# Closed copies of $\N$ in $\R^{\omega_{1}}$

Alan Dow  Department of Mathematics
UNC-Charlotte
9201 University City Blvd.
Charlotte, NC 28223-0001<EMAIL_ADDRESS>https://webpages.uncc.edu/adow ,
Klaas Pieter Hart  Faculty EEMCS
TU Delft
Postbus 5031
2600 GA Delft
the Netherlands<EMAIL_ADDRESS>https://fa.ewi.tudelft.nl/~hart ,  Jan
van Mill  KdV Institute for Mathematics
University of Amsterdam
P.O. Box 94248
1090 GE Amsterdam
The Netherlands<EMAIL_ADDRESS>https://staff.fnwi.uva.nl/j.vanmill/ and
Hans Vermeer  Faculty EEMCS
TU Delft
Postbus 5031
2600 GA Delft
the Netherlands<EMAIL_ADDRESS>
###### Abstract.

We investigate closed copies of $\N$ in powers of $\R$ with respect to
$C^{*}$\- and $C$-embedding. We show that $\R^{\omega_{1}}$ contains closed
copies of $\N$ that are not $C^{*}$-embedded.

###### Key words and phrases:

closed copy of $\N$, $C$-embedding, $C^{*}$-embedding, Aronszajn tree,
Aronszajn line, compactification, realcompactness, power of $\R$

###### 1991 Mathematics Subject Classification:

Primary 54C45; Secondary: 03E17, 03E50, 03E55, 54D35, 54D40, 54D60, 54G20

## Introduction

We continue the investigation into closed copies of $\N$ in powers of $\R$
begun in [dhvmv2022]. In that paper we constructed examples of realcompact
spaces that contain closed copies of $\N$ that are $C^{*}$-embedded but not
$C$-embedded.

Given such a space, $X$, the diagonal map of the full set $C(X)$ of real-
valued continuous functions on $X$ embeds $X$ into $\R^{C(X)}$ as a closed and
$C$-embedded subset. The closed copy of $\N$ then becomes a $C^{*}$-embedded
but not $C$-embedded in $\R^{C(X)}$.

We also found realcompact spaces with closed copies of $\N$ that were not
$C^{*}$-embedded, and hence closed copies of $\N$ in a power of $\R$ that are
not $C^{*}$-embedded.

In all cases the cardinality of the set of continuous functions is equal to
$\mathfrak{c}$, which yields the interesting result that one can find closed
copies of $\N$ in $\R^{\mathfrak{c}}$, that are not $C^{*}$-embedded, and that
are $C^{*}$-embedded but not $C$-embedded.

In the first version of [dhvmv2022] we posed two questions suggested by these
results. We repeat them here.

######  Question 1.

What is the minimum cardinal $\kappa$ such that $\R^{\kappa}$ contains a
closed copy of $\N$ that is $C^{*}$-embedded but not $C$-embedded?

######  Question 2.

What is the minimum cardinal $\kappa$ such that $\R^{\kappa}$ contains a
closed copy of $\N$ that is not $C^{*}$-embedded?

Given that $\R^{\omega_{0}}$ is metrizable and we know that in both cases we
have $\aleph_{0}<\kappa\leq\mathfrak{c}$.

The principal result of this paper answers question 2: there is a closed copy
of $\N$ in $\R^{\omega_{1}}$ that is not $C^{*}$-embedded. In later versions
of [dhvmv2022] we replaced question 2 with the answer and a reference to the
present paper.

In Sections 2 and 3 we give various examples and obtain translations of the
statement “$\R^{\omega_{1}}$ contains a closed copy of $\N$ that is not
$C^{*}$-embedded” that suggest further interesting questions.

In Section 2 we present three constructions: one directly from an Aronszajn
tree, one directly from an Aronszajn continuum, and one as the path space of
an Aronszajn tree. We decided to give all three examples because they show how
versatile these objects are.

In Section 3 we give the translations mentioned above and give a fourth
example that is of a somewhat different nature.

Section 4 deals with a class of topological spaces that feature in the
translations, and in Section 5 we take a tentative step towards an answer to
Question 1.

We should mention first that in the model obtained by adding supercompact many
Random reals to a model of $\mathsf{CH}$ the answer to Question 1 is
$\mathfrak{c}$: every $C^{*}$-embedded subspace of every space of character
less than $\mathfrak{c}$ is $C$-embedded, see [MR2823691]. The equality
$\mathfrak{d}=\mathfrak{u}=\aleph_{1}$ allows us to construct a space $\BB$ of
cardinality and weight $\aleph_{1}$, and with a closed copy of $\N$ that is
$C^{*}$-embedded but not $C$-embedded. The space $\BB$ can be embedded as a
closed set in $\R^{\omega_{1}}$, but we do not know if it can be $C^{*}$\- or
$C$-embedded in $\R^{\omega_{1}}$.

##  1\. Preliminaries

By now the reader may have guessed that by “a closed copy of $\N$” in some
space $X$ we mean a closed subspace of $X$ that is homeomorphic to the
discrete space $\N$, in other words: a countably infinite closed and discrete
subspace.

In general we say that a subspace $Y$ of a space $X$ is $C$-embedded if every
continuous function $f:Y\to\R$ has a continuous extension to all of $X$. If
this holds for all _bounded_ continuous functions then we say that $Y$ is
$C^{*}$-embedded in $X$.

The way we shall show that a closed copy of $\N$ is not $C^{*}$-embedded in
$X$ is by exhibiting disjoint subsets $A$ and $B$ of $\N$ such that are not
_completely separated_ , which means that whenever $g:X\to\R$ is bounded and
continuous the closures of $g[A]$ and $g[B]$ intersect. This then implies that
the characteristic function of $A$ has no continuous extension to $X$.

As mentioned in the introduction we shall uses Aronszajn tree and continua in
some of our constructions; Todorčević’s article [MR776625] contains all the
information that we need.

Below we shall be using a few ‘small’ cardinals from [MR776622]. These are
$\mathfrak{p}$, $\mathfrak{d}$ and $\mathfrak{u}$.

To define $\mathfrak{p}$ we first say that a family $\mathcal{A}$ of subsets
of $\N$ has the _strong finite intersection property_ if for every finite
subfamily $\mathcal{F}$ of $\mathcal{A}$ the intersection $\bigcap\mathcal{F}$
is infinite. Next we call $P$ a pseudointersection of $\mathcal{A}$ if $P$ is
infinite and $P\subseteq^{*}A$ for all $A\in\mathcal{A}$. Then $\mathfrak{p}$
is the minimum cardinality of a family of subsets of $\N$ with the strong
finite intersection property but without a pseudointersection.

The cardinal $\mathfrak{d}$ is the minimum cardinality of a subset $D$ of
$\N^{\N}$ with the property that for for every $g\in\N^{\N}$ there is an $f\in
D$ such that $g(n)\leq f(n)$ for all $n$.

Finally, $\mathfrak{u}$ is defined to be the minimum character of a free
ultrafilter on $\N$.

It is relatively easy to show that $\mathfrak{p}$ is uncountable and that all
three cardinals are not larger than $\mathfrak{c}$. As a base for an
ultrafilter is a defining family for $\mathfrak{p}$ we obtain
$\mathfrak{p}\leq\mathfrak{u}$, and [MR776622]*Theorem 3.1 shows
$\mathfrak{p}\leq\mathfrak{d}$.

Any potentially unfamiliar topological notions will be defined when needed;
definitions not given here can be found in Engelking’s book [MR1039321].

##  2\. Closed copies of $\N$ that are not $C^{*}$-embedded

In this section we show that the answer to the second question is
$\aleph_{1}$. We give three examples of closed copies of $\N$ that are not
$C^{*}$-embedded in $\R^{\omega_{1}}$. This may seem like overdoing things
somewhat but we think that this presentation is more informative.

We first give two specific constructions and extract a few translations of
“$\R^{\omega_{1}}$ contains a closed copy of $\N$ that is not
$C^{*}$-embedded” that allows us to construct a relatively simple example and
dispel a suspicion that the first two examples suggest.

###  2.1. A closed copy of $\N$ that is not $C^{*}$-embedded, from an
Aronszajn tree

The first construction uses an Aronszajn tree to guide an embedding of $\N$
into $\R^{\omega_{1}}$.

The well-known inequality $\mathfrak{p}>\aleph_{0}$ allows us to define a
family $\\{A_{t}:t\in{\vphantom{\omega}}^{<\omega_{1}}\omega\\}$ of infinite
subsets of $\N$ such that

  1. (1)

if $s\subset t$ then $A_{t}\subset^{*}A_{s}$, and

  2. (2)

for every $t$ the family $\\{A_{t*n}:n\in\omega\\}$ is a partition of $A_{s}$.

