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 $\T$ is said to converge Ramsey-uniformly to some $x$ in $\T$, if every subsequence of the positive integers has a further subsequence $(m_i)$ such that every open neighborhood $U$ of $x$ in $\T$ contains all elements $x(m_i,m_j)$ with $i < j$ except for finitely many $i$.
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 \real $ has the following property: There is a subsequence $(m_i)$ so that for all $\epsilon>0$ there is an $n \in \nat$ such that $|a_{m_im_j}-a| < \epsilon$ 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 $\bone$ on a Polish space $\Om$, 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 $\T$ converges Ramsey-uniformly to some $x \in \T$, if every subsequence of $\nat$ has a further subsequence $(m_i)$ such that for every open neighborhood $U$ of $x$ in $\T$ there is an $n \in \nat$ 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 $\Om$ be a Polish space and let $K$ be a pointwise compact subset of $\bone$. 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 $ \Om$ is Polish, if it is homeomorphic to a complete separable metric space. A real-valued function is of the first Baire-class on $\Om$, if it is the pointwise limit of a sequence of continuous functions on $\Om$. It is a fundamental result of Bourgain, Fremlin and Talagrand [BFT] that $\bone$ is an angelic space, if $\Om$ is Polish. A Hausdorff space $\T$ is angelic, if for every relatively compact subset $A$ of $\T$ each point in the closure of $A$ is the limit of a sequence in $A$ and if relatively countably compact sets in $\T$ 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 [P]. Theorem 1 strengthens - in the case of functions of the first Baire-class on a Polish space - a result of Boehme and Rosenfeld [BR], which we phrase for our purposes as follows:
Lemma 2. Let $\T$ be an angelic space, and let $x$ and $(x(i,j))_{i < j}$ be elements in a compact subset of $\T$ with $\lim_i\lim_j x(i,j)=x$. Then there is a subsequence $(m_i)$ of $\nat$ with $\lim_k x(m_{2k-1},m_{2k})=x$.
Lemma 2 was also obtained independently - in the $\bone$-setting - by Rosenthal [R]. From Theorem 1 and a result by Odell and Rosenthal [OR] 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 $\nat$, $\pinf{M}$ will denote the set of all infinite subsets of $M$. We give $\pinf{\nat}$ the topology, which is inherited by considering $\pinf{\nat}$ as a subspace of $\{0,1\}^{\nat}$ endowed with the product topology. A subset $\A \subset \pinf{\nat}$ is called a Ramsey set, if for all $L \in \pinf{\nat}$ there is an $M \in \pinf{L}$ such that either $\pinf{M} \subset \A$ or $\pinf{M} \cap \A =\emptyset$. It is known that analytic (and coanalytic) subsets of $\pinf{\nat}$ are Ramsey sets [E,S]. For a proof of this result, some history and more general results see [O].
I would like to thank D. Alspach, E. Odell and H. P. Rosenthal for useful discussions.
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 $\A=\left\{ M=(m_i) \in \pinf{\nat} :\ (x(m_{2k-1},m_{2k}))_{k=1}^{\infty} \mbox{ is pointwise convergent}\right\}$.
Lemma 4. $\A$ is coanalytic.
We postpone the proof of the lemma and proceed with the proof of the theorem. Since $\A$ is coanalytic, $\A$ is a Ramsey set. Let $L \in \pinf{\nat}$. We can thus find $M=(m_i)_{i=1}^{\infty} \in \pinf{L}$ so that $\pinf{M} \subset \A$ or $ \pinf{M} \cap \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 $M^{\prime}= (m_i^{\prime}) \in \pinf{M}$. 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 \[ x(m^{\prime\prime}_{2k-1},m^{\prime\prime}_{2k})\not\in U \mbox{ for all } k \in \nat .\] Since $M^{\prime\prime} \in \pinf{M}$, we have $M^{\prime\prime} \in \A$ and thus $\lim_k x(m^{\prime\prime}_{2k-1},m^{\prime\prime}_{2k}) =y$ for some $y \in \bone $. Note that $y \not=x$. We now construct a subsequence $N=(n_i)\in\pinf{M}$ 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 $\ell \in \nat$ 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 $\ell \in \nat$ 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 [St].
Let $Y$ be the set of all real-valued arrays
$(a(i,j))_{i < j}$, endowed with the topology of pointwise convergence.
We set $Z=\pinf{\nat}
\times Y$ and denote by $\phi:\Om \longrightarrow Y$ the canonical map
defined by $\phi(\omega)=(x(i,j)(\omega))_{i < j}$. Since $\phi$ is a
Borel-measurable map and $\Om$ is Polish, $\phi(\Om)$ is analytic in $Y$
(see [Ku]). Consequently $Z_1:=\pinf{\nat} \times \phi(\Om)$ is
analytic in $Z$.
We define a set $Z_2 \subset \pinf{\nat} \times Y$ 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 $k_1,k_2>N$ with }
\right.\\
& & \left.|a(m_{2k_1-1},m_{2k_1})-a(m_{2k_2-1},m_{2k_2})|>2^{-\ell}
\right\}
\end{eqnarray*}
is open, and that
\[Z_2= \bigcup_{\ell \in \nat}\bigcap_{N \in \nat} Z_2^{\ell,N},\]
we obtain that $Z_2$ is a $G_{\delta\sigma}$-set in $Z$.
Consequently $Z_1 \cap Z_2$ is analytic in $Z$. We let $P:Z \longrightarrow
\pinf{\nat}$ be the projection of $Z$ onto its first coordinate. One can see
easily that the complement of $\A$ is
equal to $P(Z_1 \cap Z_2)$. Thus
$\pinf{\nat} \setminus \A$ is analytic in $\pinf{\nat}$ as the continuous image of an analytic
set in $Z$ (see [Ku]).
Problem: Does Theorem 1 hold for arbitrary angelic spaces?
Lemma 2 reduces this problem to the apparently open question, whether the set $\A \subset \pinf{\nat}$, defined at the beginning of the proof, is still a Ramsey set for arbitrary angelic spaces.
In: Proceedings of the American Mathematical Society, Volume 112, Issue 2 (June 1991), 529-532.