Extracting a common convergent indexing from an uncountable family of sequences

Question

Let $\mathcal{A}$ be some uncountable index set and $X$ be some separable reflexive Banach space.

For each $\alpha \in \mathcal{A}$, let \begin{equation} \{ x_n^{\alpha} \}_{n=1}^\infty \end{equation} be a sequence bounded in the norm of $X$ by some $C_\alpha>0$ for all $n \in \mathbb{N}$. Then, Banach-Alaoglu Theorem implies that there exists a weakly convergent subsequence of $\{x_n^{\alpha} \}_{n=1}^\infty$.

Is it possible to find a "common" strictly increasing mapping $k \in \mathbb{N} \to n(k) \in \mathbb{N}$ such that \begin{equation} \{ x_{n(k)}^{\alpha} \}_{k=1}^\infty \end{equation} is weakly convergent in $X$ for all $\alpha \in \mathcal{A}$? Of course, weak limits may vary w.r.t $\alpha$.

If $\mathcal{A}$ is countable, this seems possible by inductive argument. However, I do not see what would happen when $\mathcal{A}$ is uncountable.

Answer 1

There is a simple counterexample $\mathcal{A}$ of size continuum. To see this, observe that there are only continuum many subsequence indexing functions $\alpha:k\mapsto \alpha(k)$, and for each such $\alpha$, we may pick a sequence $\{x^\alpha_n\}_n$ in such a way that $\{x^\alpha_{\alpha(k)}\}_k$ is not convergent. That is, we define the $\alpha$th sequence specifically so that $\alpha$ does not work with it. Because of this, there can be no common subsequence indexing $\alpha$ for this family of sequences, since every $\alpha$ fails with the $\alpha$th sequence $\{x^\alpha_n\}_n$.

Beyond this, in the case that the continuum hypothesis fails, one might hope for a smaller counterexample. We can introduce a natural cardinal characteristic here, namely, the smallest size of a set $\mathcal{A}$ admitting no such common subsequence indexing. Let us call it the common convergence number, denoted 𝕔. So we've established $$\aleph_1\leq 𝕔\leq 2^{\aleph_0}.$$ Perhaps this cardinal will be provably equal to one of the other better known cardinal characteristics. The fact that subsequences of convergent sequences are also convergent seems to suggest that it could be consistent with ZFC that the smallest counterexample could be less than continuum.

In light of the counterexample, we seem led naturally to the question:

Question. How large is the smallest counterexample? In other words, how big is 𝕔?

It makes sense to me to consider 𝕔 as defined with respect to real sequences only, but I am unsure whether there is any dependence on the underlying space.

(I encourage others to post further answers about this on this same question thread.)

As a first effort in this direction, let me prove the following.

Theorem. The common convergence number is at most the splitting number $\frak{s}$. $$ 𝕔\leq \frak{s}$$ Indeed, there is a family $\mathcal{A}$ of $\frak{s}$ many binary sequences with no common convergent subsequence indexing.

Proof. Suppose that $S$ is a splitting family, which means that elements of $S$ are infinite subsets of $\mathbb{N}$, such that for every infinite set $b\subseteq\mathbb{N}$ is split by some $a\in S$, which means that $b-a$ and $b\cap a$ are both infinite. The splitting number $\frak{s}$ is the size of the smallest such splitting family.

Given $S$, let $x^a_n$ be $0$ or $1$ depending on whether $n\in a$. I claim there is no common convergent subsequence indexing $\alpha$. For any proposed $\alpha$, let $b=\text{ran}(\alpha)$. So there is some $a\in S$ with $b-a$ and $b\cap a$ both infinite. It follows that the subsequence of $\{x^a_n\}_n$ defined by $\alpha$ will not converge, since the values of $x^a_{\alpha(k)}$ will be infinitely often $0$ and infinitely often $1$. So this family is a counterexample to common convergence of size $\frak{s}$. Thus, the smallest such counterexample is at most $\frak{s}$. $\Box$

It is known that the splitting number can be $\aleph_1$, even when the continuum is large.

Let $𝕔_{\{0,1\}}$ be the common convergence number, when defined for 0/1-valued sequences only.

Theorem. $𝕔_{\{0,1\}}$ is exactly the splitting number $\frak{s}$.

Proof. The counterexample provided above shows that $𝕔_{\{0,1\}}\leq\frak{s}$. For the converse, suppose that we have a family $\mathcal{A}$ of 0/1-valued sequences, with $\mathcal{A}$ of size less than $\frak{s}$. We may regard these sequences as characteristic functions of sets, and so there is an infinite set $b\subseteq\mathbb{N}$ that is either almost contained in or almost disjoint from every set arising from $\mathcal{A}$. It follows that $b$ can be used for a common convergent subsequence indexing. So every binary family of fewer than the splitting number of sequences does admit a common convergent subsequence indexing. $\Box$

Of course, $𝕔\leq 𝕔_{\{0,1\}}$, since there might be a smaller counterexample with nonbinary sequences.

