# Measuring the complexity of characterizing $[0,1]$, $S^{1}$, and
$\mathbb{R}$ up to homeomorphism

Matthew Harrison-Trainor and Eissa Haydar Harrison-Trainor was partially
supported by the National Science Foundation under Grant DMS-2153823. This
work began while Haydar was an REU student at the University of Michigan
funded under this grant.

###### Abstract

In analogy to the study of Scott rank/complexity of countable structures, we
initiate the study of the Wadge degrees of the set of homeomorphic copies of
topological spaces. One can view our results as saying that the classical
characterizations of $[0,1]$ (e.g., as the unique continuum with exactly two
non-cut points, and other similar characterizations), appropriated expressed,
are the simplest possible characterizations of $[0,1]$. Formally, we show that
the set of homeomorphic copies of $[0,1]$ is $\mathbf{\Pi}^{0}_{4}$-Wadge-
complete. We also show that the set of homeomorphic copies of $S^{1}$ is
$\mathbf{\Pi}^{0}_{4}$-Wadge-complete. On the other hand, we show that the set
of homeomorphic copies of $\mathbb{R}$ is $\mathbf{\Pi}^{1}_{1}$-Wadge-
complete. It is the local compactness that cannot be expressed in a Borel way;
the set of homeomorphic copies of $\mathbb{R}$ is
$\mathbf{\Pi}^{0}_{4}$-Wadge-complete within the locally compact spaces.

##  1 Introduction

A long-standing program in computable structure theory is the study of Scott
rank and Scott complexity. The basic question is, given a countable structure
$\mathcal{A}$, how hard is it to characterize $\mathcal{A}$ up to isomorphism
among all countable structures? This began with Scott [Sco65] who showed that
there is always a formula of the infinitary logic
$\mathcal{L}_{\omega_{1}\omega}$ characterizing $\mathcal{A}$ among countable
structures. One can then look for the simplest such sentence, with its
complexity called the Scott rank of $\mathcal{A}$.111There are numerous other
definitions of Scott rank, most using some notion of back-and-forth relations,
but for the purposes of this paper we can ignore them. See, e.g., [Mon].

The particular perspective we will take in this paper is that of Scott
complexity as introduced in [AGHTT21] by Alvir, Greenberg, Harrison-Trainor,
and Turetsky. The idea is to consider, in the Polish space
$\operatorname{Mod}(\mathcal{L})$ of $\mathcal{L}$-structures, the set

$\operatorname{Copies}(\mathcal{A})=\\{\mathcal{B}:\mathcal{B}\cong\mathcal{A}\\}$

of all isomorphic presentations of $\mathcal{A}$. By a well-known theorem of
Lopez-Escobar [LE65], $\mathcal{A}$ has a $\Sigma_{\alpha}$ Scott sentence if
and only if $\operatorname{Copies}(\mathcal{A})$ is
$\boldsymbol{\Sigma}^{0}_{\alpha}$, and similarly for other complexity classes
such as $\Pi_{\alpha}$/$\mathbf{\Pi}^{0}_{\alpha}$ and the difference
hierarchy. Thus, instead of considering sentences characterizing
$\mathcal{A}$, we can instead study the Wadge degree of
$\operatorname{Copies}(\mathcal{A})$. This Wadge degree is called the Scott
complexity of $\mathcal{A}$, and measures the complexity of charactering
$\mathcal{A}$ up to isomorphism.

In this paper, our goal is to, instead of characterising countable structures
up to isomorphism, characterize certain topological spaces up to
homeomorphism. We restrict our attention to Polish spaces, that is, separable
completely metrizable spaces. In this case, there is no syntactic perspective,
and so given a topological space $X$ we study the Wadge degree of the
homeomorphic copies of $X$,

$\operatorname{HCopies}(X)=\\{Y:Y\cong_{\operatorname{homeo}}X\\}.$

We call this the (topological) Scott complexity of $X$.222We also note that
there is an analogous program for characterizing metric spaces (or indeed,
metric structures) up to isometry, by considering
$\\{Y:Y\cong_{isometric}X\\}$. While we are not aware of any work on this
problem specifically, there is existing work along these lines, e.g., [MN13,
CMS19, BM20]. The isometry classes are known to be Borel by [BYDNT17] or
[EFP+13]. (Here, the copies $Y$ are thought of as being presented as the
completion of a countable metric space. The space of all such presentation is
itself a Polish space. See Section 2.1 for details.) The Scott complexity of
$X$ measures the difficulty of characterizing $X$ up to homeomorphism.

For a countable structure $\mathcal{A}$, $\operatorname{Copies}(\mathcal{A})$
is always Borel. This is not the case for $\operatorname{HCopies}(X)$ if $X$
is not compact; indeed, $\operatorname{HCopies}(\mathbb{R})$ is not Borel, as
shown later in this paper. However, if $X$ is compact,
$\operatorname{HCopies}(X)$ is Borel; indeed, the homeomorphism equivalence
relation is bi-reducible to a complete orbit equivalence relation of a Borel
action of a Polish group [Zie16]. Another perspective, which we do not use in
this paper, is to use the effective Gelfand duality [BEF+] to reduce the
problem to isometric isomorphism of unital abelian $C^{*}$-algebras. These are
metric structures and thus subject to the metric Scott analysis [BYDNT17], but
it is not yet clear how to fruitfully pull back the metric Scott analysis
across the Gelfand duality.

In the case of discrete structures, there are many examples for which the
Scott complexity has been computed. For example, the complexity of the groups
free groups [CHK+12, MW12]: $\mathbb{F}_{n}$ has complexity
d-$\boldsymbol{\Sigma}^{0}_{2}$ for $n$ finite and $\mathbb{F}_{\infty}$ has
complexity $\mathbf{\Pi}^{0}_{4}$. Moreover, we know [AGHTT21] that the only
possible Scott complexities are $\boldsymbol{\Sigma}^{0}_{\alpha}$,
$\mathbf{\Pi}^{0}_{\alpha}$, and d-$\boldsymbol{\Sigma}^{0}_{\alpha}$.

In the context of topological spaces not only do we not yet know any general
results about the topological Scott complexity but to our knowledge there are
as yet no non-trivial examples where we know the exact Wadge degree of
$\operatorname{HCopies}(X)$. In this paper, we study some of the first
1-dimensional topological spaces to come to mind: the compact line $[0,1]$,
the circle $S^{1}$, and the real line $\mathbb{R}$. We exactly characterize
their topological Scott complexity. For these spaces, there are classical
characterizations coming from continuum theory, e.g., $[0,1]$ is the unique
continuum with exactly two non-cut points. Our setting allows us to analyse
such characterizations and determine whether they are as simple as possible.
We show that, appropriately expressed,333For example, to express the notion of
a cut point it is helpful to first be in a locally connected space. they are
the simplest possible characterizations.

######  Theorem 1.1.

The unit interval $[0,1]$ has topological Scott complexity
$\mathbf{\Pi}^{0}_{4}$, i.e., the set

$\operatorname{HCopies}([0,1])=\\{Y:Y\cong_{\operatorname{homeo}}[0,1]\\}$

is $\mathbf{\Pi}^{0}_{4}$-Wadge-complete.

######  Theorem 1.2.

The circle $S^{1}$ has topological Scott complexity $\mathbf{\Pi}^{0}_{4}$,
i.e., the set

$\operatorname{HCopies}(S^{1})=\\{Y:Y\cong_{\operatorname{homeo}}S^{1}\\}$

is $\mathbf{\Pi}^{0}_{4}$-Wadge-complete.

All of the results in this paper are effective, i.e., these sets are actually
lightface $\Pi^{0}_{4}$ and the Wadge reductions are not just continuous but
computable. In particular we could also state theorems for index sets, e.g.,
the index set

$\\{i\mid\text{the $i$th computable Polish space is homeomorphic to
$[0,1]$}\\}$

is $\Pi^{0}_{4}$ $m$-complete.

For the real line $\mathbb{R}$, because it is not compact, the set of
homeomorphic copies is not necessarily Borel; in fact, we show that it is
$\mathbf{\Pi}^{1}_{1}$-Wadge-complete. (We conjecture that this might always
be the case for any locally compact non-compact space.) However we also show
that the set of homeomorphic copies of $\mathbb{R}$ is the intersection of the
set of locally compact spaces and a Borel set. This suggests that for working
with non-compact locally compact spaces one should work within the locally
compact spaces.

######  Theorem 1.3.

The set

$\operatorname{HCopies}(\mathbb{R})=\\{Y:Y\cong_{\operatorname{homeo}}\mathbb{R}\\}$

is $\mathbf{\Pi}^{1}_{1}$-Wadge-complete, and $\mathbf{\Pi}^{0}_{4}$-Wadge-
complete within the class of locally compact spaces.

This paper is a proof-of-concept for a program of analysing the
characterizations of topological spaces up to homeomorphism. $[0,1]$, $S^{1}$,
and $\mathbb{R}$ are the most basic 1-dimensional topological spaces. A
natural next class of examples to consider are two-dimensional topological
spaces such as $S^{2}$ and $\mathbb{R}^{2}$. In [HTM24], Melnikov and the
first author studied closed surfaces, and one of their results is that for $X$
a closed surface, $\operatorname{HCopies}(X)$ is always arithmetic. Thus
$\operatorname{HCopies}(S^{2})$ is arithmetic, but the upper bound obtained is
around $\Sigma^{0}_{27}$ which is almost certainly not the best possible.
There are no known non-trivial known lower bounds.

##  2 Background

###  2.1 Presentations of separable metric spaces

The traditional approach to studying the class of $\mathcal{L}$-structures in
computable structure theory is to consider the space of representations of
$\mathcal{L}$-structures with domain $\mathbb{N}$. Such a representation can
be thought of as an encoding of the structure in machine-readable form, and
the space of representations forms a Polish space $\text{Mod}(\mathcal{L})$.

In this work, we want to consider representations of topological (and more
specifically Polish) spaces. Since Polish spaces typically have uncountably
many points, we cannot represent each point by a natural number. Turing
[Tur36, Tur37] had already realized this long ago, and instead suggested
representing a real number by a Cauchy sequence of rational numbers. Moreover,
in order to perform any computations, we need to know how fast the Cauchy
sequence converges.

######  Definition 2.1.

A Cauchy sequence $(x_{i})_{i\in\mathbb{N}}$ in a metric space $(M,d)$ is fast
if $d(x_{i},x_{i+1})<2^{-i}$.

One thinks of the reals $\mathbb{R}$ presented as a completion of the
rationals $\mathbb{Q}$, with each real number $r$ represented by the Cauchy
sequences converging to $r$. We extend this to arbitrary Polish spaces, using
any countable dense subset of the space in place of $\mathbb{Q}$.

######  Definition 2.2.

A presentation of a Polish space $X$ is a countable set $M$ of points
(identified with $\mathbb{N}$) and a metric $d$ such that $X$ is homeomorphic
to the completion $\overline{M}$ of the metric space $(M,d)$. We think of
representing the metric $d$ by giving, for every two elements $x,y$ of $M$, a
fast Cauchy sequence converging to $d(x,y)$. Thus, a presentation of $X$ can
be identified with an infinite binary string.

Given a presentation $X=\overline{M}$ of a Polish space, we refer to the
points of $M$ as special points. We represent the non-special points of $X$
using fast Cauchy sequences from $M$. The presentation
$\mathbb{R}=\overline{\mathbb{Q}}$ is the natural presentation of $\mathbb{R}$
using $\mathbb{Q}$ as the special points, but there are, of course, other
possible presentations of $\mathbb{R}$.

When we talk about the topology on $X$, we will generally use the open balls
centered on special points with rational radii. We call these the basic open
balls and denote them by $B_{r}(x)$. We note that the basic open balls might
not even be connected, even when the space is, e.g., connected and locally
path connected. We denote the basic closed balls by $\overline{B}_{r}(x)$,
noting that this is different from the closure of the basic open ball.

We will frequently make use of $\epsilon$-paths. An $\epsilon$-path from $x$
to $y$ is a sequence of special points $x=u_{1},\ldots,u_{n}=y$ such that
$d(u_{i},u_{i+1})<\epsilon$. The beginning and ending points $x$ and $y$ need
not be special points, but for convenience we require that all the points in
between them must be special. The $\epsilon$-path is within a set $U$ if all
of the points are within $U$. If $U$ is open, then there is an $\epsilon$-path
(using special points) from $x$ to $y$ within $U$ if and only there is an
$\epsilon$-path (not neccesarily consisting of special points) from $x$ to $y$
within $U$.

