Array convergence of functions
of the first Baire-class



HELMUT KNAUST
Department of Mathematics
Oklahoma State University
Stillwater, OK 74078

Abstract
We show that every array $(x(i,j):1\leq i<j< \infty )$ of elements in a pointwise compact subset of the Baire-1 functions on a Polish space, whose iterated pointwise limit $\lim_i \lim_j x(i,j)$ exists, is converging Ramsey-uniformly. An array (x(i,j)_i<j) in a Hausdorff space ${\cal T}$ is said to converge Ramsey-uniformly to some x in ${\cal T}$, if every subsequence of the positive integers has a further subsequence (m_i) such that every open neighborhood U of x in ${\cal T}$ contains all elements x(m_i,m_j) with i<j except for finitely many i.

AMS(MOS) Subject classification (1980). Primary 46E15; Secondary 46B20.
Key words and phrases. first Baire-class, array convergence, Ramsey theory

1. Introduction

It is a well known consequence of Ramsey's Theorem that every array (a_ij)_i<j of real numbers with $\lim_i \lim_j a_{ij} =a$ for some $a\in {\rm I \hspace{-.1567em} R}$ has the following property: There is a subsequence (m_i) so that for all $\epsilon>0$ there is an $n \in {\rm I \hspace{-.1567em} N}$such that $\vert a_{m_im_j}-a\vert < \epsilon$ for all n<m_i<m_j. This result generalizes easily to Hausdorff spaces which satisfy the first countability axiom.

The purpose of our note is to show that a corresponding result holds for the space of functions of the first Baire-class ${\cal B}_1(\Omega)$ on a Polish space $\Omega$, given the topology of pointwise convergence.

Let us say that an array $(x(i,j):1\leq i<j< \infty )$ of elements in a Hausdorff space ${\cal T}$ converges Ramsey-uniformly to some $x \in {\cal T}$, if every subsequence of ${\rm I \hspace{-.1567em} N}$ has a further subsequence (m_i) such that for every open neighborhood U of x in ${\cal T}$ there is an $n \in {\rm I \hspace{-.1567em} N}$ so that $x(m_i,m_j) \in U$ for all n<m_i<m_j.

With this notation we can state our main result as follows:

Theorem 1   Let $\Omega$ be a Polish space and let K be a pointwise compact subset of ${\cal B}_1(\Omega)$. If x and (x(i,j))_i<j are elements in K with $\lim_i\lim_j x(i,j)=x$, then (x(i,j)) converges Ramsey-uniformly to x.

A topological space $\Omega$is Polish, if it is homeomorphic to a complete separable metric space. A real-valued function is of the first Baire-class on $\Omega$, if it is the pointwise limit of a sequence of continuous functions on $\Omega$.

It is a fundamental result of Bourgain, Fremlin and Talagrand [2] that ${\cal B}_1(\Omega)$ is an angelic space, if $\Omega$ is Polish. A Hausdorff space ${\cal T}$ is angelic, if for every relatively compact subset A of ${\cal T}$ each point in the closure of A is the limit of a sequence in A and if relatively countably compact sets in ${\cal T}$ are relatively compact. In angelic spaces the notions of (relative) compactness, (relative) countable compactness and (relative) sequential compactness coincide. Further basic results about angelic spaces can be found in [7].

Theorem 1 strengthens -- in the case of functions of the first Baire-class on a Polish space -- a result of Boehme and Rosenfeld [1, Theorem 1], which we phrase for our purposes as follows:

Lemma 2   Let ${\cal T}$ be an angelic space, and let x and (x(i,j))_i<j be elements in a compact subset of ${\cal T}$ with $\lim_i\lim_j x(i,j)=x$. Then there is a subsequence (m_i) of ${\rm I \hspace{-.1567em} N}$ with $\lim_k x(m_{2k-1},m_{2k})=x$.

Lemma 2 was also obtained independently -- in the ${\cal B}_1(\Omega)$-setting -- by Rosenthal [8, Theorem 3.6].

From Theorem 1 and a result by Odell and Rosenthal [6] we obtain the following Banach space corollary:

Corollary 3   Let X be a separable Banach space not containing $\ell_1$. If x^** and (x^**(i,j))_i<j are elements in a bounded subset of X^** with $\omega^*$- $\lim_i\omega^*$- $\lim_j x^{**}(i,j)=x^{**}$, then (x^**(i,j)) converges Ramsey-uniformly to x^** in the $\omega^*$-topology.

The proof of Theorem 1 utilizes Lemma 2 to extract ``nice'' converging subsequences out of the given array (x(i,j)). We use Ramsey theory to produce the subarray for which one obtains Ramsey-uniform convergence.

If M is an infinite subset of ${\rm I \hspace{-.1567em} N}$, ${\cal P}^{\infty}({M})$ will denote the set of all infinite subsets of M. We give ${\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}})$ the topology, which is inherited by considering ${\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}})$ as a subspace of $\{0,1\}^{{\rm I \hspace{-.1567em} N}}$ endowed with the product topology.

A subset ${\cal A}\subset {\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}})$ is called a Ramsey set, if for all $L \in {\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}})$ there is an $M \in {\cal P}^{\infty}({L})$ such that either ${\cal P}^{\infty}({M}) \subset {\cal A}$ or ${\cal P}^{\infty}({M}) \cap {\cal A}=\emptyset$. It is known that analytic (and coanalytic) subsets of ${\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}})$ are Ramsey sets [3,9]. For a proof of this result, some history and more general results see [5].