Next we take an Aronszajn subtree $T$ of
${\vphantom{\omega}}^{<\omega_{1}}\omega$. Say an Aronszajn subtree of the set
of all finite-to-one members of ${\vphantom{\omega}}^{<\omega_{1}}\omega$, and
such that $\\{t*n:n\in\omega\\}\subseteq T$ whenever $t\in T$.

For every non-zero $\alpha$ in $\omega_{1}$ we let $\langle
t(\alpha,n):n\in\omega\rangle$ enumerate the $\alpha$th level $T_{\alpha}$ of
$T$ in a one-to-one fashion. We abbreviate $A_{t(\alpha,n)}$ to $A(\alpha,n)$.

By construction each of the families $\\{A(\alpha,n):n\in\omega\\}$ is
pairwise almost disjoint. We can assume, after making finite modifications to
the $A(\alpha,n)$, that every family $\\{A(\alpha,n):n\in\omega\\}$ is in fact
a partition of $\N$.

We use the partitions to define a map $k\mapsto x_{k}$ from $\N$ to
$\R^{\omega_{1}}$.

First we set $x_{k}(0)=2^{-k}$ for all $k$. This ensures that
$X=\\{x_{k}:k\in\N\\}$ is a relatively discrete subspace of $\R^{\omega_{1}}$.

Second, for every non-zero $\alpha$ in $\omega_{1}$ we define

$x_{2k}(\alpha)=x_{2k+1}(\alpha)=m\text{ iff }k\in A(\alpha,m)$

This will ensure that $X$ is closed in $\R^{\omega_{1}}$ and that the sets
$\\{x_{2k}:k\in\N\\}$ and $\\{x_{2k+1}:k\in\N\\}$ are not completely separated
in $\R^{\omega_{1}}$.

To see that $X$ is closed let $x\in\operatorname{cl}X$ and let $u$ be an
ultrafilter such that $x=u\mathchar 45\relax\\!\lim x_{k}$. We claim $u$ is in
fact a fixed ultrafilter and hence that $x\in X$.

Since $u$ is a filter there is for every $\beta$ most one $n$ such that
$A(\beta,n)\in u$. Let $B=\\{\langle{\beta},{n}\rangle:A(\beta,n)\in u\\}$. If
$u$ were free then $A(\beta,n)\cap A(\gamma,m)$ would be infinite whenever
$\langle{\beta},{n}\rangle,\langle{\gamma},{m}\rangle\in B$. By the
construction of the family
$\\{A_{t}:t\in{\vphantom{\omega}}^{<\omega_{1}}\omega\\}$ this would mean that
$\\{t(\beta,n):\langle{\beta},{n}\rangle\in B\\}$ is linearly ordered in $T$,
and hence countable.

Take $\alpha$ such that
$T_{\alpha}\cap\\{t(\beta,n):\langle{\beta},{n}\rangle\in B\\}=\emptyset$, and
let $m=\bigl{\lceil}x(\alpha)\bigr{\rceil}$. Then $U=\N\setminus\bigcup_{i\leq
m}A(\alpha,i)$ belongs to $u$, and $x_{2k}(\alpha)=x_{2k+1}(\alpha)\geq
x(\alpha)+1$ for all $k\in U$. This shows that $x(\alpha)\neq u\mathchar
45\relax\\!\lim x_{k}(\alpha)$, which contradicts the assumption that
$x=u\mathchar 45\relax\\!\lim x_{k}$.

To see that $\\{x_{2k}:k\in\N\\}$ and $\\{x_{2k+1}:k\in\N\\}$ are not
completely separated in $\R^{\omega_{1}}$ let $g:\R^{\omega_{1}}\to[0,1]$ be
continuous. It is well-known, see [MR1039321]*Problem 2.7.12, that there are
$\delta<\omega_{1}$ and a continuous function $h:\R^{\delta}\to[0,1]$ such
that $g=h\circ\pi_{\delta}$.

Consider $A(\delta,0)$. By construction we know that for every non-zero
$\alpha<\delta$ there is a single $n_{\alpha}$ such
$A(\delta,0)\subset^{*}A(\alpha,n_{\alpha})$. Let $x\in\R^{\delta}$ be given
by $x(0)=0$ and $x(\alpha)=n_{\alpha}$, then the subsequences $\langle
x_{2k}:k\in A(\delta,0)\rangle$ and $\langle x_{2k+1}:k\in A(\delta,0)\rangle$
of $\langle x_{k}:k\in\N\rangle$ both converge to $x$ and so $h(x)$ is in
$\operatorname{cl}\\{g(x_{2k}):k\in\N\\}\cap\operatorname{cl}\\{g(x_{2k+1}):k\in\N\\}$.

###  2.2. Another closed copy of $\N$ that is not $C^{*}$-embedded, from an
Aronszajn line

Let $L$ be an Aronszajn continuum: a first-countable linearly ordered
continuum of weight $\aleph_{1}$ with the property that the closure of every
countable set is second-countable, see [MR776625]*Section 5. We can also
assume, without loss of generality, that $L$ has no non-trivial separable
intervals.

Let $\langle{x}_{\alpha}:\alpha\in\omega_{1}\rangle$ enumerate a dense subset
of $L$ where we assume that $x_{0}=\min L$ and $x_{1}=\max L$. Using the
first-countability of $L$ we find that
$L=\bigcup_{\alpha<\omega_{1}}\operatorname{cl}\\{x_{\beta}:\beta\leq\alpha\\}$,
that is, $L$ is the union of an increasing sequence of second-countable
compact subsets. Upon thinning out the sequence we obtain a strictly
increasing sequence $\langle{K}_{\alpha}:\alpha\in\omega_{1}\rangle$ of
second-countable compact subsets whose union is equal to $L$. The assumption
on the intervals of $L$ implies that each $K_{\alpha}$ is nowhere dense.

We claim that every $K_{\alpha}$ is a $G_{\delta}$-set of $L$. By the first-
countability of $L$ this is clear if $\alpha$ is finite, so we assume below
that $\alpha$ is infinite, and hence that $\min L$ and $\max L$ belong to
$K_{\alpha}$.

Since $K_{\alpha}$ is second-countable we can find a countable family
$\mathcal{I}$ of open intervals in $L$ such that $\\{I\cap
K_{\alpha}:I\in\mathcal{I}\\}$ is a base for the topology of $K_{\alpha}$.

Every convex component $C$ of $L\setminus K_{\alpha}$ is of the form
$(a_{C},b_{C})$, with $a_{C},b_{C}\in K_{\alpha}$. If $C$ and $D$ are two such
components then $b_{C}<a_{D}$ or $b_{D}<a_{C}$. For each $C$ take
$I_{C}\in\mathcal{I}$ such that $I_{C}\cap K_{\alpha}\subseteq[b_{C},\max L]$.
Then $a_{C}\notin I_{C}$ and so $b_{D}\notin I_{C}$ whenever $b_{D}<a_{C}$. It
follows that $I_{C}\neq I_{D}$ whenever $C\neq D$. This shows that there are
at most countably many convex components in the complement of $K_{\alpha}$.

Enumerate these components as $\langle{C}_{n}:{n}\in\omega\rangle$ and choose
for every $n\in\omega$ sequences $\langle a(n,k):k\in\omega\rangle$ and
$\langle b(n,k):k\in\omega\rangle$ in $C_{n}$ such that $a(n,k)\downarrow
a_{C_{n}}$ and $b(n,k)\uparrow b_{C_{n}}$.

Then $C_{n}=\bigcup_{k\in\omega}[a(n,k),b(n,k)]$ for all $n$. Define
$F_{k}=\bigcup_{n\leq k}[a(n,k),b(n,k)]$ for all $k$. Then
$\langle{F}_{k}:{k}\in\omega\rangle$ is a sequence of closed sets and its
union is equal to the complement of $K_{\alpha}$.

Since $L$ has weight $\aleph_{1}$ there is a compactification $\gamma\N$ of
$\N$ such that $\gamma\N\setminus\N$ is (homeomorphic to) $L$, see
[MR1039321]*Problem 3.12.18 (c). Take the quotient of
$\gamma\N\times\\{0,1\\}$ obtained by identifying $\langle{x},{0}\rangle$ and
$\langle{x},{1}\rangle$ for all $x\in L$.

The result is a new compactification $\delta\N$ of $\N$ with remainder equal
to $L$ and in which $\N$ is the union of two subsets $A$ and $B$ such that
$L=\operatorname{cl}A\cap\operatorname{cl}B$.