The space of all presentations of Polish spaces is a $\Pi^{0}_{2}$ subset of
$2^{<\omega}$ (under the identification given above) and so is itself a Polish
space in its own right. We call this space TOP. Note that TOP is homeomorphic
to Baire space $\omega^{\omega}$, and so as described in the following
subsection we can thus talk about Wadge reducibility and the Wadge degrees of
subsets of TOP.

######  Remark 2.3.

There is also an alternate perspective which is more traditional in
descriptive set theory. In this perspective we view the space of compact
Polish spaces as the space of closed subsets of the Hilbert cube
$[0,1]^{\mathbb{N}}$ under the Hausdorff metric (which makes the Hilbert cube
into a Polish space in its own right). As far as we are aware there are as yet
no formal results connecting the complexity of a class of topological spaces
in the first perspective and in the second. But it is clear that, by
effectivizing the universality of the Hilbert cube, there are low-level Borel
maps taking presentations of a compact space to a closed subset of the Hilbert
cube, and vice versa. Thus, while the exact complexity of a class of
topological spaces might change depending on whether we think of presentations
or of closed subsets of $[0,1]^{\mathbb{N}}$, morally speaking the two
approaches are the same.

###  2.2 Wadge degrees and reductions

######  Definition 2.4 (Wadge).

Let $A$ and $B$ be subsets of Baire space $\omega^{\omega}\cong\textsf{TOP}$.
We say that $A$ is _Wadge reducible_ to $B$, and write $A\leq_{W}B$, if there
is a continuous function $f$ with $A=f^{-1}[B]$, i.e.

$x\in A\Longleftrightarrow f(x)\in B.$

The equivalence classes under this pre-order are called the Wadge degrees. The
Wadge hierarchy is the set of Wadge degrees under continuous reductions. With
enough determinacy, the Wadge hierarchy is very well-behaved; it is well-
founded (as proved by Martin and Monk) and almost totally ordered (in the
sense that any anti-chain has size at most two) [Wad83].

Pointclasses such as $\boldsymbol{\Sigma}^{0}_{\alpha}$,
$\mathbf{\Pi}^{0}_{\alpha}$, $\mathbf{\Pi}^{1}_{\alpha}$, and
$\boldsymbol{\Sigma}^{1}_{\alpha}$ are all closed downwards in the Wadge
degrees. For $\Gamma$ a pointclass, and $A$ a set, we say that $A$ is
$\Gamma$-Wadge-complete (or simply $\Gamma$-complete) if it is $\Gamma$ and
also $\Gamma$-hard, which means that for every $B\in\Gamma$, $B\leq_{W}A$.

The reader may be a computability theorist used to many-one reductions of
index sets. All of our Wadge reductions will actually be computable
reductions, and so the same completeness theorems hold for many-one
reductions. But building a Wadge reduction is more general (e.g., there are
computable structures such that the set of indices of computable copies is
$\Sigma^{0}_{3}$, but the set of indices of $0^{\prime}$-computable copies is
d-$\Sigma^{0}_{2}$ relative to $0^{\prime}$). In this paper every upper bound
on the complexity of a set will be a lightface upper bound.

Sometimes the complexity of a particular set is dominated by one aspect. In
computable structure theory, one such example is finitely generated groups.
The class of all finitely generated groups is $\Sigma^{0}_{3}$, and there are
particular finitely generated groups $G$ with $\operatorname{Copies}(G)$ being
$\boldsymbol{\Sigma}^{0}_{3}$-complete [HTH18]. On the other hand for any
finitely generated group $G$ $\operatorname{Copies}(G)$ is the intersection of
a $\Pi^{0}_{3}$ set and the set of all finitely generated groups. Thus the
complexity of saying that $G$ is finitely generated “drowns out” the
complexity of saying which finitely generated group it is. Another example
from [CHK+12, MW12] is that, with $F_{\infty}$ the free group on countably
infinitely many generators, $\operatorname{Copies}(F_{\infty})$ is
$\mathbf{\Pi}^{0}_{4}$-complete, but $\mathbf{\Pi}^{0}_{3}$-complete within
the class of free groups. The standard way to separate these is to measure the
complexity of $\operatorname{Copies}(G)$ within the set of finitely generated
groups/free groups, as follows.

######  Definition 2.5.

Let $A$ and $C$ be sets in $\omega^{\omega}$.

  1. (1)

We say that $A$ is $\Gamma$ within $C$ if there is a set $A^{*}\in\Gamma$ with
$A=A^{*}\cap C$.

  2. (2)

We say that $A$ is $\Gamma$-Wadge-complete (or simply $\Gamma$-complete)
within $C$ if for every $B\in\Gamma$, there is a continuous function $f$ such
that for all $x$, $f(x)\in C$, and $x\in B\Longleftrightarrow f(x)\in A$.

The main example of this phenomenon in our paper is in Theorem 1.3 where when
characterizing $\mathbb{R}$ up to homeomorphism, the fact that it is locally
compact (which is $\Pi^{1}_{1}$) drowns out everything else. We show that
$\operatorname{HCopies}(\mathbb{R})$ is $\mathbf{\Pi}^{0}_{4}$-complete within
the class of locally compact spaces.

The second example is that $\operatorname{HCopies}([0,1])$ is
$\mathbf{\Pi}^{0}_{4}$-complete, but the most difficult part is in expressing
local connectedness. In Theorem 4.1 we show that
$\operatorname{HCopies}([0,1])$ is actually $\mathbf{\Pi}^{0}_{3}$-complete
within the compact, connected, locally connected spaces.

##  3 A $\Pi^{0}_{4}$ characterization of $[0,1]$

In this section we prove the first part of Theorem 1.1, that
$\operatorname{HCopies}([0,1])$ is $\Pi^{0}_{4}$. We prove the second half of
the theorem, that $\operatorname{HCopies}([0,1])$ is
$\mathbf{\Pi}^{0}_{4}$-hard, in the next section.

We recall the classical characterization of $[0,1]$ as the unique continuum
(non-empty compact connected metric space) with exactly two non-cut points
[Nad92, Theorem 6.17]. It is not obvious how to express the existence of non-
cut points as we are limited to quantifying over special points. Instead, we
will express an idea of “betweenness.” In the following subsections, we will
express various properties of a topological space in a $\Pi^{0}_{4}$ (or often
simpler) way, and then show that they characterize $[0,1]$.

###  3.1 Compactness

Following [MN13], compactness can be expressed in a $\Pi^{0}_{3}$ way in terms
of total boundedness: We say that a countable metric space satisfies CPCT if
and only if for every $\epsilon>0$, there is $n\in\mathbb{N}$ and $n$ special
points points $x_{1},\ldots,x_{n}$ such that for all special points $y$, $y$
is within $\epsilon$ of some $x_{i}$. We take a second here to note that
$d(y,x_{i})<\epsilon$ is a $\Sigma^{0}_{1}$ condition, because in a
presentation we are only given a fast Cauchy sequence converging to
$d(y,x_{i})$. To make CPCT $\Pi^{0}_{3}$ we must instead interpret “$y$ is
within $\epsilon$ of some $x_{i}$” as that “$d(y,x_{i})\leq\epsilon$” which is
a $\Pi^{0}_{1}$ condition. As a general principal, these types of conditions
are really about sufficient smallness and so asking for strict or non-strict
inequalities are equivalent.

###  3.2 Connectedness

Connectedness for arbitrary spaces is not Borel, even in the case of finite
dimensional Euclidean spaces, as shown in [DSR20]. However, connectedness in a
compact space can be expressed much more simply.

We say that $X=\overline{(M,d)}$ satisfies CONN if for every $\epsilon>0$ and
special points $x_{1},\ldots,x_{m}$ and $y_{1},\ldots,y_{m}$, if
$U=\overline{B}_{\epsilon}(x_{1})\cup\cdots\cup\overline{B}_{\epsilon}(x_{m})$
and
$V=\overline{B}_{\epsilon}(y_{1})\cup\cdots\cup\overline{B}_{\epsilon}(y_{m})$
cover the special points (i.e., for all special points $z$, there is some
$x_{i}$ with $d(z,x_{i})\leq\epsilon$ or some $y_{i}$ with
$d(z,y_{i})\leq\epsilon$) the sets $U^{*}=B_{2\epsilon}(x_{1})\cup\cdots\cup
B_{2\epsilon}(x_{m})$ and $V^{*}=B_{2\epsilon}(y_{1})\cup\cdots\cup
B_{2\epsilon}(y_{m})$ are not disjoint (i.e., there is a special point $z$
such that for some $x_{i}$ and $y_{j}$, $d(z,x_{i})<2\epsilon$ and
$d(z,y_{i})<2\epsilon$). Note that CONN is a $\Pi^{0}_{2}$ property.444It is
easy to see that the set of connected spaces is also $\Pi^{0}_{2}$-complete
within the compact spaces; one builds a copy of the space
$[0,1-\frac{1}{n}]\cup[1,2]$, and lets $n\to\infty$ in the infinitary outcome.
We remark again that to make this $\Pi^{0}_{2}$ we were careful to use the
closed condition in the antecedent of the implication and the open condition
in the consequence. Going forward, we will no longer explicitly comment on
such issues.

######  Lemma 3.1.

Let $X=\overline{(M,d)}$ be compact. Then $X$ is connected if and only if
$(M,d)$ satisfies CONN.

###### Proof.

We argue that a compact space $X$ is connected if and only if it satisfies
CONN. Suppose that $X$ is not connected, with $X=U\sqcup V$ being a partition.
Let $d$ be the distance between $U$ and $V$, and choose $\epsilon>0$ such that
$2\epsilon<d$. Let
$U=\overline{B}_{\epsilon}(x_{1})\cup\cdots\cup\overline{B}_{\epsilon}(x_{m})$
and
$V=\overline{B}_{\epsilon}(y_{1})\cup\cdots\cup\overline{B}_{\epsilon}(y_{m})$
be $\epsilon$-covers of $U$ and $V$ by closed balls centered at special
points. Note that $U^{*}=B_{2\epsilon}(x_{1})\cup\cdots\cup
B_{2\epsilon}(x_{m})$ and $V^{*}=B_{2\epsilon}(y_{1})\cup\cdots\cup
B_{2\epsilon}(y_{m})$ are still disjoint by choice of $\epsilon$. Thus we
witness the negation of CONN.

On the other hand, suppose that $X$ fails to satisfy CONN. Take a witness with
$U^{*}=B_{2\epsilon}(x_{1})\cup\cdots\cup B_{2\epsilon}(x_{m})$ and
$V^{*}=B_{2\epsilon}(y_{1})\cup\cdots\cup B_{2\epsilon}(y_{m})$. Then $U^{*}$
and $V^{*}$ are disjoint, since if they intersected then there would be a
special point $z$ in their intersection, as they are open. They also cover
$X$, because
$U=\overline{B}_{\epsilon}(x_{1})\cup\cdots\cup\overline{B}_{\epsilon}(x_{m})$
and
$V=\overline{B}_{\epsilon}(y_{1})\cup\cdots\cup\overline{B}_{\epsilon}(y_{m})$
cover the special points of $X$, and every point of $X$ is within $\epsilon$
of a special point of $X$. Thus $X=U^{*}\sqcup V^{*}$ witness that $X$ is not
connected. ∎

###  3.3 Local connectedness

We expect that, in general, local connectedness might be quite hard to express
(like local compactness, which is not Borel [NS15]). However, we do not have
to find a condition which is both necessary and sufficient for local
connectedness, but just a condition equivalent to local connectedness within
compact spaces. Our condition is as follows.

LC:

For all basic open sets $A=B_{r}(c)$ and $B=B_{2r}(c)$ there are special
points $x_{1},\ldots,x_{n}\in A$ and $\epsilon>0$ such that:

  1. (1)

for all special points $y\in A$, there is some $i$ such that there is an
$\epsilon$-path from $y$ to $x_{i}$ in $B$.

  2. (2)

for distinct $i,j$ there is no $\epsilon$-path from $x_{i}$ to $x_{j}$ in $B$.

  3. (3)

for all $i$ and for all special points $y\in A$ and $\epsilon^{\prime}>0$ if
there is an $\epsilon$-path from $y$ to $x_{i}$ in $B$ then there is an
$\epsilon^{\prime}$-path from $y$ to $x_{i}$ in $B$.

This is $\Pi^{0}_{4}$. First, we show that our condition LC implies local
connectedness via weak local connectedness. We recall the difference between
locally connected and weakly locally connected spaces. Let $X$ be a
topological space. $X$ is locally connected if every point has a neighbourhood
base consisting of open connected sets. $X$ is weakly locally connected if
every point has a neighbourhood base consisting of connected sets, which are
not necessarily open. However these are equivalent; see, e.g., Theorem 27.16
of [Wil70].

