Y\subset X_{3}$ and assume that $t\in X_{3}$ is a limit
point of $Y$. We will assume that $o(t)<\omega_{2}$ and leave the case when
$o(t)=\omega_{2}$ to the reader. Choose any strictly increasing cofinal
sequence $\\{\alpha_{\xi}:\xi\in\omega_{1}\\}\subset o(t)$. Note that $t\in
K_{t}$. For each $\xi<\omega_{1}$, choose $y_{\xi}\in
Y\cap[\\{t\restriction\alpha_{\xi}\\};\\{t\\}]$. Consider any finite $H\subset
H_{t}$ and finite $K\subset K_{t}$. Let $\tilde{K}=\\{s\in K_{t}:t\not\leq
s\\}$ and choose $\xi<\omega_{1}$ so that $t\restriction\alpha_{\xi}\not\leq
s$ for all $s\in\tilde{K}$. It follows easily that $y_{\eta}\in[H;K]$ for all
$\xi<\eta<\omega_{1}$ and thereby proving that $\langle
y_{\xi}:\xi\in\omega_{1}\rangle$ converges to $t$. ∎

We continue the proof that $X$ is pseudoradial. Let us note that it follows
from Claim 11 that it suffices to proof that $X\setminus D$ is pseudoradial.
For the remainder of the proof we will say that an elementary submodel $M$ is
suitable to mean that $M\prec H(\aleph_{3})$ and that
$\\{T_{0},\mathcal{L},\\{A_{t}:t\in T_{0}\cap 2^{<\omega_{2}}\\},\\{H_{t}:t\in
T\\},\\{B_{L}:L\in\mathcal{L}\\}\\}$ is an element of $M$. Here is one of the
key properties of the space.

######  Claim 13.

If $Y\subset X\setminus D$ and $t\in X_{2}\cup X_{3}$ is a limit point of $Y$,
then for any countable suitable elementary submodel $M$ such that $Y,t\in M$,
there is a converging sequence $\\{t_{n}:n\in\omega\\}\subset Y\cap M$ such
that $t$ is the limit if $t\in X_{2}$ and, if $t\in X_{3}$, the limit is
$t_{M}\in X_{2}$ where

  1. (1)

$\delta_{t_{M}}=\sup(M\cap o(t))$ and
$t_{M}=t\restriction\delta_{t_{M}}{}^{\frown}\sigma_{t_{M}}$,

  2. (2)

$H_{t_{M}}=\\{t\restriction\alpha:\alpha\in M\cap o(t)\\}$.

###### Proof of Claim:.

Let $\bar{t}$ equal to $t$ if $t\in X_{2}$ and let $\bar{t}=t_{M}$ if $t\in
X_{3}$. In either case, we have that $H_{\bar{t}}\subset M$. If $t\in X_{3}$,
then observe that $t\in K_{t}$ and $K_{t}\cap\\{t\restriction\alpha:\alpha\in
M\cap o(t)\\}$ is empty. If $t\in X_{2}$, then, by elementarity,
$K_{t}\cap\\{t\restriction\alpha:\alpha\in M\cap o(t)\\}$ is non-empty. Define
$K_{0}\in M$ to be $\\{t\\}$ if $t\in X_{3}$ and to be any finite subset of
$K_{t}\cap\\{t\restriction\alpha:\alpha\in M\cap o(t)\\}$ if $t\in X_{2}$.
Choose $H_{0}$ to be any finite subset of $H_{\bar{t}}$ so that, if $t\in
X_{2}$, then there are $\alpha<\beta$ so that $t\restriction\alpha\in K_{0}$
and $t\restriction\beta\in H_{0}$. Since $[H_{0};K_{0}]\in M$, there is a
$y_{0}\in M\cap Y\cap[H_{0};K_{0}]$.

Choose any sequence $\\{[H_{n};K_{n}]:n\in\omega\\}\subset M$ so that:

  1. (1)

$\\{H_{n}:n\in\omega\\}$ is a strictly increasing sequence of finite sets
whose union is $H_{\bar{t}}$,

  2. (2)

$\\{K_{n}:n\in\omega\\}$ is a strictly increasing sequence of finite sets
whose union equals $(K_{\bar{t}}\setminus H_{t})\cap M$,

We show that we can choose $y_{n}\in Y\cap M\cap[H_{n};K_{n}]$ for all $n$. If
$t=\bar{t}$, then $t\in[H_{n};K_{n}]$ and so $Y\cap M\cap[H_{n};K_{n}]$ is not
empty by elementarity and the assumption that $t$ is a limit point of $Y$. If
$t\neq\bar{t}$, then $H_{n}\subset H_{\bar{t}}\subset H_{t}$ and $K_{n}\subset
K_{\bar{t}}\setminus H_{t}$. Therefore, we again have that $t\in[H_{n};K_{n}]$
and so, by elementarity, there is a $y_{n}\in Y\cap M\cap[H_{n};K_{n}]$.

Now we show that the sequence $\\{y_{n}:n\in\omega\\}$ converges to $\bar{t}$.
Consider any finite $H\subset H_{\bar{t}}$ and finite $K\subset K_{\bar{t}}$.
Choose $\bar{\beta}<\delta_{\bar{t}}$ large enough so that
$o(s\wedge(t\restriction\delta_{\bar{t}}))<\bar{\beta}$ for all $s\in K$ such
that $t\restriction\delta_{\bar{t}}\not\leq s$. We may assume that
$t\restriction\bar{\beta}\in H_{\bar{t}}$. Choose $n$ so that
$H\cup\\{t\restriction\bar{\beta}\\}\subset H_{n}$, and $(K\setminus
H_{t})\cap M$ is a subset of $K_{n}$. We show that $y_{m}\in[H;K]$ for each
$m>n$. Since $H\subset H_{m}$ and $y_{m}\in[H_{m};K_{m}]$ it is clear that
$H\subset H_{y_{m}}$ and $K_{m}\subset K_{y_{m}}$. Also, $H_{y_{m}}\setminus
H_{m}$ is not empty (in the case that $H_{y_{m}}$ is finite). To show that
$y_{m}\in[H;K]$, we must simply show that $H_{y_{m}}\cap K$ is empty.