We map $\delta\N$ into $[0,1]^{\omega_{1}}$ in such a way that the image of
$\N$ will be a closed subset of $(0,1)^{\omega_{1}}$ that is not
$C^{*}$-embedded.

For every $\alpha\geq 1$ we let $f_{\alpha}:\delta\N\to[0,1]$ be continuous
such that $K_{\alpha}=f_{\alpha}^{\leftarrow}(0)$ and
$f_{\alpha}[\N]\subseteq(0,1)$. We let $f_{0}:\delta\N\to[0,1]$ be the
continuous map determined by $f_{0}(k)=\frac{1}{2}+2^{-k-2}$; it maps $L$ to
$\\{\frac{1}{2}\\}$ and $\N$ into $(\frac{1}{2},1)$.

The diagonal map $F$ of $\langle{f}_{\alpha}:\alpha\in\omega_{1}\rangle$ maps
$\delta\N$ to $[0,1]^{\omega_{1}}$ and maps $\N$ into $(0,1)^{\omega_{1}}$.

The first coordinate $f_{0}$ ensures that $F[\N]$ is relatively discrete in
$(0,1)^{\omega_{1}}$; it remains to show that it is closed and not
$C^{*}$-embedded.

To see that $F[\N]$ is closed in $(0,1)^{\omega_{1}}$ observe that for every
$x\in L$ there is an $\alpha$ such that $x\in K_{\alpha}$; but then
$f_{\beta}(x)=0$ for $\beta\geq\alpha$. It follows that
$F[\N]=F[\delta\N]\cap(0,1)^{\omega_{1}}$.

To see that $F[\N]$ is not $C^{*}$-embedded in $(0,1)^{\omega_{1}}$ let
$g:(0,1)^{\omega_{1}}\to[0,1]$ be continuous. We show that the closures of
$g[A]$ and $g[B]$ intersect.

As above there is an $\alpha$ such that $g$ factors through the first $\alpha$
coordinates, that is, there is a continuous map $h:(0,1)^{\alpha}\to[0,1]$
such that $g=h\circ\pi_{\alpha}$. Take $x\in L\setminus K_{\alpha}$. Then
$x\in\operatorname{cl}A\cap\operatorname{cl}B$, hence
$\pi_{\alpha}(x)\in\operatorname{cl}(\pi_{\alpha}[A])\cap\operatorname{cl}(\pi_{\alpha}[B])$.
But because $x\notin K_{\beta}$ for all $\beta\leq\alpha$ we find that
$\pi_{\alpha}(x)\in(0,1)^{\alpha}$ and hence we conclude that
$h(\pi_{\alpha}(x))\in\operatorname{cl}(g[A])\cap\operatorname{cl}(g[B])$.

###  2.3. A characterization

From the foregoing example we extract a characterization of there being a
closed copy of $\N$ in $\R^{\omega_{1}}$ that is not $C^{*}$-embedded.

######  Theorem 2.1.

The following two statements are equivalent:

  1. (1)

There is closed copy of $\N$ in $\R^{\omega_{1}}$ that is not
$C^{*}$-embedded.

  2. (2)

There is a compact space $X$ with a cover consisting of $\aleph_{1}$ many
zero-sets that has no countable subcover.

###### Proof.

To prove (1) implies (2) we take a countable closed and discrete subset $N$ of
$(0,1)^{\omega_{1}}$ that is not $C^{*}$-embedded. Let
$K=\operatorname{cl}N\setminus N$, where we take the closure in
$[0,1]^{\omega_{1}}$. For every $\alpha\in\omega_{1}$ and $i\in\\{0,1\\}$ we
let $A(\alpha,i)=\\{x\in K:x_{\alpha}=i\\}$. Then
$\\{A(\alpha,i):\langle{\alpha},{i}\rangle\in\omega_{1}\times 2\\}$ is a cover
of $K$ by $\aleph_{1}$ many $G_{\delta}$-sets. We show that there is no
$\alpha\in\omega_{1}$ such that
$\\{A(\beta,i):\langle{\beta},{i}\rangle\in\alpha\times 2\\}$ covers $K$.

Let $\alpha\in\omega_{1}$; we can assume that the projection
$\pi_{\alpha}:[0,1]^{\omega_{1}}\to[0,1]^{\alpha}$ is one-to-one on $N$. If
$\\{A(\beta,i):\langle{\beta},{i}\rangle\in\alpha\times 2\\}$ covers $K$ then
for every $x\in K$ there is a $\beta\in\alpha$ such that
$x_{\beta}\in\\{0,1\\}$, and hence $\pi_{\alpha}(x)\notin(0,1)^{\alpha}$. We
see that $\pi_{\alpha}[K]$ is disjoint from $(0,1)^{\alpha}$ and hence that
$\pi_{\alpha}[N]$ is closed in $(0,1)^{\alpha}$ and hence also $C$-embedded
because $(0,1)^{\alpha}$ is metrizable. But then $N$ would even be
$C$-embedded in $(0,1)^{\omega_{1}}$.

To prove that (2) implies (1) we proceed as in Section 2.2. Let $X$ be a space
as in (2) and let $\\{A_{\alpha}:\alpha\in\omega_{1}\\}$ be the cover by zero-
sets without a countable subcover. We may assume that $X$ has weight
$\aleph_{1}$, for example, by choosing a sequence
$\langle{f}_{\alpha}:\alpha\in\omega_{1}\rangle$ of continuous functions from
$X$ to $[0,1]$ such that $A_{\alpha}=f_{\alpha}^{\leftarrow}(0)$ for all
$\alpha$. The image $K$ of $X$ under the diagonal map of the sequence has the
same property of $X$ itself, where $B_{\alpha}=\\{x\in K:x_{\alpha}=0\\}$
defines the family of zero-sets.

The construction in Section 2.2 now yields a closed copy of $\N$ in
$(0,1)^{\omega_{1}}$ that is not $C^{*}$-embedded. ∎

Of course $2^{2}=4$ is also an equivalent of statement (1), as both are true,
but the characterization should be understood as a translation: to construct
the desired embedding it is necessary and sufficient to construct a particular
type of topological space.

###  2.4. Yet another closed copy of $\N$ that is not $C^{*}$-embedded, from
an Aronszajn tree

To see an application of Theorem 2.1 we create yet another closed copy of $\N$
in $\R^{\omega_{1}}$ that is not $C^{*}$-embedded, by exhibiting a space that
satisfies the properties in (2) in the theorem.

We let $T$ be an Aronszajn tree and we take its _path space_ $\sigma T$, where
a path is a linearly ordered subset $P$ that is also an initial segment: if
$t\in P$ and $s\leq t$ then $s\in P$. We view $\sigma T$, via characteristic
functions, as a subspace of the Cantor cube $\\{0,1\\}^{T}$. For more on this
construction see [MR1355064].

The condition that the paths be linearly ordered ensures that $\sigma T$ is
closed and hence compact. The weight of $\sigma T$ is at most that of
$\\{0,1\\}^{T}$, that is: $\aleph_{1}$.

For $\alpha\in\omega_{1}$ we let $K_{\alpha}$ be the set of paths that are of
length less than $\alpha$. To see that $K_{\alpha}$ is closed note that $p\in
K_{\alpha}$ iff $p\cap T_{\alpha}=\emptyset$. That is $K_{\alpha}=\sigma
T\setminus\bigcup_{t\in T_{\alpha}}O_{t}$, where $O_{t}=\\{p:t\in p\\}$. The
sets $O_{t}$ are clopen, so the union $\bigcup_{t\in T_{\alpha}}O_{t}$ is an
open $F_{\sigma}$-set.

Because $T$ is uncountable no countable subfamily of
$\\{K_{\alpha}:\alpha\in\omega_{1}\\}$ covers $\sigma T$.

Note that, as every path is countable, the space $\sigma T$ is actually
Corson-compact.

##  3\. The connection with Aronszajn trees and lines

Each of the three constructions in the previous section uses an Aronszajn tree
or line as input. The following theorem, which adds three more statements to
the list in Theorem 2.1, makes precise how these structures enter the
constructions.

######  Theorem 3.1.

The following statements are equivalent.

  1. (1)

There is closed copy of $\N$ in $\R^{\omega_{1}}$ that is not
$C^{*}$-embedded.

  2. (2)

There is a compact space $X$ with a cover consisting of $\aleph_{1}$ many
zero-sets that has no countable subcover.

  3. (3)

There is a compact space $X$ of weight $\aleph_{1}$ with a cover consisting of
$\aleph_{1}$ many zero-sets that has no countable subcover.