Answer 2

Using notation by the answer by Hamkins, I'll prove that $\mathbb c_{\mathbb R}\ge\min\{\mathfrak s,\mathfrak b\}$.

Thanks to Theorem 8.11 of Halbeisen's book Combinatorial Set Theory, we know that $\min\{\mathfrak s,\mathfrak b\}$ is equal to the smallest cardinality of any family $\mathcal P$ of maps from $[\omega]^2$ to $2$ such that there is no infinite subset $H\subseteq\omega$ that is almost homogeneous for all maps in $\mathcal P$. Let us fix $\lambda<\min\{\mathfrak s,\mathfrak b\}$.

Assume we have some bounded sequences $\langle x_n^\alpha: n<\omega\rangle$ with values in $\mathbb R$ for all $\alpha<\lambda$. We want to prove that we can make all of them converge with the same subindex set. For all $\alpha<\lambda$, let's define $f_\alpha(\{n,m\}):=0$ if $n<m$ and $x_n^\alpha<x_m^\alpha$ and $f_\alpha(\{n,m\}):=1$ if $n<m$ and $x_n^\alpha\ge x_m^\alpha$.

Then the family $\mathcal P:=\{f_\alpha: \alpha<\lambda\}$ has cardinality $\lambda$, hence there is an infinite subset $H\subseteq\omega$ that is almost homogeneous for all maps in $\mathcal P$. Let $\sigma:\omega\to H$ be the (only) order-preserving map.

We want to prove that $\langle x^\alpha_{\sigma(k)}:k<\omega\rangle$ converges for all $\alpha<\lambda$. So fix some $\alpha<\lambda$.

If $f_\alpha(\{n,m\})=0$ for almost all $\{n,m\}\in [H]^2$, then this means that for almost all $a<b<\omega$ we have $x^\alpha_{\sigma(a)}<x^\alpha_{\sigma(b)}$ and, since the sequence $\langle x^\alpha_n:n<\omega\rangle$ was bounded, the subsequence $\langle x^\alpha_{\sigma(k)}:k<\omega\rangle$ is a bounded increasing sequence, so it converges. If instead $f_\alpha(\{n,m\})=1$ for almost all $\{n,m\}\in [H]^2$, then this means that for almost all $a<b<\omega$ we have $x^\alpha_{\sigma(a)}\ge x^\alpha_{\sigma(b)}$ and, since $\langle x^\alpha_n:n<\omega\rangle$ is bounded, the subsequence $\langle x^\alpha_{\sigma(k)}:k<\omega\rangle$ is a bounded nondecreasing sequence, so it converges.

So, this used some specific properties of $\mathbb R$, like the ordering, but I guess it could be generalized to Banach spaces or, as I said in the comments to Hamkins's answer, even to generic convergence spaces. The question whether $\mathbb c=\mathbb c_{\mathbb R}$ is still open and interesting. Another interesting fact would be trying to prove $\mathbb c_{\mathbb R}=\min\{\mathfrak s,\mathfrak b\}$, that would be equivalent to $\mathfrak b\ge\mathbb c_{\mathbb R}$.

I would love to write more conclusions but where I am it is already past 2 a.m. so I should rest. But probably tomorrow or in the next few days I'll try looking more into detail.

Answer 3

We prove that $\mathbb c=\mathbb c_{\mathbb R}$: clearly $\mathbb c\le\mathbb c_{\mathbb R}$ holds, hence it is enough to prove $\mathbb c\ge\mathbb c_{\mathbb R}$. This means: for all reflexive separable Banach spaces $X$, given $\lambda$-many bounded sequences $\langle x_n^\alpha:n<\omega\rangle$ for $\alpha<\lambda$, if $\lambda<\mathbb c_{\mathbb R}$ we can make them all converge with the same index set.

So, if $X$ is separable and reflexive, the topological dual $X^\ast$ is separable as well. Let $D\subseteq X^\ast$ be dense and countable, so $D=\{d_k:k<\omega\}$. Now, we can assume that $\|x_n^\alpha\|_{X}\le1$ for all $n<\omega$ and all $\alpha<\lambda$, since the sequences are bounded so, for any fixed $\alpha$, there is a bound $b_\alpha$ such that $\|x_n^\alpha\|_{X}\le b_\alpha$ for all $n<\omega$, and notice that the sequence $\langle x_{\sigma(k)}^\alpha:k<\omega\rangle$ converges if and only if $\langle x_{\sigma(k)}^\alpha\cdot (b_\alpha)^{-1}:k<\omega\rangle$ converges. So we can assume that all the sequences are bounded by the same constant, i.e. 1.