######  Lemma 3.2.

If $X=\overline{(M,d)}$ is compact and satisfies LC, then $X$ is weakly
locally connected and hence locally connected.

###### Proof.

Let $x$ be any point, and $U$ an open set containing $x$. We can choose basic
open balls $x\in B_{r}(c)\subseteq\overline{B_{r}(c)}\subseteq
B_{2r}(c)\subseteq\overline{B_{2r}(c)}\subseteq U$. Let $A=B_{r}(c)$ and
$B=B_{2r}(c)$. We will find a connected neighbourhood $D$ of $x$ contained in
$B$. To witness that $D$ is a neighbourhood, we will have $D\cap A$ open.

By LC, there are $x_{1},\ldots,x_{n}\in A$ and $\epsilon$ satisfying (1), (2),
and (3). Define

$D_{i}=\\{y\in\overline{B}:\text{for all $\epsilon^{\prime}>0$ there is an
$\epsilon^{\prime}$-path from $y$ to $x_{i}$ in $B$}\\}.$

Clearly $D_{i}\subseteq\overline{B}\subseteq U$. We prove the following five
claims.

######  Claim 1.

Each $D_{i}$ is closed.

###### Proof.

Given $y\in\overline{B}$, suppose that $y$ is a limit point of $D_{i}$. We
argue that $y\in D_{i}$. Given $\epsilon^{\prime}>0$, there is some point
$y^{\prime}$ of $D_{i}$ with $d(y,y^{\prime})<\epsilon^{\prime}/2$. Then there
is an $\epsilon^{\prime}/2$-path from $y^{\prime}$ to $x_{i}$ in $B$, which,
by replacing $y^{\prime}$ by $y$, is an $\epsilon^{\prime}$-path from $y$ to
$x_{i}$. ∎

######  Claim 2.

Each $D_{i}$ is connected.

###### Proof.

Suppose that $D_{i}$ is not connected. Since $D_{i}$ is closed and compact,
there are disjoint relatively clopen $U,V\subseteq D_{i}$ which partition
$D_{i}$ and points $u\in U$ and $v\in V$. There is some distance
$\epsilon^{\prime}$ between $U$ and $V$. Then there is no
$\epsilon^{\prime}/2$-path from $u$ to $v$ in $B$. ∎

######  Claim 3.

The $D_{i}$ are pairwise disjoint.

###### Proof.

Suppose that there is some $y\in D_{i}\cap D_{j}$. Then there is an
$\epsilon/2$-path from $y$ to $x_{i}$ and an $\epsilon/2$-path from $y$ to
$x_{j}$. Putting them together, and deleting $y$, we get an $\epsilon$-path
from $x_{i}$ to $x_{j}$. This contradicts (2). ∎

######  Claim 4.

Each $D_{i}\cap A$ is open. Indeed, given $y\in D_{i}\cap A$ and $z\in A$ with
$d(y,z)<\epsilon/2$, $z\in D_{i}$.

###### Proof.

Choose some $\epsilon^{\prime}$. We wish to find an $\epsilon^{\prime}$-path
from $z$ to $x_{i}$ in $B$, which would place $z\in D_{i}$. Since $y\in
D_{i}$, we have an $\epsilon$-path from $y$ to $x_{i}$ in $B$. Since the
special points are dense, we may choose a path among them, other than the $y$,
which may be non-special. Call the path-elements $a_{i}$, where $a_{1}=y$ and
$a_{n}=x_{i}$. We have that $y$ is in $B_{\epsilon/2}(z)$ and
$B_{\epsilon}(a_{2})$, and since both of these are open, there is some ball
around $y$ in both of these as well. Choose in particular a special
$y^{\prime}\in B$ within these two and also within $\epsilon^{\prime}$ of $y$.
Choose also a $z^{\prime}\in A$ with $d(y,z^{\prime})<\epsilon/2$ and
$d(z,z^{\prime})<\epsilon^{\prime}$. Then we have an $\epsilon$-path
$z^{\prime},y^{\prime},a_{2},\ldots,a_{n}=x_{i}$ among the special points,
which by (3) gives us an $\epsilon^{\prime}$-path from $z^{\prime}$ to $x_{i}$
in $B$, which by our choice of $z^{\prime}$ gives us an
$\epsilon^{\prime}$-path from $z$ to $x_{i}$ in $B$. Thus $z\in D_{i}$.

Together with the fact that $A$ is open, this shows that $D_{i}\cap A$ is
open. ∎

######  Claim 5.

The $D_{i}$ cover $A$.

###### Proof.

For any $y\in A$, choose some special $y^{\prime}$ with
$d(y,y^{\prime})<\epsilon/2$. By the “Indeed” statement in the above claim, we
have that $y^{\prime}\in D_{i}\Rightarrow y\in D_{i}$ for any $i$. Since
$y^{\prime}$ is special, (1) and (3) give us that $y\in D_{i}$ for some $i$. ∎

Now since the $D_{i}$ cover $A$, $x\in D_{i}$ for some $i$. Then $D_{i}$ is a
connected neighbourhood of $x$ contained in $U$. This completes the proof of
the lemma. ∎

Second, we will show that any compact and locally connected space satisfies
the condition LC.

######  Lemma 3.3.

If $X=\overline{(M,d)}$ is compact and locally connected then it satisfies LC.

###### Proof.

Firstly, take basic open sets with $A=B_{r}(c)$ and $B=B_{2r}(c)$ for some $c$
and $r$. We want to show that there is a finite set of $x_{i}$ together with
an $\epsilon$ that satisfy (1)-(3) of LC.

Let $C_{1},C_{2},\ldots$ be the connected components of $B$, which are open
sets because $B$ is locally connected. Define an equivalence relation $\sim$
on the $C_{i}$ saying that $C_{i}\sim C_{j}$ if for every
$\epsilon^{\prime}>0$ and $x\in C_{i}$ and $y\in C_{j}$, there is an
$\epsilon^{\prime}$-path from $x$ to $y$ in $B$. (Note that this is true of
any two points $x,y$ in the same $C_{i}$.) For each equivalence class, define
an open set $D$ which is the union of the $C_{i}$ in that equivalence class.
Let $D_{1},D_{2},\ldots$ be these sets. Then:

  1. (1)

For each $i$, $\overline{D_{i}}$ is connected. If not, then there is a clopen
splitting of $\overline{D}_{i}$ with points $x$ and $y$ on different halves.
Though the sets in this splitting are only clopen within $\overline{D}_{i}$,
they are closed in $X$. Since there is some distance between these two sets,
there is some $\epsilon^{\prime}$ smaller than this distance such that there
is no $\epsilon^{\prime}$-path from $x$ to $y$ in $B$.

  2. (2)

Each pair of distinct $\overline{D_{i}}$ and $\overline{D_{j}}$ are disjoint.
Otherwise, they would have shared a limit point and would then have been of a
distance less than any $\epsilon$, contradicting that they come from different
equivalence classes.

  3. (3)

The $D_{i}$ are an open cover of $B$, and hence of the compact set
$\overline{A}$. So there are finitely many of them, say $D_{1},\ldots,D_{n}$,
which cover $\overline{A}$.

For each $i,j$, choose $\epsilon_{i,j}>0$ such that there is no
$\epsilon_{i,j}$-path from any point $x\in D_{i}$ to any point $y\in D_{j}$.
Let $\epsilon=\min_{i,j}\epsilon_{i,j}$, and choose a special point $x_{i}$ in
each $D_{i}$.

######  Claim 1.

$\epsilon$ and $x_{1},\ldots,x_{n}$ satisfy (1).

###### Proof.

Since the $D_{i}$ cover $\overline{A}$, they cover $A$, and so since $y\in A$,
$y\in D_{i}$ for some $i$. By the construction of the $D_{i}$ and since
$x_{i}\in D_{i}$, this means that there is an $\epsilon^{\prime}$-path from
$y$ to $x_{i}$ for all $\epsilon^{\prime}>0$, including $\epsilon$. ∎

######  Claim 2.

$\epsilon$ and $x_{1},\ldots,x_{n}$ satisfy (2).

###### Proof.

Suppose that the converse of (2) was true. Then for some $i\neq j$, there
would be an $\epsilon$-path from $x_{i}$ to $x_{j}$ in $B$. But this
contradicts the choice of $\epsilon$ in the construction above. ∎

######  Claim 3.

$\epsilon$ and $x_{1},\ldots,x_{n}$ satisfy (3).

###### Proof.

This follows from the construction of the $D_{i}$ as the union of connected
components which have $\epsilon$-paths between any two elements for all
$\epsilon$. By the choice of $\epsilon$ in the above construction, $y$ having
an $\epsilon$-path to some $x_{i}$ places $y$ in $D_{i}$, which then, for any
$\epsilon^{\prime}$, has an $\epsilon^{\prime}$-path to any other point within
$D_{i}$, including $x_{i}$. ∎

The claims prove the lemma. ∎

If $X$ is locally connected, compact, and connected, then it is path-connected
and locally path-connected [Eng89, 6.3.11].

######  Corollary 3.4.

If $X=\overline{(M,d)}$ satisfies CPCT, CONN, and LC then it is path-connected
and locally path-connected.

###  3.4 Betweenness

In a standard proof of the classification of $[0,1]$ as the unique continuum
with exactly two non-cut points, the separation ordering plays an important
role. We will deal not with the ordering, but with the corresponding notion of
betweenness where one point is between two others if removing it separates the
two points.

######  Definition 3.5.

Define a ternary betweenness relation $\textup{{BTW}}(x,y,z)$ by setting
$\textup{{BTW}}(x,y,z)$ if $x,y,z$ are all distinct and for all $\delta>0$,
there are $\delta^{\prime},\epsilon<\delta$ such that for all $\epsilon$-paths
$x=a_{1},\ldots,a_{n}=z$ from $x$ to $z$, there is some $i$ such that
$d(y,a_{i})\leq\delta^{\prime}$.

If $\textup{{BTW}}(x,y,z)$ holds, then we say that $y$ is between $x$ and $z$.
It is easy to see that there is a symmetry between the first and third
arguments, i.e., $\textup{{BTW}}(x,y,z)$ if and only if
$\textup{{BTW}}(z,y,x)$. Note that the betweenness relation is $\Pi^{0}_{3}$.

######  Definition 3.6.

Let ORD express that for all distinct special points $x,y,z$, either
$\textup{{BTW}}(x,y,z)$, $\textup{{BTW}}(y,x,z)$, or $\textup{{BTW}}(z,x,y)$.

This expresses totality of the betweennes relation, and is $\Pi^{0}_{3}$.

######  Lemma 3.7.

If $X=\overline{(M,d)}$ is homeomorphic to $[0,1]$ then $X$ satisfies ORD.

###### Proof.

Take distinct $x,y,z\in X$. Since $\overline{X}\cong[0,1]$, we can relabel
$x$, $y$, and $z$ so $y$ is between $x$ and $z$. We wish to show that for all
rational $\delta$, there exists $\delta^{\prime}<\delta$ and
$\epsilon<\delta^{\prime}$ such that any $\epsilon$-path from $x$ to $z$ goes
within $\delta^{\prime}$ of $y$. Take any $\delta^{\prime}<\delta$ such that
$x,z\notin B_{\delta^{\prime}}(y)$. Then in $X-B_{\delta^{\prime}}(y)$, the
points $x$ and $z$ are in disjoint closed sets. Let $\epsilon<\delta$ be
smaller than the distance between these two closed sets. Then there is no
$\epsilon$-path from $x$ to $z$ in $X-B_{\delta^{\prime}}(y)$, hence any
$\epsilon$-path from $x$ to $z$ goes within $\delta^{\prime}$ of $y$. ∎

###  3.5 Putting everything together: arcs and cutpoints

So far, none of the properties we have introduced stipulate that a space must
be non-trivial, so we introduce one final property. Recall that a continuum is
said to be non-degenerate if it has more than one point. We say that
$X=\overline{(M,d)}$ satisfies NDEGEN if it has at least two distinct points.
This is $\Sigma^{0}_{2}$.

Combining various previous lemmas, we have the following.

######  Lemma 3.8.

If $X=\overline{(M,d)}$ is homeomorphic to $[0,1]$, then it satisfies NDEGEN,
CPCT, CONN, LC, and ORD.

We must now show the converse.

######  Lemma 3.9.

If $X=\overline{(M,d)}$ satisfies NDEGEN, CPCT, CONN, LC, and ORD, then it is
homeomorphic to $[0,1]$.

