In a metrizable TVS $E$ a point $x$ is an accumulation point of a sequence $S$ if and only if $S$ contains a subsequence which converges to $x$.

Question

Let $E$ be a metrizable space, that is, $E$ is a Hausdorff space and if there is a countable basis of neighborhoods of $0 \in E$ in E. I want to prove: a point $x \in E$ is a an accumulation point of a sequence $S:=(x_n)_{n \in\mathbb{N}} \subset E$ if and only if $S$ contains a subsequence which converges to $x$.

I know that, if we consider the set, for each $n \in \mathbb{N}$, $$S_n:=\{x_n,x_{n+1},x_{n+2},\cdots\},$$ the the family $\mathcal{B}:=\{S_n \subset E\; ; \; n \in \mathbb{N}\}$ is a basis of the filter $$\mathcal{F}_S=\{A \subset E\; ; \; S_n \subset A, \text{for some}\: n \in \mathbb{N}\}.$$ In addition, the sequence $S$ has a accumulation point if $x \in E$, and only if, $x$ is a accumulation point of $\mathcal{F}_S$, where $x$ is accumulation point of $\mathcal{F}_S$ means that $x \in \overline{V}$, for all $V \in \mathcal{F}_S.$

From these facts, how do I proceed?

Answer 1

If $x$ is an accumulation point of $S$, then $B(x,\epsilon)\cap S_n\ne\varnothing$ for each $\epsilon>0$ and $n\in\Bbb N$. Let

$$k_1=\min\{k\in\Bbb N:x_k\in B(x,1)\cap S_1\}\;.$$

Suppose that we already have $k_1<\ldots<k_m$ such that $x_{k_i}\in B\left(x,\frac1i\right)\cap S_i$ for $i=1,\ldots,m$; then we can let

$$k_{m+1}=\min\left\{k\in\Bbb N:k>k_m\text{ and }x_k\in B\left(x,\frac1{m+1}\right)\cap S_{m+1}\right\}$$

and continue the recursive construction to get a strictly increasing sequence $\langle k_i:i\in\Bbb Z^+\rangle$ in $\Bbb N$ such that $x_{k_i}\in B\left(x,\frac1i\right)\cap S_i$ for each $i\in\Bbb Z^+$. Then $\langle x_{k_i}:i\in\Bbb Z^+\rangle$ is a subsequence of $S$ converging to $x$.

Answer 2

The vector spaces info is not needed: if $X$ is a first countable (topological) space and the sequence $(x_n)_n$ has an accumulation point $x$ then there is a subsequence of $(x_n)_n$ that converges to $x$.

We simply construct it by recursion: let $O_n, n \in \Bbb N$ be a countable local base for $x$. WLOG we can assume that $O_{n+1} \subseteq O_n$ for all $n$ (a decreasing local base, in metric spaces we just use $B(x, \frac{1}{n})$, e.g.).

As $x$ is an accumulation point of the sequence for each open neighbourhood $O$ of $x$ the set $N(O):= \{n: x_n \in O\}$ is infinite.

So let $n_1 = \min(N(O_1))$ and if we have constructed $n_1 < n_2 < \ldots < n_k$ such that $x_{n_i} \in O_i$ for all $ i \le k$, we note that $N(O_{k+1})$ is infinite and we pick an element in it that is $> n_k$. This preserves the increasingness of the indices and the fact that always $x_{n_i} \in O_i$.

Because we have a decreasing local base, it's easy to see that $(x_{n_k})_k \to x$.