In the case that $t=\bar{t}$, then $K_{\bar{t}}\setminus H_{t}=K_{\bar{t}}$,
and so $H_{y_{m}}\cap K\subset H_{y_{m}}\cap K_{m}=\emptyset$. So now we
assume that $\bar{t}\neq t\in X_{3}$ and recall that $t\in K_{0}$. In this
case, $K_{\bar{t}}\cap M$ is also disjoint from $H_{t}$, hence $K\cap M\subset
K_{n}$. If $y_{m}\notin X_{3}$, then $H_{y_{m}}\subset M$ and so
$H_{y_{m}}\cap K\subset H_{y_{m}}\cap K\cap M\subset H_{y_{m}}\cap
K_{m}=\emptyset$. Now suppose that $y_{m}\in X_{3}$ and assume that $s\in
H_{y_{m}}\cap K$. Note that we have that $t\restriction\bar{\beta}\in
H_{y_{m}}$ and so $t\restriction\delta_{\bar{t}}\leq s\leq y_{m}$. Also,
$\\{t\restriction\alpha:\alpha<\delta_{\bar{t}}\\}$ is a subset of
$H_{y_{m}}$. Since $t\in K_{0}$ implies that $o(t\wedge y_{m})\in M\cap
o(t)=\delta_{\bar{t}}$, this contradicts that
$t\restriction\delta_{\bar{t}}\leq s$. ∎

######  Claim 14.

If $Y\subset X\setminus D$ and $t\in X_{3}$ is a limit point of $Y$, then
there is a sequence $\\{y_{\alpha}:\alpha\in\omega_{1}\\}$ converging to a
point in $X_{3}$ where each $y_{\alpha}$ is in the sequential closure of $Y$.

Before we prove this Claim, we observe that this completes the proof that $X$
is pseudoradial. If $Y$ is a subset of $X\setminus D$ then, by Claim 13 every
limit point in $X_{2}$ is in the sequential closure, and by Claim 14, the
points of $X_{3}$ in the radial closure of $Y$ are dense in $\overline{Y}\cap
X_{3}$. Since $X_{3}$ is radial (Claim 12), this implies that the radial
closure of $Y$ is closed.

###### Proof of Claim:.

Fix any increasing sequence $\\{M_{\alpha}:\alpha\in\omega_{1}\\}$ of suitable
countable elementary submodels satisfying that $Y,t$ are elements of $M_{0}$.
There is an ordinal $\theta\in S_{1}$ such that
$\theta=\bigcup\\{M_{\alpha}\cap\omega_{2}:\alpha\in\omega_{1}\\}$ (see
[Kunen]). If $o(t)<\omega_{2}$, let $\lambda=o(t)\in S_{1}$. If
$o(t)=\omega_{2}$, let $\lambda=\theta$.

For each $\alpha\in\omega_{1}$, let
$\delta_{\alpha}=\sup(M_{\alpha}\cap\lambda)$. Following Claim 13, let
$y_{\alpha}=t\restriction\delta_{\alpha}{}^{\frown}\sigma_{\alpha}$ where
$\sigma_{\alpha}$ is chosen so that
$H_{y_{\alpha}}=\\{t\restriction\gamma:\gamma\in M_{\alpha}\cap\lambda\\}$. It
follows from Claim 13, that $y_{\alpha}$ is in the sequential closure of $Y$.

Now we simply check that $\\{y_{\alpha}:\alpha\in\omega_{1}\\}$ converges to
$t\restriction\lambda$. Fix any basic open set $[H;K]$ of
$t\restriction\lambda$. Choose any $\beta<\omega_{1}$ so that $H\in
M_{\beta}$. Observe that $H\subset H_{y_{\alpha}}$ for all
$\beta<\alpha\in\omega_{1}$. Similarly, $H_{y_{\alpha}}\subset
H_{t\restriction\delta_{\alpha}}\subset H_{t\restriction\lambda}$ for all
$\alpha\in\omega_{1}$. Therefore $K\cap H_{y_{\alpha}}=\emptyset$, and so
$y_{\alpha}\in[H;K]$, for all $\beta<\alpha\in\omega_{1}$. This completes the
proof of the Claim. ∎

This completes the proof of the Theorem in the case that
$2^{\aleph_{1}}=\aleph_{2}$. The only modification that is needed for the case
$2^{\aleph_{1}}>\aleph_{2}$ is to pass to the subtree $T_{0}\cap
2^{\leq\theta}$ where $\theta\in S_{1}$ is the minimum level of $T_{0}$ which
has cardinality greater than $\aleph_{2}$. This ensures that $T_{0}\cap
2^{<\theta}$ has cardinality $\aleph_{2}$. Establish the new enumeration
$\\{r_{t}:t\in T_{0}\cap 2^{<\theta}\\}$ of $2^{\omega}\setminus\\{0\\}$ and
the proof proceeds exactly as above. ∎

######  Question 3.

Is it consistent that there is no separable pseudoradial space of cardinality
$2^{\mathfrak{c}}$?

######  Question 4.

Do separable regular pseudoradial spaces of cardinality greater than
$\mathfrak{c}$ exist?

We also do not know if large separable Hausdorff pseudoradial spaces of
cardinality greater than $\mathfrak{c}$ exist.

## References