###### Proof.

Since $X=\overline{(M,d)}$ satisfies CPCT, CONN, and LC we know that it is
compact, connected, locally connected, path connected, and locally path
connected. In particular, $X$ is a continuum. We will show that $X$ has
exactly two non-cut points, which implies that is is homeomorphic to $[0,1]$.
(See [Nad92, Theorem 6.17].)

######  Claim 1.

Suppose that there is an arc from $x$ to $z$ that does not pass through $y$.
Then $\neg\textup{{BTW}}(x,y,z)$

###### Proof.

Let $f$ be the arc from $x$ to $z$ and suppose that $y$ is not on this arc.
There is some distance $d$ between $y$ and $f$. For sufficiently small
$\epsilon$, an $\epsilon$-path from $x$ to $z$ approximating $f$ avoids
$B_{d/2}(y)$. Thus $\neg\textup{{BTW}}(x,y,z)$. ∎

######  Claim 2.

For $x,y\in M$, there is a unique arc from $x$ to $y$.

###### Proof.

Much of the technical difficulty with proving this claim is the fact that ORD
only tells us about special points, but to prove the claim we need to argue
for all points $x,y$ not necessarily special.

The existence of such an arc follows from path-connectedness (which is
equivalent to arc-connectedness in Hausdorff spaces). To prove uniqueness,
suppose we have two distinct arcs, $f$ and $g$, from $x$ to $y$. Take a point
$z$ on $f$ that is not on $g$. Take a path-connected set around $z$ that is
disjoint from $g$, say $U$, and take a special point $z^{\prime}\in U$. Then
there is an arc $h$ from $z$ to $z^{\prime}$ within $U$, and thus disjoint
from $g$. We may assume by moving $z^{\prime}$ and shortening $h$ that $z$ is
the only point on both $h$ and $f$. Break up $f$ at the point $z$ into two
segments $f_{1}$ from $x$ to $z$ and $f_{2}$ from $z$ to $y$. Choose also
special points $x^{\prime}$ and $y^{\prime}$ close to $x$ and $y$,
respectively, with arcs from $x$ to $x^{\prime}$ near $x$ and from $y$ to
$y^{\prime}$ near $y$. The picture is something like this:

$x$$x^{\prime}$$y$$y^{\prime}$$z$$z^{\prime}$$f_{2}$$f_{1}$$g$$U$

Thus we have arcs containing $x^{\prime}$ and $y^{\prime}$ but not
$z^{\prime}$, $y^{\prime}$ and $z^{\prime}$ but not $x^{\prime}$, and
$x^{\prime}$ and $z^{\prime}$ but not $y^{\prime}$. By Claim 1 we have that
$\neg\textup{{BTW}}(y^{\prime},z^{\prime},x^{\prime})$,
$\neg\textup{{BTW}}(x^{\prime},y^{\prime},z^{\prime})$, and
$\neg\textup{{BTW}}(z^{\prime},x^{\prime},y^{\prime})$. This contradicts ORD
and proves the lemma. ∎

######  Claim 3.

There are at most two non-cut points.

###### Proof.

Suppose that there were three non-cut points $a,b,c$. Without loss of
generality, we may assume that the arc $f$ from $a$ to $b$ does not contain
$c$. (If it did, then take the shorter arc from $a$ to $c$ and swap $b$ and
$c$.) We may also assume that the arc $g$ from $b$ to $c$ does not contain
$a$, as if it did, then we could take the arc from $a$ to $c$ and swap $a$ and
$b$.

Now we argue that $b$ is in fact a cut point. Suppose not; then $X-\\{b\\}$ is
connected, and since it is still locally path connected, it is path connected.
Thus there is an arc from $a$ to $c$ avoiding $b$, a contradiction. ∎

It is a theorem of Moore that any non-degenerate continuum must have at least
two non-cut points (see [Nad92, Theorem 6.6]). Thus we have exactly two non-
cut points. This proves the lemma.∎

Putting together all of the lemmas, we have one half of Theorem 1.1.

######  Theorem 3.10.

$X=\overline{(M,d)}$ satisfies NDEGEN, CPCT, CONN, LC, and ORD if and only if
$X$ is homeomorphic to $[0,1]$. In particular, $\operatorname{HCopies}([0,1])$
is $\Pi^{0}_{4}$.

Note that of all of these conditions, it is only LC which is $\Pi^{0}_{4}$.
Thus, in some sense, the hard part of saying that $X=\overline{(M,d)}$ is
homeomorphic to $\operatorname{HCopies}([0,1])$ is in saying that $X$ is
locally connected.

######  Corollary 3.11.

$\operatorname{HCopies}([0,1])$ is $\Pi^{0}_{3}$ within the locally connected
spaces.

In Theorem 4.1 to follow, we will show that $\operatorname{HCopies}([0,1])$ is
$\Pi^{0}_{3}$-complete within the locally connected spaces.

##  4 A warmup: $\mathbf{\Pi}^{0}_{3}$-hardness within the compact, connected,
locally connected spaces

We will prove the following theorem as a warmup.

######  Theorem 4.1.

$\operatorname{HCopies}([0,1])$ is $\mathbf{\Pi}^{0}_{3}$-Wadge-complete
within the compact, connected, locally connected spaces.

###### Proof.

It follows from Corollary 3.11 that $\operatorname{HCopies}([0,1])$ is
$\Pi^{0}_{3}$ within the compact locally connected spaces. So it remains to
prove a hardness result. Consider the $\mathbf{\Pi}^{0}_{3}$-Wadge-complete
set $S=\\{W\subseteq\omega\times\omega\;\mid\;\forall n\;\;|W_{n}|<\infty\\}$
where $W_{n}=\\{(n,m)\mid(n,m)\in W\\}$ is the $n$th column of $W$. Given $W$
we must build a compact, connected, locally connected space $X=X_{W}$ such
that $W\in S$ if and only if $X_{W}\cong[0,1]$. Fix $W$ for which we must
define $X=X_{W}$. We will work somewhat dynamically by letting
$W_{n}[i]=W_{n}\\!\\!\upharpoonright i=\\{m\in W_{n}\mid m<i\\}$. Note that
$W_{n}[0]=\varnothing$.

We construct $X$ inside the unit square $[0,1]\times[0,1]$. We will always
include in $X$ the line $[0,1]\times\\{0\\}=\\{(x,0):x\in[0,1]\\}$. We will
also include in $X$ the graph $\Gamma_{f}$ of a continuous function
$f\colon[0,1]\to[0,1]$. Thus

$X=\Gamma_{f}\;\cup\;[0,1]\times\\{0\\}.$

The function $f$ will have the property that $f(0)=0$ and $f(1)=1$. Thus $X$
will always be connected, locally connected, and compact. $X$ will be
homeomorphic to $[0,1]$ if and only if $f(x)>0$ for all $x>0$.

It just remains to define $f$. Pick some sequence of elements, alternating
between $a_{i}$ and $b_{i}$,

$0<\cdots<b_{2}<a_{2}<b_{1}<a_{1}<b_{0}<a_{0}=1$

converging to $0$. In each interval $[a_{i+1},a_{i}]$ choose an increasing
sequence

$a_{i+1}=\ell_{i}^{0}<\ell_{i}^{1}<\ell_{i}^{2}<\cdots<b_{i}$

converging to $b_{i}$, and a decreasing sequence

$b_{i}<\cdots<r_{i}^{2}<r_{i}^{1}<r_{i}^{0}=a_{i}$

converging to $b_{i}$. Thus we have divided the interval $[0,1]$ into
intervals $[a_{i+1},a_{i}]$, and we have divided each of these intervals into
a left interval $[a_{i+1},b_{i}]$ and a right interval $[b_{i},a_{i}]$. The
left and right intervals are in turn divided into intervals
$[\ell_{i}^{j},\ell_{i}^{j+1}]$ on the left and $[r_{i}^{j+1},r_{i}^{j}]$ on
the right:

$a_{i+1}=\ell_{i}^{0}<\ell_{i}^{1}<\ell_{i}^{2}<\cdots<b_{i}<\cdots<r_{i}^{2}<r_{i}^{1}<r_{i}^{0}=a_{i}.$

Every $t\in[0,1]$ is included in exactly one interval
$[\ell_{i}^{j},\ell_{i}^{j+1}]$ or $[r_{i}^{j+1},r_{i}^{j}]$, except the
endpoints which are included in two of these intervals, and the points $0$ and
$b_{i}$ are not included in any of these intervals.

The idea is that $X$ will look somewhat like the following:

The straight horizontal line is $[0,1]\times\\{0\\}$ and the squiggly line is
the graph of $f$. Each of the downward spikes in the graph of $f$ happens
between some $a_{i}$ and $a_{i+1}$, with the points at $b_{i}$. The larger
$|W_{i}|$, the lower the spike gets, so that if $|W_{i}|=\infty$, then the
spike touches the horizontal line. This makes the space $X$ not homeomorphic
to $[0,1]$. If no spike touches the horizontal line, then $X$ will be
homeomorphic to $[0,1]$.

Given $t\in[0,1]$, define $f(t)$ as follows:

  1. (1)

If $t=0$, then $f(t)=f(0)=0$.

  2. (2)

If $t=b_{i}$, then $f(t)=f(b_{i})=t2^{-|W_{i}|}=b_{i}2^{-|W_{i}|}$ (which is
equal to $0$ if $W_{i}$ is infinite).

  3. (3)

If $t\in[\ell_{i}^{j},\ell_{i}^{j+1}]$ then

$f(t)=t\left[\frac{t-\ell_{i}^{j}}{\ell_{i}^{j+1}-\ell_{i}^{j}}\left(2^{-|W_{i}[j+1]|}-2^{-|W_{i}[j]|}\right)+2^{-|W_{i}[j]|}\right].$

So $f(\ell_{i}^{j})=t2^{-|W_{i}[j]|}=\ell_{i}^{j}2^{-|W_{i}[j]|}$ and
$f(\ell_{i}^{j+1})=t2^{-|W_{i}[j+1]|}=\ell_{i}^{j+1}2^{-|W_{i}[j+1]|}$. In
between, $f(t)$ is $t$ times a linear interpolation.

  4. (4)

If $t\in[r_{i}^{j+1},r_{i}^{j}]$ then

$f(t)=t\left[\frac{t-r_{i}^{j+1}}{r_{i}^{j}-r_{i}^{j+1}}\left(2^{-|W_{i}[j]|}-2^{-|W_{i}[j+1]|}\right)+2^{-|W_{i}[j+1]|}\right].$

This is essentially (3) in reverse. We have
$f(r_{i}^{j})=t2^{-|W_{i}[j]|}=r_{i}^{j}2^{-|W_{i}[j]|}$ and
$f(r_{i}^{j+1})=t2^{-|W_{i}[j+1]|}=r_{i}^{j+1}2^{-|W_{i}[j+1]|}$.

We must check that $f(t)$ is well-defined; that is, for any point which is in
two of these intervals, the definitions of $f$ agree on those common points.
This is easy to see for the points $\ell_{i}^{j}$ with $j>0$ which are parts
of the intervals $[\ell_{i}^{j-1},\ell_{i}^{j}]$ and
$[\ell_{i}^{j},\ell_{i}^{j+1}]$. Similarly, it is easy to see for the points
$r_{i}^{j}$ with $j>0$. The non-trivial case is the points
$a_{i}=\ell_{i}^{0}=r_{i-1}^{0}$. But in this case,
$|W_{i}[0]|=|W_{i-1}[0]|=0$ and so $f(a_{i})=a_{i}$.

We also need to check that the function is continuous, which means checking
that (1) and (2) agree with the limiting behaviour of $f$. For (1), we have
$f(t)\leq t$ and so $\lim_{t\to 0}f(t)=0$. For (2), for
$t\in[\ell_{i}^{j},\ell_{i}^{j+1}]$ or $t\in[r_{i}^{j+1},r_{i}^{j}]$ we have
$t2^{-|W_{i}[j+1]|}\leq f(t)\leq t2^{-|W_{i}[j]|}$. Thus
$f(b_{i})=b_{i}2^{-|W_{i}|}$ is the limit as $t\to b_{i}$.

We also need to be able to produce $X=X_{W}$ computably and uniformly from
$W$. Note that we cannot compute $|W_{i}|$ uniformly from $W$, so we cannot
compute $f(b_{i})$. But we can compute $f$ on all other inputs, and this gives
a dense set to produce a presentation of $X_{W}$.

Finally, note that if each $W_{i}$ is finite, then $f(t)>0$ when $t>0$, but if
$W_{i}$ is infinite, then $f(b_{i})=0$. As discussed before, $X_{W}$ is
homeomorphic to $[0,1]$ if and only if $f(t)>0$ for all $t>0$. This completes
the argument. ∎

##  5 $\mathbf{\Pi}^{0}_{4}$-hardness for $[0,1]$

We now turn to proving that $\operatorname{HCopies}([0,1])$ is
$\mathbf{\Pi}^{0}_{4}$-complete.