  4. (4)

The space $\N^{*}$ has a cover by $\aleph_{1}$ many zero-sets that has no
countable subcover.

  5. (5)

There is an $\omega_{1}\times\omega$-matrix $\langle
A(\alpha,n):\langle{\alpha},{n}\rangle\in\omega_{1}\times\omega\rangle$ of
infinite subsets of $\N$ such that

    1. (a)

for every countable subset $C$ of $\omega_{1}$ there is a function
$f:C\to\omega$ such that $\\{A(\alpha,f(\alpha)):\alpha\in F\\}$ has the
strong finite intersection property, and

    2. (b)

there is _no function_ $f:\omega_{1}\to\omega$ such that
$\\{A(\alpha,f(\alpha)):\alpha\in\omega_{1}\\}$ has the strong finite
intersection property.

###### Proof.

We already proved that (1) implies (2). The proof that (2) implies (1) in
Theorem 2.1 establishes implicitly that that (2) implies (3) and (3) implies
(1).

Clearly (4) implies (2).

To prove that (3) implies (4) we take a continuous map $f$ from $\N^{*}$ onto
$X$ and take the preimages of the members of the given cover. This yields the
desired cover of $\N^{*}$.

It remains to show that (4) and (5) are equivalent. This follows from the
strong zero-dimensionality of $\N^{*}$: is $Z$ is a zero-set in $\N^{*}$ then
one can cover $\N^{*}\setminus Z$ by a countable pairwise disjoint family of
clopen sets. This family can be expressed as $\\{A_{n}^{*}:n\in\omega\\}$,
where each $A_{n}$ is an infinite subset of $\N$.

Conversely if $\\{A_{n}:n\in\omega\\}$ is a family of infinite subsets of $\N$
then $\N^{*}\setminus\bigcup_{n\in\omega}A_{n}^{*}$ is a zero-set.

Thus a family $\\{Z_{\alpha}:\alpha\in\omega_{1}\\}$ of zero-sets of $\N^{*}$
can be represented by a matrix $\langle
A(\alpha,n):\langle{\alpha},{n}\rangle\in\omega_{1}\times\omega\rangle$ of
infinite subsets of $\N$ such that
$Z_{\alpha}=\N^{*}\setminus\bigcup_{n\in\omega}A(\alpha,n)^{*}$.

Then condition (a) expresses that no countable subfamily covers $\N^{*}$, and
condition (b) expresses that the family does cover $\N^{*}$. ∎

The matrix $\langle
A(\alpha,n):\langle{\alpha},{n}\rangle\in\omega_{1}\times\omega\rangle$ of
sets from Section 2.1, that resulted from enumerating the levels of the
Aronszajn tree as $\langle t(\alpha,n):n\in\omega\rangle$, satisfies the
conditions in item (5) of Theorem 3.1.

It would seem natural to call such a matrix an Aronszajn matrix and a compact
space with a cover of cardinality $\aleph_{1}$ by closed $G_{\delta}$-sets
without a countable subcover an Aronszajn compactum. This usage would conflict
with that of Hart and Kunen in [MR2516236]; and, more importantly, it would
not be quite correct, as we show next.

###  3.1. A matrix and space that are not derived from an Aronszajn tree

The three examples constructed in Section 2 all have in common that they have
an _increasing_ cover of length $\omega_{1}$ by closed $G_{\delta}$-sets.

Here we construct a compact space of weight $\aleph_{1}$ with an
$\aleph_{1}$-sized cover by closed $G_{\delta}$-sets that has no countable
subcover, and that is definitely not increasing. The space is a variation of
Example 7 in [MR584666].

To begin we take an injective map $f:\omega_{1}\to\R$ with the property that
for every $\alpha$ the image of the interval
$I_{\alpha}=\bigl{[}\omega\cdot\alpha,\omega\cdot(\alpha+1)\bigr{)}$ under $f$
is dense in $\R$. This is easily arranged, for example by taking $\aleph_{1}$
many cosets of the subgroup of rationals and mapping each interval
$I_{\alpha}$ onto one of these cosets.

We let $X$ be the set of all subsets of $\omega_{1}$ on which $f$ is
monotonically increasing; we identify $X$, via characteristic functions, with
a subset of $2^{\omega_{1}}$ and give it the subspace topology.

The complement of $X$ is open: if $x\notin X$ then there are two ordinals
$\alpha$ and $\beta$ such that $x_{\alpha}=x_{\beta}=1$, $\alpha\in\beta$, and
$f(\beta)<f(\alpha)$. Then $\\{y:y_{\alpha}=y_{\beta}=1\\}$ is an open set
disjoint from $X$. It follows that $X$ is compact.

As subsets of $\R$ that are well-ordered by the normal order are countable the
space $X$ is Corson compact.

It remains to exhibit a cover of $X$ by closed $G_{\delta}$-sets that has no
countable subcover.

To this end we let $G_{\alpha}=\\{x\in X:(\forall\beta\in
I_{\alpha})(x_{\beta}=0)\\}$. This is a closed $G_{\delta}$-set; it is the
intersection of countably many basic clopen sets:
$G_{\alpha}=\bigcap_{\beta\in I_{\alpha}}\\{x:x_{\beta}=0\\}$.

To see that $\\{G_{\alpha}:\alpha\in\omega_{1}\\}$ is a cover of $X$, let
$x\in X$. Then, because $S=\\{\beta:x_{\beta}=1\\}$ is countable, there is an
$\alpha$ such that $S\subset\alpha$; then $S\cap I_{\alpha}=\emptyset$ and so
$x\in G_{\alpha}$.

To see that no countable subfamily covers $X$ we let $\delta\in\omega_{1}$. We
take a subset $A$ of $\R$ that is ordered in order-type $\delta+2$ by the
normal order of $\R$ and so we list $A$ as $\langle
a_{\alpha}:\alpha<\delta+2\rangle$ in increasing order. Next take a sequence
$\langle\gamma_{\alpha}:\alpha<\delta+2\rangle$ of ordinals such that
$\gamma_{\alpha}\in I_{\alpha}$ and
$a_{\alpha}<f(\gamma_{\alpha})<a_{\alpha+1}$ for all $\alpha$. Then the set
$\\{\gamma_{\alpha}:\alpha<\delta+2\\}$ determines a point in $X$ that is not
in $\bigcup_{\alpha\in\delta}G_{\alpha}$.

The same argument enables one to show that the sets $G_{\alpha}$ are quite
independent: given two disjoint countable sets of ordinals $A$ and $B$ one can
find points in $\bigcap_{\alpha\in A}G_{\alpha}\setminus\bigcup_{\beta\in
B}G_{\beta}$.

Via a map from $\N^{*}$ onto $X$ we can then create a matrix that is quite
different from the ones derived from Aronszajn trees.

##  4\. Pseudo-Aronszajn compacta

Let us, for the nonce, call a compact space a _pseudo-Aronszajn compactum_ if
it has a cover of cardinality $\aleph_{1}$ by closed $G_{\delta}$-sets that
has no countable subcover. We let $\mathcal{A}$ denote the class of these
compacta.

It is readily seen that $\mathcal{A}$ is closed under taking (compact)
preimages: simply pull back the cover.

We have established that every Aronszajn continuum is in $\mathcal{A}$, and
hence that a Souslin continuum is a ccc compactum in $\mathcal{A}$.

The ordinal space $\omega_{1}+1$ does not belong to $\mathcal{A}$ as every
$G_{\delta}$-set that contains the point $\omega_{1}$ is co-countable.

Somewhat surprisingly, uncountable compact metrizable spaces may or may not
all be pseudo-Aronszajn compacta. They all are under $\mathsf{CH}$ and they
all are not under $\mathsf{MA}+\neg\mathsf{CH}$.

######  Proposition 4.1 ($\mathsf{CH}$).

If $X$ is compact and admits a continuous map $f:X\to\R$ such that $f[X]$ is
uncountable, then $X\in\mathcal{A}$.

###### Proof.

The image $f[X]$ is in $\mathcal{A}$, as witnessed by the family of singleton
subsets. ∎

######  Proposition 4.2 ($\mathsf{MA}+\neg\mathsf{CH}$).

If $X$ is compact, uncountable and hereditarily Lindelöf, then
$X\not\in\mathcal{A}$.

###### Proof.

Let $\mathcal{Z}$ be a witness of the fact that the uncountable compact
hereditarily space $X$ is in $\mathcal{A}$. We will derive a contradiction.