(This is basically equivalent, in the $\mathbb R$eal case, at assuming that all the sequences are contained in $[0,1]$. It's not a necessary hypothesis, probably the proof still works without assuming it, but... life is already complicated, we can simplify something sometimes).

Now, for all $k<\omega$, $n<\omega$ and $\alpha<\lambda$ it holds $|d_k(x_n^\alpha)|\le \|d_k\|_{X^\ast}\cdot \|x_n^\alpha\|_{X} \le \|d_k\|_{X^\ast}$. Now we prove something by induction on $k<\omega$. Let us call $\delta_k:=\|d_k\|_{X^\ast}>0$ to be shorter.

Consider the sequences $\langle d_0(x_n^\alpha):n<\omega\rangle$ for $\alpha<\lambda$. These are $\lambda$-many sequences of real numbers in $[-\delta_0,\delta_0]$, so, since $\lambda<\mathbb c_{\mathbb R}$, there is some infinite subset $H_0\subseteq\omega$ such that $\langle d_0(x^\alpha_{n}):n\in H_0\rangle$ converges for all $\alpha<\lambda$. Now consider the sequences $\langle d_1(x^\alpha_{n}):n\in H_0\rangle$ for $\alpha<\lambda$. These are $\lambda$-many sequences of real numbers in $[-\delta_1,\delta_1]$, hence from $\lambda<\mathbb c_{\mathbb R}$ we find that there is some infinite subset $H_1\subseteq H_0$ such that $\langle d_1(x^\alpha_{n}):n\in H_1\rangle$ converges for all $\alpha<\lambda$. Note that $\langle d_0(x^\alpha_{n}):n\in H_1\rangle$ is still converging, since it is a subsequence of $\langle d_0(x^\alpha_{n}):n\in H_0\rangle$.

Now, assume by inductive hypothesis that for some fixed $k<\omega$ we found an infinite subset $H_k\subseteq\omega$ such that $\langle d_i(x^\alpha_{n}):n\in H_k\rangle$ converges for all $\alpha<\lambda$ and for all $0\le i\le k$. Then consider the sequences $\langle d_{k+1}(x^\alpha_{n}):n\in H_k\rangle$: those are $\lambda$-many sequences in $[-\delta_{k+1},\delta_{k+1}]$, so since $\lambda<\mathbb c_{\mathbb R}$ we can find some infinite $H_{k+1}\subseteq H_k$ such that $\langle d_{k+1}(x^\alpha_{n}):n\in H_{k+1}\rangle$ converges for all $\alpha<\lambda$, and by hypothesis we also get that $\langle d_{i}(x^\alpha_{n}):n\in H_{k+1}\rangle$ converges for all $0\le i\le k$ and all $\alpha<\lambda$.

Let now $p_0$ be the $0$-th element of $H_0$, let $p_1$ be the $1$-st element of $H_1$ and so on let $p_k$ be the $k$-th element of $H_k$ (to make it more formal, just consider the fact that the order type of $H_k$ is $\omega$, fix the only order-preserving bijection $\sigma_k:H_k\to \omega$ and let $p_k:=\sigma_k(k)$). Consider now the set

$$H_\omega:=\{p_k:k<\omega\}.$$

We notice that $H_\omega\subseteq^\ast H_k$ (contained modulo finite) for all $k<\omega$. Indeed, all $p_i$ with $i\ge k$ belong to $H_k$. Hence we get that, for all $i<\omega$ and all $\alpha<\lambda$, the sequence $\langle d_{i}(x^\alpha_{n}):n\in H_{\omega}\rangle$ converges.

We now proceed to show that the sequence $\langle d(x^\alpha_{n}):n\in H_{\omega}\rangle$ converges, for all $\alpha<\lambda$ and all $d\in X^\ast$. Fix $\alpha$ and $d$. Note that this is just a sequence of real numbers. So we can consider

$$|d(x_n^\alpha)-d(x_m^\alpha)| \le |d(x_n^\alpha)-d_i(x_n^\alpha)|+|d_i(x_n^\alpha)-d_i(x_m^\alpha)|+|d_i(x_m^\alpha)-d(x_m^\alpha)|$$

for all $i<\omega$, and since $\|x_n^\alpha\|_X\le 1$ by hypothesis, this implies that

$$|d(x_n^\alpha)-d(x_m^\alpha)| \le 2 \|d-d_i\|_{X^\ast} + |d_i(x_n^\alpha)-d_i(x_m^\alpha)|.$$

Now, given any $\varepsilon>0$ we can find by density some $i_0<\omega$ such that $\|d-d_i\|_{X^\ast}\le \varepsilon/4$.