###  5.1 The main idea

We begin by showing that $\operatorname{HCopies}([0,1])$ is
$\boldsymbol{\Sigma}^{0}_{3}$-complete. By attaching infinitely many instances
of this construction end-to-end, we will get a $\mathbf{\Pi}^{0}_{4}$
completeness result.

Looking at the characterization above, we see that the $\mathbf{\Pi}^{0}_{4}$
complexity comes from the local connectedness, and so we should try to make a
line in the $\boldsymbol{\Sigma}^{0}_{3}$ outcome, and a non-locally-connected
space in the $\mathbf{\Pi}^{0}_{3}$ outcome. The non-locally-connected spaces
we build will be based off of the following:

$\vdots$$\vdots$$\vdots$$\vdots$$\cdots$

The space is built up from infinitely many tendrils. Above we drew each
tendril with right-angled turns, like the following:

In the actual construction, we will have the tendrils consist of two straight
lines, as follows:

But for clarity of the pictures it will be easier to draw the tendrils using
horizontal and vertical lines.

Each of the tendrils originates from somewhere on the left half, and goes all
the way to the rightmost side. The rightmost point has no small connected
neighbourhoods, as each neighbourhood contains some tendril coming from the
left side.

At each stage $s$, we will have a “target” topological space which we are
building towards. This target space will be a non-locally-connected space
looking something like the space above with many tendrils. However, at each
stage of the construction we will have only actually put finitely many points
from the target space into the metric space being constructed. For example,
suppose that the target space has a tendril

At a finite stage, we might have only placed points configured as

We might then decide to collapse the tendril, changing the target space to

Since we are already committed to the finitely many points we have placed, and
what we have said about the distances between them, they must all stay in the
same positions in the new target space. But by changing the target space, we
can incorporate the tendril into a (new) homeomorphic copy of the unit
interval.

We will then introduce new tendrils, which may themselves get collapsed, after
which we introduce more new tendrils, and so on. If we collapse all but
finitely many tendrils infinitely often, then in the limit our space will be
locally connected and homeomorphic to the unit interval. Otherwise, there will
be infinitely many tendrils and our space will not be locally connected.

One remaining technical point is to explain why, in the case that we collapse
all but finitely many tendrils, we end up with a space homeomorphic to the
unit interval. As illustrated above, we end up with a sawtooth function even
after collapsing a single tendril. We need to ensure that in the process of
introducing and collapsing new tendrils, this sawtooth is not too wild, and
instead comes to a nice limit. To do this, when we introduce new tendrils,
they will closely follow the existing sawtooth space. In this way, even though
the sawtooth gets more and more bumpy, it comes to a limit.

###  5.2 Construction for $\boldsymbol{\Sigma}^{0}_{3}$-completeness

Consider the $\Sigma^{0}_{3}$ set

$A=\\{W\subseteq\omega\times\omega\;\mid\;\exists n\;\;|W_{n}|=\infty\\}$

where, as before, $W_{n}$ is the $n$th column of $W$.

Given $W$, we must build a Polish space $X_{W}=\overline{M_{W}}$ such that
$W\in A$ if and only if $X_{W}$ is homeomorphic to $[0,1]$. The main idea is
that whenever an element enters $W_{n}$ we will collapse the $n$th tendril
along with every greater tendril. Thus if for every $n$ the set $W_{n}$ is
finite then for each $n$ there is some point after which we stop collapsing
the $n$th tendril. In that case, we have infinitely many tendrils, and $X_{m}$
is not homeomorphic to $[0,1]$. Otherwise, if $W_{n}$ is infinite for some
$n$, then we collapse the $n$th tendril and every greater tendril infinitely
often, and in the end have only finitely many tendrils. So $X_{W}$ is
homeomorphic to $[0,1]$.

For the following, fix the $W$ for which we will construct $X_{W}$. For
simplicity, we suppress the subscript $W$, writing, e.g., $X$ for $X_{W}$ and
$M$ for $M_{W}$. The construction will be a dynamic construction, and as
before we use $W_{n}[s]=W_{n}\\!\\!\upharpoonright s$ for the restriction of
$W_{n}$ to elements $<s$. We can view $W_{n}[s]$ as the portion of $W_{n}$
that we have seen by stage $s$, just like if we were doing a many-one
reduction in computability theory. (For the computability theorist, one can
think of $W_{n}$ as the set of stages at which a new element enters a c.e.
set.)

We will work inside $\mathbb{R}^{2}$ and indeed inside $[-1,1]\times[0,1]$; we
call these the $x$ and $y$-coordinates, respectively. To simplify a later part
of the construction, we will ensure that there is only one special point for
each particular value of the $x$ coordinate. Each special point of $M$ will be
a rational tuple.

####  5.2.1 The data

At each stage $s$, we will have the following data:

  1. (1)

The main line: A function $\mathfrak{m}:[-1,1]\to[0,1]$ with
$\mathfrak{m}(-1)=0$ and $\mathfrak{m}(1)=0$.

  2. (2)

Infinitely many tendrils where the $n$th tendril consists of the data:

    1. (a)

a left attachment point $\ell_{n}$, given as an $x$-value;

    2. (b)

a right attachment point $r_{n}$, given as an $x$-value;

    3. (c)

a top arc $\mathfrak{t}_{n}\colon[\ell_{n},1]\to[0,1]$;

    4. (d)

a bottom arc $\mathfrak{b}_{n}\colon[r_{n},1]\to[0,1]$.

The functions $\mathfrak{m}$, $\mathfrak{t}_{n}$, and $\mathfrak{b}_{n}$
should be thought of as representing the points on their graphs, e.g., the
actual points in $\mathbb{R}^{2}$ making up the $n$th tendril will be the
union of the graphs of $\mathfrak{t}_{n}$ and $\mathfrak{b}_{n}$. These
functions will change stage-by-stage, and for the value at stage $s$ we will
write $\mathfrak{m}[s]$, $\mathfrak{t}_{n}[s]$, and $\mathfrak{b}_{n}[s]$.
When the stage is understood, this may be omitted.

The data should satisfy the following conditions:

  1. (1)

Each tendril is attached to the main line on the left side, and is above the
main line. For the $n$th tendril, we have:

    1. (a)

$\ell_{n}<r_{n}<0$: the left attachment point is to the left of the right
attachment point, and both are to the left of the midpoint.

    2. (b)

$\mathfrak{m}(\ell_{n})=\mathfrak{t}_{n}(\ell_{n})$: the top arc of the
tendril attaches to the main line at the left attachment point.

    3. (c)

$\mathfrak{m}(r_{n})=\mathfrak{b}_{n}(r_{n})$: the bottom arc of the tendril
attaches to the main line at the right attachment point.

    4. (d)

$\mathfrak{t}_{n}(1)=\mathfrak{b}_{n}(1)$: the top and bottom arcs of the
tendril meet at $x=1$.

    5. (e)

For $x\in(\ell_{n},r_{n}]$ we have $\mathfrak{m}(x)<\mathfrak{t}_{n}(x)$, and
for $x\in(r_{n},1)$ we have
$\mathfrak{m}(x)<\mathfrak{b}_{n}(x)<\mathfrak{t}_{n}(x)$.

  2. (2)

The tendrils are attached to the main line in order from left to right, and
are non-intersecting:

    1. (a)

$\ell_{1}<r_{1}<\ell_{2}<r_{2}<\cdots$.

    2. (b)

for $m<n$ and values where they are both defined, i.e., $x\in(\ell_{n},1)$, we
have $\mathfrak{t}_{n}(x)<\mathfrak{b}_{m}(x)$.

Given the above data, the target space is the closed subspace which consists
of the main line with the portion between $\ell_{n}$ and $r_{n}$ replaced by
the $n$th tendril. That is, the following points:

  1. (1)

the points $(x,\mathfrak{m}(x))$ for $x\notin(\ell_{n},r_{n})$ for any $n$.

  2. (2)

for each $n$,

    1. (a)

the points $(x,\mathfrak{t}_{n}(x))$ for $x\in[\ell_{n},1]$;

    2. (b)

the points $(x,\mathfrak{b}_{n}(x))$ for $x\in[r_{n},1]$.

Recall that the target space is not homeomorphic to $[0,1]$, as it is the
target in $\Pi^{0}_{3}$ outcome.

####  5.2.2 The construction at each stage

Recall that we are building a countable metric space $M=M_{W}$ as a subspace
of $\mathbb{R}^{2}$ with the induced metric. We begin at stage $0$ with $M$
being almost empty, and at each stage we may put some points into $M$. Then
the topological space $X=X_{W}$ will be the completion of $M$.

Let $\\{2,p_{n,c},q_{n,c}:n,c\in\mathbb{N}\\}$ be a computable listing of all
the primes with no repetitions.

Construction at stage $0$: We begin the construction at stage $0$ with the
main line $\mathfrak{m}(x)=0$ and the tendrils given by
$\ell_{n}=-\frac{1}{2}-\frac{1}{2n}$, $r_{n}=-\frac{1}{2}-\frac{1}{2n+1}$, and

$\displaystyle\mathfrak{t}_{n}:[\ell_{n},1]$ $\displaystyle\to[0,1]$
$\displaystyle x$
$\displaystyle\mapsto\mathfrak{m}(x)+\frac{1}{2^{n}}\left(\frac{x-\ell_{n}}{1-\ell_{n}}\right)$

and

$\displaystyle\mathfrak{b}_{n}:[r_{n},1]$ $\displaystyle\to[0,1]$
$\displaystyle x$
$\displaystyle\mapsto\mathfrak{m}(x)+\frac{1}{2^{n}}\left(\frac{x-r_{n}}{1-r_{n}}\right)$

Put $(-1,0)$ and $(1,0)$ into the metric space $M=M_{W}$.

Construction at stage $s+1$: Suppose that at stage $s+1$, a new element enters
$W_{n}[s+1]$ that was not in $W_{n}[s]$. Then:

  * •

First, we collapse all tendrils numbered $n$ onward. We must incorporate the
special points on these tendrils that we have already included in $M$ into the
main line. We define the new main line $\mathfrak{m}[s+1]$ as follows. For
$x\in(-1,\ell_{n}]$, we keep the main line the same:
$\mathfrak{m}[s+1](x):=\mathfrak{m}[s](x)$. Now consider all of the special
points $(x,y)$ which we have already put in $M$ which are either on the main
line (with $x\in[\ell_{n},1]$) or on the $n$th tendril or a greater one.
Namely, these are all of the points $(x,y)$ with $x\in[\ell_{n},1]$ and
$y<\mathfrak{b}_{n-1}[s](x)$. Order these points from left to right as
$(x_{1},y_{1}),(x_{2},y_{2}),\ldots,(x_{k},y_{k})$. Recall that there is no
repetition among the $x_{i}$, as no two special points share the same
$x$-coordinate. Now on $[\ell_{n},1]$ let $\mathfrak{l}$ be the piecewise
linear function which linearly interpolates between the points

$(\ell_{n},\mathfrak{m}[s](\ell_{n})),(x_{1},y_{1}),(x_{2},y_{2}),\ldots,(x_{k},y_{k}),(1,0).$

Let $\mathfrak{m}[s+1](x)=\max(\mathfrak{m}[s](x),\mathfrak{l}(x))$. (As we
will see later, for each $i$, $y_{i}\geq\mathfrak{m}[s](x_{i})$, so that each
of the points $(x_{i},y_{i})$ is on the graph of $\mathfrak{m}[s+1]$. Also,
$(-1,0)$ and $(1,0)$ were put into $M$ at stage $0$, and so we have
$\mathfrak{m}[s+1](-1)=\mathfrak{m}[s+1](1)=0$.)

  * •

Second, we must create new tendrils. Choose new values
$\ell_{n},r_{n},\ell_{n+1},r_{n+1},\ldots$ such that
$r_{n-1}<-\frac{1}{2s+2}<\ell_{n}<r_{n}<\ell_{n+1}<r_{n+1}<\cdots<-\frac{1}{2s+4}<0$
and such that there are no special points between $\ell_{n}$ and $r_{n}$,
between $\ell_{n+1}$ and $r_{n+1}$, and so on. Let

$g:=\sup_{x\in[\ell_{n},1]}\mathfrak{b}_{n-1}(x)-\mathfrak{m}[s+1](x).$

Think of $g>0$ as the gap between the main line $\mathfrak{m}$ and the
remaining tendrils above $\mathfrak{m}$ (which have not been collapsed). Then
define

$\displaystyle\mathfrak{t}_{n}:[\ell_{n},1]$ $\displaystyle\to[0,1]$
$\displaystyle x$
$\displaystyle\mapsto\mathfrak{m}[s+1](x)+\frac{g}{2^{n+s}}\left(\frac{x-\ell_{n}}{1+\ell_{n}}\right)$

and

$\displaystyle\mathfrak{b}_{n}:[r_{n},1]$ $\displaystyle\to[0,1]$
$\displaystyle x$
$\displaystyle\mapsto\mathfrak{m}[s+1](x)+\frac{g}{2^{n+s}}\left(\frac{x-r_{n}}{1+r_{n}}\right)$

It is somewhat tedious but not hard to see that the new main line and tendrils
satisfy all of the conditions required.