Let $X_{0}=X$ and
$U_{0}=\bigcup_{Z\in\mathcal{Z}}\operatorname{int}_{X_{0}}Z$. There is a
countable subfamily $\mathcal{Z}_{0}$ of $\mathcal{Z}$ such that
$U_{0}=\bigcup_{Z\in{\mathcal{Z}_{0}}}\operatorname{int}_{X_{0}}Z$.

Assume that for some $\alpha<\omega_{1}$, we defined closed sets $X_{\beta}$,
open sets $U_{\beta}$, and subfamilies $\mathcal{Z}_{\beta}$ of $\mathcal{Z}$,
for all $\beta<\alpha$.

Let $V=\bigcup_{\beta<\alpha}U_{\beta}$,
$X_{\alpha}=(\bigcap_{\beta<\alpha}X_{\beta})\setminus V$, and
$\mathcal{S}=\bigcup_{\beta<\alpha}\mathcal{Z}_{\beta}$.

Inside $X_{\alpha}$ let
$W=\bigcup_{Z\in\mathcal{Z}}\operatorname{int}_{X_{\alpha}}(Z\cap
X_{\alpha})$. Then $U_{\alpha}=V\cup W$ is open in $X$, and there is a
countable subcollection $\mathcal{T}$ of $\mathcal{Z}$ such that
$W=\bigcup_{Z\in\mathcal{T}}\operatorname{int}_{X_{\alpha}}(Z\cap
X_{\alpha})$. We let $\mathcal{Z}_{\alpha}=\mathcal{S}\cup\mathcal{T}$.

There is a first $\alpha\in\omega_{1}$ such that $U_{\alpha}=U_{\alpha+1}$. If
$Y=X\setminus U_{\alpha}$ is countable, then we are clearly done. If $Y$ is
uncountable, then for every $Z\in\mathcal{Z}$, the intersection $Z\cap Y$ is
nowhere dense in $Y$. But this contradicts $\mathsf{MA}+\neg\mathsf{CH}$, for
$Y$ is an uncountable compact ccc space with a cover by fewer than
$\mathfrak{c}$ many nowhere dense sets. ∎

One may wonder whether $\mathsf{MA}+\neg\mathsf{CH}$ prevents more compact
spaces from being pseudo-Aronszajn. We have seen that a Souslin line is a
pseudo-Aronszajn compactum and we also know that $\mathsf{MA}+\neg\mathsf{CH}$
implies there are no Souslin lines. Thus we may conjecture that it implies
that there are no pseudo-Aronszajn compacta that are ccc.

However, as there are pseudo-Aronszajn compacta of weight $\aleph_{1}$ one can
construct a compactification $\gamma\N$ of $\N$ with a pseudo-Aronszajn
remainder. That compactification is itself also pseudo-Aronszajn: simply add
the isolated points to the cover of the remainder. Thus we see that
$\mathcal{A}$ contains separable spaces.

We can strengthen the ccc assumption by making it hereditary; it is well known
that having the hereditary ccc is equivalent to every relatively discrete
subspace being countable, see [MR1039321]*Problem 2.7.9(b) for example. Thus,
the hereditary ccc is also a weakening of the hereditary Lindelöf property and
a positive answer to the following question would yield a strengthening of
Proposition 4.2.

######  Question 3.

Does $\mathsf{MA}+\neg\mathsf{CH}$ imply that uncountable compact hereditarily
ccc spaces are not pseudo-Aronszajn?

We remark in passing that it is also unknown whether compact hereditarily ccc
spaces are continuous images of $\N^{*}$, see [hvm2022]*Question 44.

##  5\. Closed copies of $\N$ that are $C^{*}$-embedded but not $C$-embedded

In the model obtained by adding supercompact many Random reals to a model of
$\mathsf{CH}$ every $C^{*}$-embedded subspace of $\R^{\omega_{1}}$ is
$C$-embedded; actually, [MR2823691] establishes much more but this suffices
for the discussion in this section.

We try to construct a closed copy of $\N$ in $\R^{\omega_{1}}$ that is
$C^{*}$-embedded but not $C$-embedded. We know that this is possible when
$\mathsf{CH}$ holds but we are interested in the case where
$\aleph_{1}<\mathfrak{c}$. In the light of the result mentioned above we shall
need some extra assumption besides just $\neg\mathsf{CH}$.

The following assumption allows us to take some steps towards the stated goal.

$None$ $\mathfrak{d}=\mathfrak{u}=\aleph_{1}$

This assumption is consistent with $\neg\mathsf{CH}$; it holds, for example,
in both the iterated and side-by-side Sacks models. Also note that
$\mathfrak{d}=\aleph_{1}<\mathfrak{u}=\mathfrak{c}$ in the Random real model.

We have the following theorem.

######  Theorem 5.1.

Equality $(*)$ implies that there is a closed subset $\BB$ of
$\R^{\omega_{1}}$, of cardinality $\aleph_{1}$, that contains a closed copy of
$\N$ that is $C^{*}$-embedded but not $C$-embedded in $\BB$.

This is not quite what we want: the closed copy of $\N$ is $C^{*}$-embedded in
$\BB$ only.

The rest of this section is devoted to a proof of Theorem 5.1.

The following result should be known and is valid for arbitrary cardinals but
we could not find it in the literature.

######  Theorem 5.2.

Let $X$ be a Tychonoff space. Then $X$ can be embedded into $\R^{\omega_{1}}$
as a closed subspace if and only if it has weight $\aleph_{1}$ and it has a
compactification $K$ such that $K\setminus X$ is the union of $\aleph_{1}$
many $G_{\delta}$-subsets of $K$.

###### Proof.

We work with $s=(0,1)^{\omega_{1}}$ rather than $\R^{\omega_{1}}$ and treat
$s$ as a subspace of the Tychonoff cube $[0,1]^{\omega_{1}}$.

Necessity is easily established: embed $X$ as a closed subspace in $s$ and let
$K$ be its closure in the cube $[0,1]^{\omega_{1}}$. The faces of the cube
provide the cover of $K\setminus X$.

Sufficiency will take some more effort.

Let $\\{A_{\alpha}:\alpha\in\omega_{1}\\}$ be a family of $\aleph_{1}$ many
$G_{\delta}$-sets such that $K\setminus
X=\bigcup_{\alpha\in\omega_{1}}A_{\alpha}$. For all $\alpha$ take
corresponding continuous functions $f_{\alpha}:K\to[\frac{1}{2},1]$ such that
$A_{\alpha}=f^{\leftarrow}(1)$ and let $F:K\to[0,1]^{\omega_{1}}$ be the
diagonal map. Then $F^{\leftarrow}[s]=X$ and so $Y=F[X]$ is closed in $s$, in
addition the restriction $f:X\to Y$ of $F$ to $X$ is perfect.

As $X$ has weight $\aleph_{1}$ we can assume it is already embedded in $s$ and
we can consider $h:X\to s\times s$ defined by $h(x)=\langle{f(x)},{x}\rangle$.
Then $h$ is an embedding, and we shall show that $h[X]$ is closed.

Take $\langle{a},{b}\rangle$ outside $h[X]$. If $a\notin f[X]$ then
$(S\setminus f[X])\times s$ is a neighbourhood of $\langle{a},{b}\rangle$ that
is disjoint from $h[X]$.

If $a\in f[X]$ then $a\neq f(b)$ because $\langle{f(b)},{b}\rangle\in h[X]$.
Because $f$ is perfect we can take a neighbourhood $U_{a}$ of $a$ such that
$b\notin\operatorname{cl}f^{\leftarrow}[U_{a}]$. Let $U$ and $V$ be open in
$s$ such that $U\cap f[X]=U_{a}$, and $b\in V$ and
$V\cap\operatorname{cl}f^{\leftarrow}[U_{a}]=\emptyset$.

Then $\langle{a},{b}\rangle\in U\times V$, and $U\times V$ is disjoint from
$h[X]$. For if $p\in X$ and $\langle{f(p)},{p}\rangle\in U\times V$ then we
get $f(p)\in U_{a}$ and at the same time $p\notin f^{\leftarrow}[U_{a}]$, so
$f(p)\notin U_{a}$. ∎

For the next lemma we recall that $\mathfrak{d}$ can be characterized as the
minimum number of compact sets needed to cover the space of irrational
numbers, see [zbMATH03161747]*3.6 or [MR776622]*Theorem 8.2.

######  Lemma 5.3 ($\mathfrak{d}=\aleph_{1}$).

There are a subset $E$ and a family $\mathcal{K}$ of compact subsets of the
Cantor set $2^{\omega}$ such that $E$ and $\mathcal{K}$ both have cardinality
$\aleph_{1}$, and $2^{\omega}\setminus E=\bigcup\mathcal{K}$.

