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 $\{x_n^{\alpha} \}_{n=1}^\infty$.

My question is that: 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.

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.

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 for all $n \in \mathbb{N}$. Then, Banach-Alaoglu Theorem implies that there exists a weakly convergent subsequence of $\{x_n^{\alpha} \}_{n=1}^\infty$.

My question is that: 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.

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 $\{x_n^{\alpha} \}_{n=1}^\infty$.

My question is that: 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.

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 for all $n \in \mathbb{N}$. Then, Banach-Alaoglu Theorem implies that there exists a weakly convergent subsequence of $\{x_n^{\alpha} \}_{n=1}^\infty$.

My question is that: 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 $\alpha \in \mathcal{A}$?

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

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 for all $n \in \mathbb{N}$. Then, Banach-Alaoglu Theorem implies that there exists a weakly convergent subsequence of $\{x_n^{\alpha} \}_{n=1}^\infty$.

My question is that: 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.