If no element enters any $W_{n}$ at stage $s+1$, then we leave everything the
same (e.g., $\mathfrak{m}[s+1]=\mathfrak{m}[s]$).

We may now have a new target space (though if no elements entered any $W_{n}$,
then our target space remains the same). We must add new points to our space.
For each $n$, let $c_{n}$ be the number of times that the $n$th tendril has
been collapsed. Add to $M$ the following points from the target space:

  1. (1)

for each $x\in(-1,1)$, $x\neq 0$, of the form $t/2^{s}$ for any
$t\in\mathbb{N}$, and such that $x\notin(\ell_{n},r_{n})$, add
$(x,\mathfrak{m}(x))$ to $M$.

  2. (2)

for each $x\in[t_{n},1)$, $x\neq 0$, of the form $t/p_{n,c_{n}}^{s}$ for any
$t\in\mathbb{N}$, add $(x,\mathfrak{t}_{n}(x))$ to $M$.

  3. (3)

for each $x\in[b_{n},1)$, $x\neq 0$, of the form $t/q_{n,c_{n}}^{s}$ for any
$t\in\mathbb{N}$, add $(x,\mathfrak{b}_{n}(x))$ to $M$.

Note that $p_{n,c_{n}}$ and $q_{n,c_{n}}$ are unique primes not only for the
$n$th tendril, but unique for this instance of the $n$th tendril (i.e.,
different than the primes used for previous instances of the $n$th tendril
that have now been collapsed). Also, for each point $(x,y)$ that we have added
to $M$, $y\geq\mathfrak{m}[s+1](x)$. Thus the main line is a lower bound on
all of the points in $M$.

This completes the construction.

###  5.3 Verification

Fix $W$ for which we have constructed $X=W$. We begin with a list of
properties of the construction.

######  Remark 5.1.

We have the following:

  1. (1)

If at stage $s$ we collapse the $n$th, $n+1$st, …, tendril, then
$\mathfrak{t}_{n}[s-1]\geq\mathfrak{l}$ where $\mathfrak{l}$ is as defined at
stage $s$.

Note that $\mathfrak{t}_{n}[s-1]$ is a line, and $\mathfrak{l}$ is a linear
interpolation of points all of which are on or below $\mathfrak{t}_{n}[s-1]$.

  2. (2)

At any stage $s$, the first tendril is above the second tendril, the second is
above the third, and so on. All of the tendrils are above the main line.

Both points are easy to see by following the construction. The point of the
gap $g$ is to ensure that this is the case; also, note that the line
$\mathfrak{l}$ is under any of the tendrils that are not collapsed.

  3. (3)

If a special point $p$ is put in $M$ at stage $s$, then $p$ is above the main
line $\mathfrak{m}[s]$.

This is because all of the tendrils are above the main line, and the points
that are added to $M$ are either on a collapsed tendril or on the main line.

  4. (4)

For each fixed $x$, the sequence $\mathfrak{m}[s](x)$ is increasing.

We define $\mathfrak{m}[s+1](x)=\max(\mathfrak{m}[s](x),\mathfrak{l}(x))$ or
$\mathfrak{m}[s+1](x)=\mathfrak{m}[s](x)$ depending on what happens at stage
$s+1$.

  5. (5)

If, at a stage $s$, a special point $p=(x,y)$ is in $M$ and is on the main
line, then it is on the main line at any greater stage. Moreover, from stage
$s$ on, $x$ is never between any pair of left and right endpoints $\ell_{i}$
and $r_{i}$.

At stage $s$, suppose that we collapse the $n$th, $n+1$st, …, tendrils. If
$p=(x,y)$ with $x\in[-1,\ell_{n}]$ then
$\mathfrak{m}[s+1](x)=\mathfrak{m}[s](x)=y$ and so $p$ is still on the main
line. Otherwise, if $x\in[\ell_{n},1]$, then $p$ is on the line $\mathfrak{l}$
defined at stage $s$, and
$\mathfrak{m}[s+1](x)=\max(\mathfrak{m}[s](x),\mathfrak{l}(x))$ but
$\mathfrak{m}[s](x)=\mathfrak{l}(x)=y$.

For the moreover clause, note that if $p$ is added directly onto the main
line, then it is not between any pair of left and right endpoints. If on the
other hand $p$ is put into $M$ as part of a tendril and becomes part of the
main line due to collapse, then any pair of left and right endpoints
containing it are redefined. Finally, in either case, when any pair of left
and right endpoints are redefined at any later stage, they do not contain
between them any special points of $M$.

  6. (6)

If the $n$th tendril was first defined at stage $t$, and was never collapsed
between stage $t$ and stage $s>t$, then
$\mathfrak{t}_{n}(x)-\mathfrak{m}[s](x)\leq 2^{-t}$.

At stage $t$, we would have defined
$\mathfrak{t}_{n}(x)=\mathfrak{t}_{n}[t](x)$ such that
$\mathfrak{t}_{n}(x)-\mathfrak{m}[t](x)\leq 2^{-t}$. But
$\mathfrak{m}[t](x)\leq\mathfrak{m}[s](x)\leq\mathfrak{t}_{n}(x)$.

  7. (7)

If, at stage $s$, a new element enters $W_{n}$ for the $k$th time, then
$\|\mathfrak{m}[s]-\mathfrak{m}[s+1]\|\leq 2^{-n-k}$.

  8. (8)

If $\ell_{n}$ or $r_{n}$ was defined at stage $s$, then
$\ell_{n}<r_{n}<-\frac{1}{2s+2}$.

This is because at stage $0$ we initially define $\ell_{n}<r_{n}<-\frac{1}{2}$
and if, at stage $s+1$ we redefine them, it is with
$-\frac{1}{2s+2}<\ell_{n}<r_{n}<\ell_{n+1}<r_{n+1}<\cdots<-\frac{1}{2s+4}<0$.

First, we will argue that the main line comes to a limit
$\mathfrak{m}_{\infty}$.

######  Lemma 5.2.

The limit $\mathfrak{m}_{\infty}(x)=\lim_{s\to\infty}\mathfrak{m}[s](x)$
exists, and $\mathfrak{m}_{\infty}\colon[-1,1]\to[0,1]$ is continuous.
$\mathfrak{m}_{\infty}(-1)=\mathfrak{m}_{\infty}(1)=0$.

###### Proof.

For a fixed $x$, the sequence $(\mathfrak{m}[s](x))_{s\in\mathbb{N}}$ is
increasing and bounded above (by $1$), and hence converges. Thus
$\mathfrak{m}[s]$ converges pointwise to a function
$\mathfrak{m}_{\infty}\colon[-1,1]\to[0,1]$. Moreover, by Remark 5.1 (7), for
each $k$ there can only be finitely many stages $s$ with
$\|\mathfrak{m}[s]-\mathfrak{m}[s+1]\|>2^{-k}$. Thus the sequence of functions
$(\mathfrak{m}[s])$ is a Cauchy sequence under the uniform metric, and so
converges uniformly to $\mathfrak{m}_{\infty}$ which is therefore a continuous
function. The final claim of the lemma follows because for each $s$,
$\mathfrak{m}[s](-1)=\mathfrak{m}[s](1)=0$. ∎

####  5.3.1 The case $W\in A$

Suppose that $W\in A$. We must argue that $X_{W}$ is homeomorphic to $[0,1]$.
Let $n$ be least such that $W_{n}$ is infinite, and let $s^{*}$ be a stage
after which no element ever enters $W_{1},\ldots,W_{n-1}$.

For $i=1,\ldots,n-1$, let $\ell_{i},r_{i},\mathfrak{b}_{i},\mathfrak{t}_{i}$
be the data of the $i$th tendril at stage $s^{*}$. The tendril is never
collapsed at any later stage, so this is the final data associated with the
tendril. Let $Y$ consist of:

  1. (1)

the points $(x,\mathfrak{m}_{\infty}(x))$ for $x\in[-1,\ell_{1}]$;

  2. (2)

for each $1\leq i<n-1$,

    1. (a)

the points $(x,\mathfrak{m}_{\infty}(x))$ for $x\in[r_{i},\ell_{i+1}]$;

  3. (3)

the points $(x,\mathfrak{m}_{\infty}(x))$ for $x\in[r_{n-1},1]$;

  4. (4)

for each $i<n$,

    1. (a)

the points $(x,\mathfrak{t}_{i}(x))$ for $x\in[\ell_{i},1]$;

    2. (b)

the points $(x,\mathfrak{b}_{i}(x))$ for $x\in[r_{i},1]$.

Each of these is the graph of a continuous function over a closed interval and
hence homeomorphic to $[0,1]$. Thus $Y$ is the union of finitely many closed
arcs, which are connected end-to-end and otherwise disjoint, and so $Y$ is
homeomorphic to $[0,1]$. Now we must argue that $X=X_{W}$ is equal to $Y$,
that is, that $M=M_{W}$ is a subset of $Y$ and that $\overline{M}=Y$.

First we argue that $M\subseteq Y$. Suppose that a point $(x,y)$ is put into
$M$ at stage $s$. There are three possible ways that it might be put into $M$.
If it is put into the main line, then by Remark 5.1 (5) $(x,y)$ will be part
of the main line at every later stage (and will never be between any pair of
left and right endpoints). If $(x,y)$ is added to $M$ as part of the $i$th
tendril, then either that tendril is never later collapsed, in which case it
remains in that tendril, or the tendril is at some later point collapsed to
the main line, and $(x,y)$ remains in the main line at all stages after that.

Now we will argue that the closure $X=\overline{M}$ covers all of $Y$. Given
$(x,y)\in Y$ and $\epsilon>0$, we break into cases. Suppose that $(x,y)$ is a
point of the form $(x,\mathfrak{m}_{\infty}(x))$ on the main line, with
$x\notin(\ell_{i},r_{i})$ for any $i\leq n$. Choose $\delta>0$ and a stage $s$
such that

  1. (1)

if $|x-x^{\prime}|<\delta$ then
$|\mathfrak{m}_{\infty}(x)-\mathfrak{m}_{\infty}(x^{\prime})|<\epsilon/3$;

  2. (2)

at stage $s$, either $(x-\delta,x]$ or $[x,x+\delta)$ is disjoint from any
interval $(\ell_{i},r_{i})$;

  3. (3)

$2^{-s}<\delta$;

  4. (4)

$2^{-s}<\epsilon/3$;

  5. (5)

$\|\mathfrak{m}_{\infty}-\mathfrak{m}[s]\|<\epsilon/3$.

It takes some argument to see that we can do this. We can choose $\delta$
satisfying (1) since $\mathfrak{m}_{\infty}$ is uniformly continuous (as it is
continuous on a compact set). Moreover, (1) is maintained if we shrink
$\delta$. We obtain (2) as follows. If $x\geq 0$, then (2) is clear for
$[x,x+\delta)$. Otherwise, if $x<0$, we argue as follows. By shrinking
$\delta$, we may assume that for any $s\geq s^{*}$ (2) holds restricted solely
to $i\leq n$. By taking $s$ sufficiently large that $x+\delta<-\frac{1}{2s}$,
and moreover taking $s$ to be a sufficiently large stage such that an element
enters $W_{n}$ at stage $s$, by Remark 5.1 (8) at stage $s$ we have that
$x+\delta<-\frac{1}{2s}<\ell_{n}<r_{n}<\cdots$ and moreover this remains true
at any stage $s^{\prime}\geq s$. Increasing $s$ further, we can obtain (3),
(4), and (5).