###### Proof.

Let $D$ be countable and dense in $2^{\omega}$; then its complement is
homeomorphic to the space of irrational numbers. Using the assumption
$\mathfrak{d}=\aleph_{1}$ we can find a family
$\\{L_{\alpha}:\alpha\in\omega_{1}\\}$ of closed sets such that, say,
$\\{L_{n}:n\in\omega_{0}\\}=\\{\\{d\\}:d\in D\\}$ and
$\bigcup\\{L_{\alpha}:\alpha\in\omega_{1}\setminus\omega_{0}\\}=2^{\omega}\setminus
D$. All sets $L_{\alpha}$ are nowhere dense, so no countable subfamily covers
the Cantor set.

Thus $G_{\alpha}=L_{\alpha}\setminus\bigcup_{\beta<\alpha}L_{\beta}$
recursively defines a partition of $2^{\omega}$ into $G_{\delta}$-sets,
uncountably many of which are nonempty. We assume that all $G_{\alpha}$ are
nonempty and take a choice function
$\langle{c}_{\alpha}:\alpha\in\omega_{1}\rangle$ for the family
$\\{G_{\alpha}:\alpha\in\omega_{1}\\}$.

We let $E=\\{c_{\alpha}:\alpha\in\omega_{1}\\}$. Furthermore each set
$G_{\alpha}\setminus\\{x_{\alpha}\\}$ is a $G_{\delta}$-set that is disjoint
from $E$ and, as an image of the space of irrational numbers, a union of
$\aleph_{1}$ many compact sets. These families together form $\mathcal{K}$. ∎

We shall adapt the construction of the plank $\A$ from [dhvmv2022]*Section 5
to construct our space $\BB$.

As in that construction we let $D$ be the tree $2^{<\omega}$ with the discrete
topology and topologize $D\cup 2^{\omega}$ so as to obtain a compactification
$cD$ of $D$ with the Cantor set $2^{\omega}$ as its remainder. We do this by
letting $\\{s\in cD:x\restr n\subseteq s\\}$ be the $n$th basic neighbourhood
of $x\in 2^{\omega}$. We let $e:\beta D\to cD$ be the extension of the
identity map.

The other factor in our plank is the ordinal space $\omega_{1}+1$. We use $I$
to denote the set of isolated points of $\omega_{1}+1$.

We also re-enumerate $E$ as $\langle d_{\alpha}:\alpha\in I\rangle$. Since
$e^{\leftarrow}(d_{\alpha})$ has nonempty interior in $D^{*}$ we can choose a
sequence $\langle x_{\alpha}:\alpha\in I\rangle$ of points in $D^{*}$ such
that $e(x_{\alpha})=d_{\alpha}$ and $\chi(x_{\alpha},\beta D)=\aleph_{1}$ for
all $\alpha\in I$ (here is where we use $\mathfrak{u}=\aleph_{1}$).

We let $P=(\omega_{1}+1)\times\beta D$ and take the subspace

$\BB=\bigl{(}(\omega_{1}+1)\times
D\bigr{)}\cup\bigl{\\{}\langle{\alpha},{x_{\alpha}}\rangle:\alpha\in
I\bigr{\\}}$

As in [dhvmv2022], we write
$T=\bigl{\\{}\langle{\alpha},{x_{\alpha}}\rangle:\alpha\in I\bigr{\\}}$ (the
top line) and $R=\bigl{\\{}\langle{\omega_{1}},{d}\rangle:d\in D\bigr{\\}}$
(the right-hand side).

The right-hand side $R$ is a closed copy of $\N$ in $\BB$.

To see that it is $C^{*}$-embedded we do as in [dhvmv2022]: extend a given
bounded function $g:R\to\R$ horizontally by $g(\alpha,d)=g(\omega_{1},d)$ and
use the unique extension of $d\mapsto g(\omega_{1},d)$ to $\beta D$ to extend
$f$ vertically.

To see that it is not $C$-embedded we define $f:R\to\R$ by
$f(\omega_{1},d)=|d|$. For every potential extension $F$ of $f$ there is an
$\alpha$ such that $F(\beta,d)=f(\omega_{1},d)$ for all $\beta\geq\alpha$ and
all $d\in D$. But then $F$ is unbounded on every neighbourhood of
$\langle{\beta},{x_{\beta}}\rangle$ whenever $\beta\geq\alpha$.

To see that $\BB$ is realcompact and, more precisely, embeddable as a closed
set in $\R^{\omega_{1}}$ we verify that $\BB$ satisfies the conditions in
Theorem 5.2.

The cardinality of $\BB$ is $\aleph_{1}$ and all points have character at most
$\aleph_{1}$, so the weight of $\BB$ is indeed equal to $\aleph_{1}$.

The product $P$ is a compactification of $\BB$. We verify that $P\setminus\BB$
has a cover by $\aleph_{1}$ many $G_{\delta}$-sets of $P$. We let
$\pi_{0}:P\to\omega_{1}+1$ and $\pi_{1}:P\to\beta D$ be the projections.

For each limit ordinal $\alpha$ the set $\\{\alpha\\}\times\beta D$ is a
$G_{\delta}$-set contained in $P\setminus\BB$; these $\aleph_{1}$ many sets
cover $(\omega_{1}\setminus I)\times\beta D$. It remains to cover
$\bigl{(}(\omega_{1}+1)\times D^{*}\bigr{)}\setminus T$.

First we take $\\{(e\circ\pi_{1})^{\leftarrow}[K]:K\in\mathcal{K}\\}$. This
family of $G_{\delta}$-sets covers the complement of
$(e\circ\pi_{1})^{\leftarrow}[E]$ in $(\omega_{1}+1)\times D^{*}$.

Second, for each $\alpha\in I$ we take
$(e\circ\pi_{1})^{\leftarrow}(d_{\alpha})\setminus(\\{\alpha\\}\times\beta
D)$. These $G_{\delta}$-sets cover $(e\circ\pi_{1})^{\leftarrow}[E]$ _except
for_ the union $\bigcup_{\alpha\in I}\bigl{(}\\{\alpha\\}\times
e^{\leftarrow}(d_{\alpha})\bigr{)}$.

Third, for each $\alpha\in I$ we use a local base of cardinality $\aleph_{1}$
at $x_{\alpha}$ consisting of clopen sets to write
$e^{\leftarrow}(d_{\alpha})\setminus\\{x_{\alpha}\\}$ as the union of
$\aleph_{1}$ many $G_{\delta}$-sets.

These four families of $G_{\delta}$-sets cover $P\setminus\BB$ exactly and
their union has cardinality $\aleph_{1}$.

Thus, upon embedding $\BB$ as a closed subset of $\R^{\omega_{1}}$ the right-
hand side $R$ becomes a closed copy of $\N$ in $\R^{\omega_{1}}$ that is not
$C$-embedded, but that is (non-trivially) $C^{*}$-embedded in a closed subset.

This leaves the question whether $\BB$ can be embedded as a closed
$C^{*}$-embedded subset in $\R^{\omega_{1}}$, or at least in such a way that
$R$ becomes $C^{*}$-embedded in the product.

It is clear that we could use a version of Theorem 5.2 that characterizes, or
at least ensures, $C^{*}$-embeddability as well.

## References


A RENORMALIZATION GROUP APPROACH

TO THE FREE CENTRAL LIMIT THEOREM

Jad<EMAIL_ADDRESS>
Mathematical Institute, University of Oxford

Abstract. This short note presents a proof of the Free Central Limit Theorem
for measures with finite third moment using the renormalization group
approach. We construct a contractive renormalization group map over the space
of probability measures with finite third moment, equipped with an
$R-$transform–based metric which has the semi–circle distribution as a fixed
point.

## . ntroduction

Following a recent work of Ott [7] which presents a renormalization group
proof of the classical Central Limit Theorem (CLT), this note shows how the
argument can be adapted to prove the free CLT. As in [7], the proof presented
here is neither claimed to be new nor optimal, and is instead meant as yet
another illustration of the renormalization group approach. The latter was
first applied to the classical CLT in [5], and its use of Fourier–based
metrics preceded the work of Gabetta, Toscani and Wennberg [4], where they
were believed to have been introduced.

The renormalization group approach consists of defining a metric $d$ on a
subspace $\mathcal{Q}_{3}$ of the space of probability measures on
$\mathbf{R}$, for which the renormalization map $T$ fixes the semicircle
distribution $\rho_{\text{SC}}$ and is a contraction on $(\mathcal{Q}_{3},d)$.
A thorough discussion of this method can be found in [7] (along with relevant
bibliography) where it is referred to as the “Banach method”, and the
structure of the proof below is identical to that in said article.

