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.

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