Notice also that since the sequence $\langle d_{i_0}(x_n^\alpha):n<\omega\rangle$ is convergent in $\mathbb R$, it is a Cauchy sequence, hence there is some $k_0<\omega$ such that for all $n$ and $m$ greater than $k_0$ it holds $|d_{i_0}(x_n^\alpha)-d_{i_0}(x_m^\alpha)|\le \varepsilon/2$. By combining everything, we proved that for all $\varepsilon>0$ there is $k_0<\omega$ such that $|d(x_n^\alpha)-d(x_m^\alpha)|\le \varepsilon$ for all $n$ and $m$ greater than $k_0$. So the sequence $\langle d(x^\alpha_{n}):n\in H_{\omega}\rangle$ is a Cauchy sequence in $\mathbb R$, hence it converges to some value $x^\alpha(d)$ that depends only on $d$ and $\alpha$. It's easy to check that, for $\alpha$ fixed, the value of $x^\alpha(d)$ changes linearly with $d$, so $x^\alpha: X^\ast\to\mathbb R$ is a linear map and $\|x^\alpha\|_{X^{\ast\ast}}\le 1$, hence $x^\alpha\in X^{\ast\ast}=X$ and the sequences $\langle x^\alpha_n:n\in H_\omega\rangle$ converge weakly to $x^\alpha$ for all $\alpha<\lambda$.

So, putting together all results by me and professor Hamkins: $$\min\{\mathfrak s,\mathfrak b\}\le \mathbb c=\mathbb c_{\mathbb R}\le\mathbb c_{\{0,1\}}=\mathfrak s.$$

I guess it's something! I'm curious if we could prove something better about $\mathbb c_{\mathbb R}$, now that (as Hamkins correctly anticipated) we know that that one is the "really important case"

The proof clearly works also for $X^\ast$ a Banach (dual) space, if the pre-dual $X$ is separable, with weak$^\ast$ convergence instead of weak convergence. That is the usual setting of Banach-Bourbaki-Alaoglu theorem, of which this fact is a generalization.

Answer 4

(Here are details re comment above, to show that $c_{\mathbb{R}}=c_{0,1}$.)

Suppose that $c=c_{\mathbb{R}}<c_{0,1}$. Let $\left<x_\alpha\right>_{\alpha<c}$ be a sequence of bounded $\mathbb{R}$-valued sequences (that is, $x_\alpha:\mathbb{N}\to\mathbb{R}$ and it is bounded in $\mathbb{R}$), such that there is no subsequence on which they all converge. By rescaling and translating each one individually, we may assume that they all take values in the interval $[0,1]$.

Associated to each $\alpha<c$, define an $\omega$-sequence $\left<x^n_\alpha\right>_{n<\omega}$ of functions $x^n_\alpha:\mathbb{N}\to\{0,1\}$ as follows:

-- $x^0_\alpha(m)=0$ if $x_\alpha(m)<\frac{1}{2}$, and $x^0_\alpha(m)=1$ otherwise.

-- $x^1_\alpha(m)=0$ if $$x_\alpha(m)\in\Big[0,\frac{1}{4}\Big)\cup\Big[\frac{1}{2},\frac{3}{4}\Big),$$ and $x^1_\alpha(m)=1$ otherwise.

-- Etc. That is, $x^{n+1}_\alpha(m)=0$ if $$x_\alpha(m)\in\Big[0,\frac{1}{2^{n+2}}\Big)\cup\Big[\frac{2}{2^{n+2}},\frac{3}{2^{n+2}}\Big)\cup\ldots\cup\Big[\frac{2^{n+2}-2}{2^{n+2}},\frac{2^{n+2}-1}{2^{n+2}}\Big),$$ and $x_\alpha(m)=1$ otherwise.

Since $c<c_{0,1}$, we can and do fix an infinite set $I_0\subseteq\mathbb{N}$ such that $\left<x^0_\alpha\upharpoonright I_0\right>_{\alpha<c}$ consists of eventually constant functions.

Again since $c<c_{0,1}$, fix an infinite $I_1\subseteq I_0$ such that $\left<x^1_\alpha\upharpoonright I_1\right>_{\alpha<c}$ consists of eventually constant functions, and we may and do take $I_1$ with $\min(I_1)=\min(I_0)$.

Continue in this way, getting $I_{n+1}\subseteq I_n$, and these two sets having the same first $n+1$ elements.

Now let $I_\infty=\cap_{n\in\omega}I_n$, which is infinite, and note that the original functions all converge along $I_\infty$, a contradiction.

Answer 5

As noted in my commen, this has a one line counter-example, where the sequence takes its values in a two point set (so the whole separable Banach space, weak topology stuff is a red herring). The index set is the family of subsets of the integers and the sequence $(x_n^A)$ is just the characteristic function of $A$, regarded as a sequence in the usual way. This fails your condition by a standard diagonalisation argument.