Our definition of $d$ here is inspired by that of the Fourier–based metric
used in the latter, of which an in–depth discussion can be found in [3],
except that ours uses the $R$–transform. Fittingly, our definition of $T$
(first used by Anshelevich in [1], where it is referred to as the central
limit operator) replaces the convolution of measures by their free additive
convolution.

To the best of our knowledge, Anshelevich’s is the only previous work to
interpret the Free CLT as a fixed point theorem, but their approach differs
significantly from the one below. Instead of introducing a metric, they focus
on the problem of linearizing $T$ (computing its Gâteaux derivative), from
which they compute its eigenvalues and show that its eigenfunctions are
absolutely continuous with respect to the Lebesgue measure, with densities
equalling multiples of Chebyshev polynomials of the first kind. As a
corollary, they show that in a subspace of compactly supported measures, the
differential of $T$ evaluated at $\rho_{\text{SC}}$ has spectrum inside the
unit disk (we would then expect $\rho_{\text{SC}}$ to be an attracting fixed
point on that subspace). By contrast, the proof below only relies on a finite
third moment assumption.

## . enormalization Proof of the Free CLT

### 2.1. Preliminaries from Free Probability Theory

Given a non-commutative probability space $(\mathcal{A},\varphi)$ and a (non-
commutative) random variable $a\in\mathcal{A}$, recall that we can define its
algebraic distribution as $\mu_{a}=\int pd\mu_{a}=\varphi(p(a))$ for any
polynomial $p$ of $a$ with complex coefficients. If $\mathcal{A}$ is a
$C^{*}-$algebra, $a$ is self-adjoint and $\varphi$ is positive and has norm 1,
we can extend $\mu_{a}$ from $\mathbf{C}[x]$ to $\mathbf{C}_{b}(\mathbf{R})$
(the space of bounded continuous functions supported in $\mathbf{R}$), making
it into a bona fide probability measure. We say that $a$ is of law $\mu_{a}$
if the moments $\\{m_{n}\\}_{n\geq 1}:=\\{\varphi(a^{n})\\}_{n\geq 1}$ of $a$
coincide with those of $\mu_{a}$ (as a probability measure on $\mathbf{R}$).
The so–called free cumulants $\\{\kappa_{n}\\}_{n\geq 1}$ of $a$ are then
given by the following formula

$m_{k}=\sum_{\pi\in\text{NC}(k),\pi=\\{B_{1},...,B_{n}\\}}\prod_{j=1}^{n}c_{|B_{j}|}$

noting the similarity with the classical moment–cumulant formula, the set of
all partitions having been replaced by that of non–crossing partitions
$\text{NC}(k)$. Here, the $B_{j}$ denote the classes of a partition $\pi$ and
$|B_{i}|$ the number of elements in $|B_{i}|$. Lastly, we define a
semi–circular element to be an element $s\in\mathcal{A}$ whose moments are
those of the semi–circle law, and denote the latter by $\rho$. That is,

$d\rho(x)=\frac{1}{2\pi}\sqrt{(4-x^{2})_{+}}dx$

where $(x)_{+}$ denotes $\max(x,0)$). This distribution can be thought of as
the “free analogue” of the standard normal distribution, as all of its free
cumulants other than the second one equal $0$.

If $\mathcal{A}_{1},\dots,\mathcal{A}_{n}$ is a collection of unital
subalgebras of $\mathcal{A}$, they are said to be freely independent (or
simply free) with respect to $\varphi$ if for any $r\geq 2$ and
$a_{1},\dots,a_{r}\in\mathcal{A}$, we have $\varphi(a_{1}\dots a_{r})=0$
whenever the following conditions are satisfied

  * $\bullet$

$\varphi(a_{i})=0$ for $1\leq i\leq r$

  * $\bullet$

$a_{i}\in\mathcal{A}_{{j_{i}}}$, $1\leq j_{i}\leq n$ for $1\leq i\leq r$

  * $\bullet$

$j_{1}\neq j_{2},j_{2}\neq j_{3},...,j_{r-1}\neq j_{r}$.

In short, the alternating product of centered elements is centered.
Non–commutative random variables are then said to be free if the respective
subalgebras they generate are free.

The key takeaway here is that, much like independence in the classical case,
freeness gives us a rule to evaluate mixed moments of non–commutative random
variables in terms of the individual moments of those variables. We now have
the sufficient background to introduce the Free Central Limit Theorem.

######  Theorem 1 (Free Central Limit Theorem).

Let $(\mathcal{A},\varphi)$ be a non-commutative probability space where
$\mathcal{A}$ is a $C^{*}$–algebra and $\varphi$ is positive and has norm $1$.
Let $\\{a_{i}\\}_{i\geq 1}$ form a sequence of freely independent, self-
adjoint, non-commutative random variables with
$\varphi(a_{i})=0,\varphi(a_{i}^{2})=1$, then

$\frac{1}{\sqrt{N}}\sum_{i=1}^{N}a_{i}\xrightarrow[N\to\infty]{\text{dist}}s$

where $s$ is a semi-circular element and the convergence is in distribution.

######  Remark 1.

Assuming that $\varphi(a_{i}^{2})=\sigma^{2}$ instead of one gives convergence
to a semi–circular element $s_{\sigma}$ of variance $\sigma^{2}$, meaning that
$\varphi(s_{\sigma}^{n})=\sigma^{n}\varphi(s^{n})$.

This theorem was first proved by Voiculescu [9] using operator–theoretic
techniques, and a proof utilizing the combinatorics of non–crossing partitions
was later discovered by Speicher [8].

### 2.2. The Cauchy and $R-$transform of a probability measure

For any self–adjoint $a\in\mathcal{A}$ of law $\mu_{a}$ and any
$z\in\mathbf{C}\setminus\mathbf{R}$, define its Cauchy transform to be the
function $G_{\mu_{a}}=G_{a}(z)=\int_{\mathbf{R}}1/(z-t)d\mu_{a}(t)$, which is
analytic on $\mathbf{C}^{+}$ with range contained in $\mathbf{C}^{-}$ (where
$\mathbf{C}^{\pm}$ denote the upper and lower half-planes). It follows from
the Stieltjes inversion formula that a measure is uniquely determined by its
Cauchy transform. The $R-$transform $R_{\mu_{a}}=R_{a}$ of $\mu_{a}$ is then
defined as the analytic solution to the equation

$G_{a}\left(R_{a}(z)+\frac{1}{z}\right)=z,$

which satisfies the following properties.

######  Theorem 2 (c.f. Ch.3, Th. 17 & 26, [6]).

Let $\mu$ be a probability measure on $\mathbf{R}$ and $R_{\mu}=R$ be its
$R$-transform. Then

  * $\bullet$

If $\mu$ is compactly supported on $[-r,r]$, $R$ is analytic on the disk
$|z|<1/(6r)$ where its has the power series expansion $\sum_{n\geq
1}\kappa_{n}z^{n-1}$ ($\kappa_{n}$ being the $n$-th free cumulant of $\mu$).

  * $\bullet$

If $\mu$ has finite variance $\sigma^{2}$, then $R$ is analytic in the open
disc with centre $-i/(4\sigma)$ and radius $1/(4\sigma)$ (which doesn’t
include $0$). If $\mu$ has moments of all order, its free cumulants
$\kappa_{n}$ are given by an asymptotic expansion at $0$ of $R$.

######  Theorem 3 (c.f. Ch.3, Ex. 16 & Th. 25 [6]).

Let $\mu$ be a probability measure on $\mathbf{R}$ with mean 0, unit variance
and finite third moment $\alpha_{3}$, and denote by $G$ its Cauchy transform.
Then for large enough $y\in\mathbf{R}^{+}$, if $z=iy$, we have

$z-\frac{1}{G(z)}=\frac{1}{z}+\frac{\alpha_{3}}{z^{2}}+o\left(\frac{1}{z^{2}}\right).$

######  Corollary 4.

Let $\mu$ and $\alpha_{3}$ be as in the statement of the previous theorem and
denote by $R$ and $G$ the Stieltjes and $R-$transform of $\mu$, respectively.
Let $F(z):=1/G(z)$. Then for small enough $y>0$ and $z=-iy$,

$R(z):=F^{\langle-1\rangle}\left(\frac{1}{z}\right)-\frac{1}{z}=z+\alpha_{3}z^{2}+\omega(z^{2})$

where the asymptotic notation $\omega$ refers to a function $f(z)$ satisfying
$|f(z)/z^{2}|\to\infty$ when $y\to\infty$.

We also state the following scaling property, which follows directly from the
definition of the $R$–transform.