I would like to thank D. Alspach, E. Odell and H. P. Rosenthal for useful discussions.

2. Proof

Proof of Theorem 1: Let x and x(i,j) with $1 \leq i<j< \infty$ be elements in K such that $\lim_i\lim_j x(i,j)=x$. We let

$${\cal A}=\left\{ M=(m_i) \in {\cal P}^{\infty}({{\rm I \hspac... ..._{2k}))_{k=1}^{\infty} \mbox{ is pointwise convergent}\right\}.$$

Lemma 4   ${\cal A}$ is coanalytic.

We postpone the proof of the lemma and proceed with the proof of the theorem. Since ${\cal A}$ is coanalytic, ${\cal A}$ is a Ramsey set. Let $L \in {\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}})$. We can thus find $M=(m_i)_{i=1}^{\infty} \in {\cal P}^{\infty}({L})$ so that ${\cal P}^{\infty}({M}) \subset {\cal A}$ or ${\cal P}^{\infty}({M}) \cap {\cal A}=\emptyset$. Lemma 2 shows that the first alternative holds. Moreover, Lemma 2 asserts that $\lim_k x(m_{2k-1}^{\prime},m_{2k}^{\prime})=x$ for some $M^{\prime}= (m_i^{\prime}) \in {\cal P}^{\infty}({M})$.

Suppose now the conclusion of Theorem 1 fails for $M^{\prime}$. Then there is an open neighborhood U of x and a subsequence $M^{\prime\prime} \subset M^{\prime}$ with

$$x(m^{\prime\prime}_{2k-1},m^{\prime\prime}_{2k})\not\in U \mbox{ for all } k \in {\rm I \hspace{-.1567em} N}.$$

Since $M^{\prime\prime} \in {\cal P}^{\infty}({M})$, we have $M^{\prime\prime} \in {\cal A}$ and thus $\lim_k x(m^{\prime\prime}_{2k-1},m^{\prime\prime}_{2k}) =y$ for some $y \in {\cal B}_1(\Omega)$. Note that $y \not=x$.

We now construct a subsequence $N=(n_i)\in{\cal P}^{\infty}({M})$ inductively as follows: Let $n_1=m_1^{\prime}$ and $n_2=m_2^{\prime}$. Once $n_1,n_2,\ldots,n_{2k}$ have been chosen, we define n_2k+1 and n_2k+2 as follows: If k is odd, we choose an $\ell \in {\rm I \hspace{-.1567em} N}$ so that $m^{\prime\prime}_{2\ell -1}> n_{2k}$ and let $n_{2k+1}=m^{\prime\prime}_{2\ell -1},n_{2k+2}=m^{\prime\prime}_{2 \ell}$. If k is even, we can find an $\ell \in {\rm I \hspace{-.1567em} N}$ with $m_{2\ell -1}^{\prime}>n_{2k}$ and then let $n_{2k+1}=m_{2 \ell -1}^{\prime},n_{2k+2}=m_{2\ell}^{\prime}$. On the one hand the sequence (x(n_2k-1,n_2k)) is pointwise convergent, on the other hand it contains two subsequences converging to x and y respectively. This yields a contradiction.
Proof of Lemma 4: The proof of Lemma 4 uses techniques similar to those employed in [10].

Let Y be the set of all real-valued arrays (a(i,j))_i<j, endowed with the topology of pointwise convergence. We set $Z={\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}}) \times Y$ and denote by $\phi:\Omega\longrightarrow Y$ the canonical map defined by $\phi(\omega)=(x(i,j)(\omega))_{i<j}$. Since $\phi$ is a Borel-measurable map and $\Omega$ is Polish, $\phi(\Omega)$ is analytic in Y (see [4, §38]). Consequently $Z_1:={\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}}) \times \phi(\Omega)$ is analytic in Z.

We define a set $Z_2 \subset {\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}}) \times Y$ as follows:

$$Z_2=\left\{(M,(a(i,j))):\ (a(m_{2k-1},m_{2k}))_{k=1}^{\infty} \mbox{ is not Cauchy } \right\}$$

Observing that the set

$$\begin{eqnarray*}Z_2^{\ell,N}&:=&\left\{ (M,(a(i,j))):\mbox{ there are $k_1,k_2>... ...2k_1-1},m_{2k_1})-a(m_{2k_2-1},m_{2k_2})\vert>2^{-\ell} \right\} \end{eqnarray*}$$

is open, and that

$$Z_2= \bigcup_{\ell \in {\rm I \hspace{-.1567em} N}}\bigcap_{N \in {\rm I \hspace{-.1567em} N}} Z_2^{\ell,N},$$

we obtain that Z₂ is a $G_{\delta\sigma}$-set in Z.

Consequently $Z_1 \cap Z_2$ is analytic in Z. We let $P:Z \longrightarrow {\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}})$ be the projection of Z onto its first coordinate. One can see easily that the complement of ${\cal A}$ is equal to $P(Z_1 \cap Z_2)$. Thus ${\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}}) \setminus {\cal A}$ is analytic in ${\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}})$ as the continuous image of an analytic set in Z (see [4, §38]).
Problem: Does Theorem 1 hold for arbitrary angelic spaces? Lemma 2 reduces this problem to the apparently open question, whether the set ${\cal A}\subset {\cal P}^{\infty}({{\rm I \hspace{-.1567em} N}})$, defined at the beginning of the proof, is still a Ramsey set for arbitrary angelic spaces.

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