Then at each stage $s$, there is a point of the form $x^{\prime}=t/2^{s}$
within $2^{-s}$ of $x$ and not in any interval $(\ell_{i},r_{i})$. Then
$(x^{\prime},\mathfrak{m}[s](x^{\prime}))$ is added to $M$ at stage $s$. Since
$|x-x^{\prime}|<2^{-s}<\delta$,
$|\mathfrak{m}(x)-\mathfrak{m}(x^{\prime})|<\epsilon/3$. Also,
$|\mathfrak{m}(x^{\prime})-\mathfrak{m}[s](x^{\prime})|<\epsilon/3$. Thus the
distance between $(x,\mathfrak{m}(x))$ and
$(x^{\prime},\mathfrak{m}[s](x^{\prime}))$ is at most

$|x-x^{\prime}|+|\mathfrak{m}(x)-\mathfrak{m}[s](x^{\prime})|\leq|x-x^{\prime}|+|\mathfrak{m}(x)-\mathfrak{m}(x^{\prime})|+|\mathfrak{m}(x^{\prime})-\mathfrak{m}[s](x^{\prime})|<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon.$

If $(x,y)$ is a point of the form $(x,\mathfrak{t}_{i}(x))$ or
$(x,\mathfrak{b}_{i}(x))$, $i<n$, a simpler version of the above argument
works, where the stage $s$ is taken to be large enough that the $i$th tendril
has stabilized. (Also, $2^{-s}$ is replaced by $p_{i,c_{i}}^{-s}$ or
$p_{i,c_{i}}^{-s}$.)

######  Remark 5.3.

The fact that $\mathfrak{m}(-1)=\mathfrak{m}(1)=0$ means that the
homeomorphism $[0,1]\to X$ maps $0$ to $(-1,0)$ and $1$ to $(1,0)$.

####  5.3.2 The case $W\notin A$

Suppose that $W\notin A$, so that for each $n$, $W_{n}$ is finite. We will
argue that $X_{W}$ is not locally connected, whence it is not homeomorphic to
$[0,1]$.

For each $n$, there is a stage $s_{n}$ after which no element enters
$W_{1},\ldots,W_{n}$. Thus, after stage $s_{n}$, the $n$th tendril remains the
same. It is easy to see that the $n$th tendril—that is, the points
$(x,\mathfrak{t}_{n}(x))$ for $x\in[\ell_{n},1]$ and $(x,\mathfrak{b}_{n}(x))$
for $x\in[r_{n},1]$—are in $X=X_{W}$. There are infinitely many tendrils and
we argue that $X$ is not locally connected.

Indeed, let $B$ be any open set containing $(1,0)$ and of diameter at most,
say, $1/2$. Then for sufficiently large $n$, say $n\geq N$, the tip of the
$n$th tendril, $(1,\mathfrak{t}_{n}(1))=(1,\mathfrak{b}_{n}(1))$ is contained
within $B$. However the left and right endpoints of the tendril lie outside of
$B$ since the diameter of $B$ is at most $1/2$. The tip of each tendril
($n\geq N$) is in a different connected component of $B$. Thus $B$ is not
connected.

###  5.4 The construction for $\mathbf{\Pi}^{0}_{4}$-completeness

Let $B$ be the $\mathbf{\Pi}^{0}_{4}$-Wadge-complete set

$B=\\{U\in\omega\times\omega\times\omega\;\mid\;\forall m\;\;U_{m}\in A\\}.$

Given $U$, for each $m$, let $X_{U_{m}}$ be the metric space produced by the
construction just given when applied to $U_{m}$; that is, $X_{U_{m}}$ is not
locally connected if $U_{m}\notin A$, and $X_{U_{m}}$ is homeomorphic to
$[0,1]$ if $U_{m}\in A$. Moreover, in the latter case, the homeomorphism maps
$0\mapsto(-1,0)$ and $1\mapsto(1,0)$. We may assume, by an easy rescaling,
that $X_{U_{m}}\subseteq[0,1]\times[0,1]$ with $0\mapsto(0,0)$ and
$1\mapsto(0,1)$.

Let $Y_{U}$ be the space which contains, for each $m$, a copy of $X_{U_{m}}$
which is scaled down by a factor of $1/2^{m}$ and shifted so that the point
corresponding to $(1,0)$ in $X_{U_{m}}$ matches up with the point
corresponding to $(0,0)$ in $X_{U_{m+1}}$. Moreover, we “twist” each
subsequent copy in a new direction so that there is no other intersection.
Without the twists, this looks like:

$X_{n,1}$$X_{n,2}$$X_{n,3}$$\cdots$

We work in $c_{0}$, the space of sequences in $\mathbb{R}^{\mathbb{N}}$ which
converge to $0$. Let $\mathbf{u}$ be $(1,0,0,0,\ldots)$ and let
$\mathbf{w}_{1},\mathbf{w}_{2},\ldots$ be a list of the other standard basis
elements of $c_{0}$. These $\mathbf{w}_{i}$ will be the direction for the
twists. For each $m$, $Y_{U}$ will contain the following points:

$\left\\{\left(\frac{1}{2}+\cdots+\frac{1}{2^{m-1}}+\frac{1}{2^{m}}x\right)\mathbf{u}+\frac{1}{2^{m}}y\mathbf{w}_{m}:(x,y)\in
X_{U_{m}}\right\\}$

$Y_{U}$ will also contain $(1,0)$ as the limit point of the above points. It
is not hard to see that $Y_{U}$ is homeomorphic to $[0,1]$ if and only if each
$X_{U_{m}}$ is homeomorphic to $[0,1]$. Thus $Y_{U}$ is homeomorphic to
$[0,1]$ if and only if $U\in B$, as desired.

######  Remark 5.4.

In the next section, we will use the fact that $Y_{U}$ always includes
$(0,0,0,\ldots)$ and $(1,0,0,\ldots)$ and, in the case that it is homeomorphic
to $[0,1]$, then these are the two endpoints.

##  6 The circle $S^{1}$

We now show that $\operatorname{HCopies}(S^{1})$ is $\Pi^{0}_{4}$. The
strategy is similar to that for $[0,1]$, and we reuse conditions such as CPCT
and CONN from Section 3. Our new condition describes the circle by saying that
any four points can be arranged in order around the circle, where we use the
same ideas of betweenness as before to identify the order in which they are
arranged.

######  Definition 6.1.

$X=\overline{(M,d)}$ satisfies CIRC if for any special points
$x_{0},x_{1},x_{2},x_{3}$ (where we consider
$0,1,2,3\in\mathbb{Z}/4\mathbb{Z}$) there is some reordering of them such that

  1. (1)

for all $i\in\mathbb{Z}/4\mathbb{Z}$ and $\delta>0$ there is $\rho<\delta$
such that for all $\epsilon$ there is an $\epsilon$-path
$u_{1}=x_{i},\ldots,u_{n}=x_{i+1}$ with no $u_{j}$ in
$\overline{B}_{\rho}(x_{i+2})\cup\overline{B}_{\rho}(x_{i+3})$.555Recall that
we use $\overline{B}_{\rho}(x)$ for the closed ball rather than the closure of
the open ball.

  2. (2)

for all $i\in\mathbb{Z}/4\mathbb{Z}$ and $\delta>0$ there is $\epsilon$ such
that for any $\epsilon$-path $u_{1}=x_{i},\ldots,u_{n}=x_{i+2}$, for some $j$
either $u_{j}\in B_{\delta}(x_{i+1})$ or $u_{j}\in B_{\delta}(x_{i-1})$.

Note that this is $\Pi^{0}_{4}$.

######  Lemma 6.2.

If $X=\overline{(M,d)}$ is homeomorphic to $S^{1}$, then it satisfies CIRC (as
well as NDEGEN, CPCT, CONN, and LC).

###### Proof.

All of these properties other than CIRC follow from Section 3. So we now
verify CIRC. Given any four points on $S^{1}$, list them in order around the
circle as $x_{0},x_{1},x_{2},x_{3}$.

For (1), given $i$ and $\delta>0$, choose $\rho<\delta$ as follows. Let $C$ be
the segment of $S^{1}$ between $x_{i}$ and $x_{i+1}$ not containing $x_{i+2}$
and $x_{i+3}$. Choose $\rho<\delta$ such that $2\rho$ is smaller than the
distance between this segment and either of $x_{i+2}$ or $x_{i+3}$. Then for
any $\epsilon$, we can choose an $\epsilon$-path approximating the arc $C$
between $x_{i}$ and $x_{i+1}$ (in the sense that each element of the
$\epsilon$-path is within $\rho$ of $C$).

For (2), given $i$ and $\delta>0$, if either $x_{i}$ or $x_{i+1}$ are in
$B_{\delta}(x_{i-1})$ or $B_{\delta}(x_{i+1})$, then (2) holds trivially.
Otherwise, $X-B_{\delta}(x_{i+1})-B_{\delta}(x_{i-1})$ is closed and
disconnected, and $x_{i}$ and $x_{i+2}$ are in different connected components.
Choosing $\epsilon$ to be smaller than the distance between clopen (in
$X-B_{\delta}(x_{i+1})-B_{\delta}(x_{i-1})$) separating sets containing each
of them, there is no $\epsilon$-path from $x_{i}$ to $x_{i+2}$ in
$X-B_{\delta}(x_{i+1})-B_{\delta}(x_{i-1})$. ∎

######  Lemma 6.3.

If $X=\overline{(M,d)}$ satisfies CPCT, CONN, LC, and CIRC then for any four
special points, there is some ordering $x_{0},x_{1},x_{2},x_{3}$ and arcs
$x_{0}\to x_{1}\to x_{2}\to x_{3}\to x_{0}$ homeomorphic to $S^{1}$. Moreover,
any path from $x_{i}$ to $x_{i+2}$ must pass through $x_{i+1}$ or $x_{i-1}$.

###### Proof.

Since $X=\overline{(M,d)}$ satisfies CPCT, CONN, and LC it is compact,
connected, locally connected, path connected, and locally path connected.
Given any four special points, order them as $x_{0},x_{1},x_{2},x_{3}$ to
satisfy CIRC. By (1), for each $i$, $x_{i}$ and $x_{i+1}$ are in the same
connected component, and so there is a path $C_{i}$ from $x_{i}$ to $x_{i+1}$.

Next we must argue that any two of these arcs do not intersect except possibly
at their starting and ending points. If not, and, say, $C_{i}$ and $C_{i+1}$
intersect, let $y$ be the point of intersection. (There are other
combinatorial possibilities for the intersection, but a similar argument works
in each case.) If $y$ was a special point, then $y,x_{i+1},x_{i-1},x_{i+2}$
will not satisfy (2) of CIRC, because there are paths between any two of
$y,x_{i+1},x_{i-1}$ avoiding the other and $x_{i+2}$. If $y$ is not a special
point, then because $X$ is locally path connected we can choose some special
point $y^{\prime}$ sufficient close to $y$ such that there is a path from $y$
to $y^{\prime}$ disjoint from any of the $x$’s and disjoint from $C_{i-1}$ and
$C_{i+2}$. Then a similar argument applies to show that
$y^{\prime},x_{i+1},x_{i-1},x_{i+2}$ will not satisfy (2) of CIRC.

The same argument works for the moreover clause. ∎

######  Lemma 6.4.

If $X=\overline{(M,d)}$ satisfies NDEGEN, CPCT, CONN, LC, and CIRC, then $X$
is homeomorphic to $S^{1}$.

###### Proof.

Since $X=\overline{(M,d)}$ satisfies CPCT, CONN, and LC it is non-degenerate,
compact, connected, locally connected, path connected, and locally path
connected.

Let $x_{0},x_{1},x_{2},x_{3}$ be four special points, as in Lemma 6.3, and for
each $i$, let $C_{i}$ be the arc from $x_{i}$ to $x_{i+1}$ with the four arcs
forming a homeomorphic copy $C$ of $S^{1}$. We argue that any other point $a$
must lie on $C$. Suppose not, that $a$ does not lie on $C$.

Now there is a special point $y$ near $a$, and an arc from $y$ to $a$ disjoint
from $C$. There must also be an arc from $y$ to $C$. There are several
possibilities, in all of which we get a contradiction:

  1. (1)

The arc from $y$ to $C$ first meets $C$ at a point other than the $x_{i}$. Say
it meets $C_{i}$. Then $y,x_{i},x_{i+1},x_{i+2}$ contradict Lemma 6.3.
(Indeed, there is an arc between any two of $y,x_{i},x_{i+1}$ avoiding the
third and $x_{i+2}$).

  2. (2)

Otherwise, suppose that the arc from $y$ to $C$ first meets $C$ at $x_{i}$,
and that any arc from $y$ to $C$ must pass through $x_{i}$. Then
$y,x_{i},x_{i-1},x_{i+1}$ contradict Lemma 6.3.

  3. (3)

Otherwise, suppose that in addition to the arc from $y$ to $x_{i}$, there is
also an arc from $y$ that first meets $C$ at $x_{i+2}$. Then
$y,x_{i-1},x_{i+1},x_{i+2}$ contradicts Lemma 6.3 because there is a path
between any two of $y,x_{i-1},x_{i+1}$ avoiding the third and $x_{i+2}$.