######  Lemma 5.

Let $a\in\mathcal{A}$ and $\lambda\in\mathbf{C}$, then $R_{\lambda
a}(z)=\lambda R_{a}(\lambda z)$ for any $z$ in the domain of $R$.

If $a,b$ are freely independent random variables of law $\mu_{a},\mu_{b}$, the
law of $a+b$ is the so-called free additive convolution of these two measures
which is usually denoted by $\mu_{a}\boxplus\mu_{b}$. The primary interest of
the $R$–transform is that it linearizes the latter, much like the logarithm of
the Fourier transform in the classical case, meaning that
$R_{a+b}(z)=R_{a}(z)+R_{b}(z)$. Using this fact with lemma 5 and equating free
cumulants (which we recall are $\kappa_{n}=\mathbf{1}_{[n=1]}$ for the semi-
circle law), we obtain the following.

######  Lemma 6.

Let $s_{1}$ and $s_{2}$ be free semi-circular elements of $\mathcal{A}$. Then

$R_{(s_{1}+s_{2})/\sqrt{2}}(z)=R_{\rho}(z)=z$

for any $z\in\mathbf{C}$.

Lastly, we note that pointwise convergence of Cauchy transforms of probability
measures implies weak convergence of the measures in question. For a proof of
this result, details regarding the notions discussed above and an excellent
introduction to free probability, we refer the reader to the first three
chapters of [6].

### 2.3. Choice of Metric and Renormalization Transform

We denote by $\mathcal{P}_{3}=\mathcal{P}_{3}(\mathbf{R})$ the space of
probability measures on $\mathbf{R}$ with finite second moment, and by
$\mathcal{Q}_{3}$ the subspace of those measures which are centered and
reduced:

$\mathcal{Q}_{3}:=\\{\mu\in\mathcal{P}_{3}\,:\,\varphi(a)=0,\,\,\varphi(a^{2})=1\\}.$

We equip $\mathcal{Q}_{3}$ with the distance

$d(\nu,\mu)=\sup_{0<y\leq 1/4}\frac{|R_{\mu}(iy)-R_{\nu}(iy)|}{|y|^{2}}.$

######  Lemma 7.

$d$ is a finite metric on $\mathcal{Q}_{3}$, where convergence in the metric
topology implies weak convergence of measures.

###### Proof.

The symmetry of $d$ is clear, and separation follows from the identity theorem
and the fact that probability measures are uniquely determined by their
$R-$transform. The triangle inequality follows from $\sup f+g\leq\sup f+\sup
g$ and the triangle inequality for the complex norm.

By theorem 2, the $R$–transform of a probability measure with finite variance
is bounded on $\\{iy:y\in(\epsilon,1/4)\\}$ for any $\epsilon>0$. To show that
$d$ is finite, it therefore suffices to show that

$\lim_{y\to 0^{+}}\frac{|R_{\mu}(-iy)-R_{\nu}(-iy)|}{|y|^{2}}<\infty$

for any $\mu,\nu\in\mathcal{Q}_{3}$, which follows immediately from 4.

Lastly, $d$ metrizes weak convergence since pointwise convergence of Cauchy
transforms implies weak convergence, and the former follows from the pointwise
convergence of $R$–transforms by continuity. ∎

Let $a\in\mathcal{A}$ be such that $\mu_{a}\in\mathcal{Q}_{3}$, and define our
renormalization map $T$ to be the map
$\mu_{a}\mapsto\mu_{(a+a^{\prime})/\sqrt{2}}$, where $a^{\prime}$ is a
(freely) independent copy of $a$. In other words, $T$ maps $\mu_{a}$ to the
renormalized free additive convolution of $\mu_{a}$ with itself. Then the
following is clear.

######  Lemma 8.

$T\mu\in\mathcal{Q}_{3}$ if $\mu\in\mathcal{Q}_{3}$.

Furthermore, lemma 6 can be stated equivalently as $T\rho=\rho$ (recall that
$\rho$ is the semi–circle law), and for any $\mu\in\mathcal{Q}_{3}$,
$R_{T\mu}(z)=\sqrt{2}R_{\mu}(z/\sqrt{2})$ by lemma 5 and the additivity of the
$R$–transform.

######  Theorem 9.

$T$ is a contraction on $(\mathcal{Q}_{3},d)$ with contraction constant
$2^{-1/2}$, and

$d(T^{n}\nu,\rho)\leq 2^{-n/2}d(\nu,\rho).$

###### Proof.

Recall that the $R-$transform of $T\nu$ is
$R_{T\nu}(z)=\sqrt{2}R_{\nu}(z/\sqrt{2})$, thus

$\displaystyle d(T\nu,T\mu)$ $\displaystyle=\sup_{0<y\leq
1/4}\frac{\sqrt{2}}{2}\frac{|R_{\mu}(-iy/\sqrt{2})-R_{\nu}(-iy/\sqrt{2})|}{|y/\sqrt{2}|^{2}}$
$\displaystyle=\sup_{0<x\leq
1/(4\sqrt{2})}\frac{1}{\sqrt{2}}\frac{|R_{\mu}(-ix)-R_{\nu}(-ix)|}{|x|^{2}}$
$\displaystyle\leq\frac{d(\nu,\mu)}{\sqrt{2}}.$

The second part of the claim then follows from the fact that $T$ fixes $\rho$.
∎

######  Remark 2.

The metric $d$ is said to be 3–ideal [10], meaning that for any
$\mu_{a},\mu_{b}\in\mathcal{Q}_{3}$ and constant $\lambda\in\mathbf{C}$,
$d(\mu_{\lambda a},\mu_{\lambda b})\leq\lambda^{3}d(\mu_{a},\mu_{b})$. More
generally, if we let $d_{s}$ be the generalization of $d$ with $|z|^{s}$ in
the denominator, then $d_{s}$ is $(s+1)$–ideal. This is in contrast with the
analogous Fourier metric (see [7]) which is only $s$–ideal, as we are aided
here by the additional factor of $\lambda$ incurred by lemma 5. Indeed, 9
still holds with $s=1+\epsilon$ instead of $2$, but we chose the latter for
aesthetic reasons.

######  Remark 3.

Nothing is said here about the completeness of $(\mathcal{Q}_{3},d)$, which is
not needed for our proof of the Free CLT. To the author’s best knowledge, such
$R$–transform based metrics have yet to be studied, in contrast with the
family of Fourier–based metrics in [7] which are well–understood [3].

### 2.4. Proof of the Free CLT (with finite third moment assumption)

Let $(A,\varphi)$, $\\{a_{i}\\}_{i}$ and $\mu$ be as in the statement of
theorem 1, and assume additionally that $\mu$ has finite third moment. Then
$\mu\in\mathcal{Q}_{3}$, and noting that

$S_{n}=\frac{1}{2^{n/2}}\sum_{i=1}^{2^{n}}a_{i}\equiv\frac{1}{\sqrt{N}}\sum_{i=1}^{N}a_{i}$

has law $T^{n}\mu$, the claim along any geometric subsequence follows from
theorem 9, lemma 8 and the fact that $d$ is finite. As explained in [7], we
can extend this to arbitrary sequences using the following properties of $d$,
which follow directly from its definition: if $a,a^{\prime},b$ and
$b^{\prime}$ are free with laws in $\mathcal{Q}_{c}$ and
$\lambda\in\mathbf{C}$, then

$\displaystyle\,d(\mu_{a+a^{\prime}},\mu_{b+b^{\prime}})$ $\displaystyle\leq
d(\mu_{a},\mu_{b})+d(\mu_{a^{\prime}},\mu_{b^{\prime}}),$ (1)
$\displaystyle\,d(\mu_{\lambda a},\mu_{\lambda b})$
$\displaystyle\leq\lambda^{3}d(\mu_{a},\mu_{b}),$ (2)
$\displaystyle\,d(\mu_{S},\rho)$
$\displaystyle\leq\frac{d(\mu_{a},\rho)}{n^{1/2}}$ (3)

where $S=n^{-1/2}\sum_{i=1}^{n}a$.

## cknowledgements

The author would like to thank his advisor, Prof. Jon Keating, for bringing
Ott’s article to his attention and for raising the question of whether the
renormalization group approach could be used for the Free CLT. Many thanks are
also due to Dr. Ott himself for his valuable feedback on this work, as well as
Profs. Roland Speicher and Michael Anshelevich for their comments (in
particular for referring me to the latter’s work).

The author is supported by the EPSRC Centre for Doctoral Training in
Mathematics of Random Systems: Analysis, Modelling and Simulation
(EP/S023925/1)

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