  4. (4)

Finally, suppose that in addition to the arc from $y$ to $x_{i}$, there is
also an arc from $y$ that first meets $C$ at $x_{i+1}$ (or the symmetric case
of $x_{i-1}$). Choose a special point $z$ such that there is an arc from $z$
to the middle of $C_{i}$ that is disjoint from any of the other arcs. Then
$y,z,x_{i},x_{i+2}$ contradicts Lemma 6.3 because there is a path between any
two of $y,z,x_{i}$ avoiding the third and $x_{i+2}$.

This completes the proof that any other point $a$ must lie on $C$; that is,
$C$ is in fact all of $X$ and so $X$ is homeomorphic to $S^{1}$. ∎

######  Theorem 6.5.

The set

$\operatorname{HCopies}(S^{1})=\\{Y:Y\cong_{\operatorname{homeo}}S^{1}\\}$

is $\mathbf{\Pi}^{0}_{4}$-Wadge-complete.

###### Proof.

The other lemmas of this section show that this set is $\Pi^{0}_{4}$. To show
$\mathbf{\Pi}^{0}_{4}$-completeness, we use the same $\Pi^{0}_{4}$ set $B$
from Section 5.4. Let $Y_{U}$ be the topological space constructed there, with
$U\in B$ if and only if $Y_{U}$ is homeomorphic to $[0,1]$. By Remark 5.4,
$Y_{U}\subseteq\mathbb{R}^{\mathbb{N}}$ always contains $(0,0,0,\ldots)$ and
$(1,0,0,\ldots)$ and, in the case that it is homeomorphic to $[0,1]$, these
are the endpoints. We can easily construct from $Y_{U}$ a space $Z_{U}$ such
that $Z_{U}$ is homeomorphic to $S^{1}$ if and only if $Y_{U}$ is homeomorphic
to $[0,1]$: In $\mathbb{R}\times\mathbb{R}^{\mathbb{N}}$, let

$Z_{U}=\\{(t,0,0,\ldots):t\in[0,1]\\}\cup\\{(1,t,0,\ldots):t\in[0,1]\\}\cup\\{(t,1,0,\ldots):t\in[0,1]\\}\cup\\{(0,\mathbf{u}):\mathbf{u}\in
Y_{U}\\}.$

Essentially, to construct $Z_{U}$, we have taken a segment of $S^{1}$ and
replaced it by $Y_{U}$. ∎

##  7 The real line $\mathbb{R}$

In this section we will prove that $\mathbb{R}$ has topological Scott
complexity $\mathbf{\Pi}^{1}_{1}$.

######  Theorem 7.1.

$\operatorname{HCopies}(\mathbb{R})$ is $\mathbf{\Pi}^{1}_{1}$-Wadge-complete.

The proof is in two parts; first we give a $\Pi^{1}_{1}$ characterisation of
$\mathbb{R}$, and second we prove a $\mathbf{\Pi}^{1}_{1}$-hardness result.

###  7.1 A $\Pi^{1}_{1}$-characterization of $\mathbb{R}$

######  Theorem 7.2.

$\operatorname{HCopies}(\mathbb{R})$ is $\Pi^{0}_{4}$ within the locally
compact spaces. In particular, $\operatorname{HCopies}(\mathbb{R})$ is
$\Pi^{1}_{1}$.

###### Proof.

We begin by working within the locally compact spaces. Given a locally compact
but non-compact metric space $(X,d)$, we can form the one-point
compactification by adding a new point $\infty$. We can form a metric
$\hat{d}$ on the compactification as follows [Man89]. Fix a point $p\in X$,
and let $h(x)=\frac{1}{1+d(p,x)}$. Let

$\hat{d}(x,y)=\min(d(x,y),h(x)+h(y)),\text{ and }\hat{d}(\infty,x)=h(x).$

This construction lets us form, from a presentation of a locally non-compact
space, a presentation of the one-point compactification. Moreover, this is
computable.

Then a locally compact space is homeomorphic to $\mathbb{R}$ if and only if it
is non-compact and its one-point compactification is homeomorphic to $S^{1}$.
(This is because $S^{1}$ is homogeneous, and $S^{1}$ minus any point is
homeomorphic to $\mathbb{R}$.) The former is a $\Sigma^{0}_{3}$ condition and
the latter is $\Pi^{0}_{4}$ as proved earlier in this paper.

The set of locally compact spaces is $\Pi^{1}_{1}$ [MN13], and thus
$\operatorname{HCopies}(\mathbb{R})$ is $\Pi^{1}_{1}$. ∎

Using the fact that $\operatorname{HCopies}([0,1])$ is
$\mathbf{\Pi}^{0}_{4}$-Wadge-complete, it is easier to prove that
$\operatorname{HCopies}(\mathbb{R})$ is $\mathbf{\Pi}^{0}_{4}$-Wadge-complete
within the set of locally compact spaces. The proof is essentially the same
idea as Theorem 6.5 where we take a standard copy of $\mathbb{R}$ and replace
a subinterval by the construction of Section 5.4

######  Proposition 7.3.

$\operatorname{HCopies}(\mathbb{R})$ is $\mathbf{\Pi}^{0}_{4}$-Wadge-complete
within the set of locally compact spaces.

###  7.2 $\mathbf{\Pi}^{1}_{1}$-completeness

Now we must show that it is $\mathbf{\Pi}^{1}_{1}$-hard. Given a tree
$T\subseteq\omega^{<\omega}$, we will produce a space $X=X_{T}$ such that $T$
is well-founded if and only if $X$ is homeomorphic to $\mathbb{R}$.

Let $(\sigma_{n})$ be a computable listing of $\omega^{<\omega}$ with the
property that if $\sigma\prec\tau$ then $\sigma$ appears before $\tau$ in this
list.

We work within the ambient space $c_{0}$ of sequences in
$\mathbb{R}^{\mathbb{N}}$ that converge to $0$, with the norm
$\|\cdot\|_{\infty}$. Let $\mathbf{u}_{\sigma}$ (for
$\sigma\in\omega^{<\omega}$) and $\mathbf{v}_{n}$ (for $n\in\omega$) and
$\mathbf{w}$ be distinct standard basis elements. For convenience, let

$\chi_{\sigma}=\begin{cases}\frac{1}{|\sigma|}&\sigma\in T\\\
1&\text{otherwise}\end{cases}$

Given $\tau$, let

$\mathbf{x}_{\tau}=\sum_{\sigma\preceq\tau}\chi_{\sigma}\mathbf{u}_{\sigma}.$

Given $\pi\in\omega^{\omega}$, note that
$\lim_{n\to\infty}\mathbf{x}_{\pi\upharpoonright n}$ converges if and only if
$\pi$ is a path through $T$. If it does converge, then it converges to

$\mathbf{y}_{\pi}=\sum_{n}\frac{1}{n}\mathbf{u}_{\pi\upharpoonright n}.$

Let

$\psi_{n}(t)=\begin{cases}0&t\leq n\\\ 2(t-n)&n\leq t\leq n+\frac{1}{2}\\\
2(n+1-t)&n+\frac{1}{2}\leq t\leq n+1\\\ 0&n\geq n+1\end{cases}$

That is, $\psi_{n}(t)$ is $0$ at $t=n$, increases linearly to $1$ at
$t=n+\frac{1}{2}$, and then decreases linearly back to $0$ at $t=n+1$.

Define a continuous function $p:\mathbb{R}\to c_{0}$ as follows:

  * •

For $t\leq 0$, let $p(t)=e^{-t}\mathbf{w}$.

  * •

For $0\leq t\leq 1$, let

$p(t)=t\mathbf{x}_{\sigma_{1}}+e^{-t}\mathbf{w}$

  * •

For $1\leq n\leq t\leq n+1$, let

$p(t)=(n+1-t)\mathbf{x}_{\sigma_{n}}+(t-n)\mathbf{x}_{\sigma_{n+1}}+\psi_{n}(t)\mathbf{v}_{n}+e^{-t}\mathbf{w}.$

Note that $p(n)=\mathbf{x}_{\sigma_{n}}$. Let $X=X(T)$ be the closure of the
image of $p$. That is, let $M=p[\mathbb{Q}]$ be the countable metric space
that is the image of the rationals, and let
$X=\overline{M}=\overline{p[\mathbb{R}]}$.

It is easy to see that $p$ is injective, continuous (even uniformly
continuous), and computable from $T$. In fact, $p$ is a homeomorphism onto its
image (but its image is not necessarily closed in $c_{0}$).

######  Lemma 7.4.

$p$ is a homeomorphism onto its image.

###### Proof.

We must show given a sequence $t_{i}\in\mathbb{R}$, and $s\in\mathbb{R}$, that
$t_{i}\to s$ if and only if $p(t_{i})\to p(s)$. One direction follows from the
fact that $p$ is continuous, and the other—that if $p(t_{i})\to p(s)$ then
$t_{i}\to s$—can be seen using the term $e^{-t}\mathbf{w}$ that always appears
in $p(t)$. ∎

We now argue that $p$ is a homeomorphism onto the closure of its image if and
only if its image is closed if and only if $T$ is well-founded.

######  Lemma 7.5.

Suppose that $T$ is well-founded. Then the image of $p$ is closed.

###### Proof.

Suppose that $p(t_{i})\to\mathbf{z}$, and $t_{i}$ does not converge. Write

$\mathbf{z}=\sum_{\sigma}a_{\sigma}\mathbf{u}_{\sigma}+\sum_{n}b_{n}\mathbf{v}_{n}+c\mathbf{w}.$

We must have $c=\lim_{t\to\infty}e^{-t_{i}}$ and so, since the $t_{i}$ do not
converge, $c=0$ and $t_{i}\to\infty$. Since $t_{i}\to\infty$, we must have
that $b_{n}=0$ for all $n$.

Thus

$\mathbf{z}=\sum_{\sigma}a_{\sigma}\mathbf{u}_{\sigma}.$

By passing to a subsequence, we may assume that for each $i$ there is
$n_{i}\in\mathbb{N}$ such that
$t_{i}\in[n_{i}-\frac{1}{2},n_{i}+\frac{1}{2}]$, and moreover we may assume
that the $n_{i}$ form a strictly increasing sequence. Also, since $\mathbf{z}$
does not involve any of the $\mathbf{v}_{n}$, it must be that

$\lim_{i\to\infty}|n_{i}-t_{i}|=0.$

So in fact we may assume that $t_{i}=n_{i}$, and
$p(n_{i})=\mathbf{x}_{\sigma_{n_{i}}}$.

Now it is not hard to see that the only way to have
$\lim_{i\to\infty}p(n_{i})$ exist is to have the $\sigma_{n_{i}}$ lie on a
path $\pi$ through $T$, and $\lim_{i\to\infty}p(n_{i})=\mathbf{y}_{\pi}$. Thus
$T$ has a path.

So we conclude that if $T$ is well-founded, then $\lim_{i\to\infty}p(t_{i})$
converges if and only if $t_{i}$ converges; in particular, $p$ is a
homeomorphism of $\mathbb{R}$ onto its image and its image is closed in
$c_{0}$. ∎

Thus if $T$ is well-founded,
$X=\overline{M}=\overline{p[\mathbb{R}]}=p[\mathbb{R}]$, and this is
homeomorphic to $\mathbb{R}$ by Lemma 7.4.

######  Lemma 7.6.

Suppose that $T$ has an infinite path $\pi$. Then there is an increasing
integer sequence $n_{i}$ such that $\lim_{i\to\infty}p(n_{i})$ converges.

###### Proof.

Let $\pi$ be a path through $T$, and let $n_{i}$ be an increasing integer
sequence such that $\sigma_{n_{i}}=\pi\\!\\!\upharpoonright i$. Then
$p(n_{i})=\mathbf{x}_{\pi\upharpoonright i}+e^{-n_{i}}\mathbf{w}$, and

$\lim_{i\to\infty}p(n_{i})=\mathbf{y}_{\pi}=\sum_{n}\frac{1}{n}\mathbf{u}_{\pi\upharpoonright
n}.$

Thus $p$ maps a non-convergent sequence to a convergent sequence, and so
cannot be a homeomorphism onto its image. ∎

If $T$ is not well-founded, then $X=\overline{M}=\overline{p[\mathbb{R}]}$
contains a homeomorphic image of $\mathbb{R}$ as a proper dense subset. But
$\mathbb{R}$ does not contain a homeomorphic image of $\mathbb{R}$ as a proper
dense subspace (indeed, $\mathbb{R}$ has no proper connected dense subsets).
Thus $X$ is not homeomorphic to $\mathbb{R}$.

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