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Suppose we have a topological space $X$, let's say at least Hausdorff. Let $A$ be a nonempty subset of $X$. and $\overline{x}\in \overline{A}$. In general, can we always find a sequence $(x_n)_{n\in \mathbb{N}}$ of elements from $A$ such that $$x_n\to\overline{x}\,?$$

zorro47
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    Check out this post for some counterexamples:

    https://math.stackexchange.com/questions/521936/find-a-point-l-in-the-closure-of-a-so-that-no-sequence-with-values-in-a-co

     – SlipEternal Jun 04 '18 at 14:22

2 Answers2

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In general, no. However, it is true if you assume that your space is first countable (that is, that for each point $x$ there is a contable set $N$ of neighbourhoods of $x$ such that every neighbourhood of $x$ contains an element of $N$). In particular, it is true for all metric spaces.

José Carlos Santos
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  • But it does not need to be true for weak topologies (usually they are not metrizable) and this is my biggest concern. Would mind explaining my why the answer is in general no or where to look for it? – zorro47 Jun 04 '18 at 14:21
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    @zorro47 there are some useful results in this direction for weak topologies. For instance, the weak* topology is metrizable when restricted to the unit ball of a space whose predual is separable. If the space is reflexive, then the same is true for the weak topology. Moreover, for a general (non-reflexive) Banach space you have the Eberlein-Smulian theorem (https://en.wikipedia.org/wiki/Eberlein%E2%80%93%C5%A0mulian_theorem), which states that weak compactness and sequential weak compactness are equivalent. – Lorenzo Q Jun 04 '18 at 14:24
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    @zorro47 You will find an example here. – José Carlos Santos Jun 04 '18 at 14:25
  • @Lorenzo Q Thank you! – zorro47 Jun 04 '18 at 14:35
  • @José Carlos Santos thank you! – zorro47 Jun 04 '18 at 14:36
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No. In general, the topology of a set is not determined by the behavior of its sequences alone. The topological spaces for which this is true are called sequential spaces. All metric spaces are sequential, and more generally, all first countable spaces are (i.e. those where each point has a countable neighborhood basis).

In order to obtain a result like the one you claim for arbitrary topological spaces, one need to replace sequences with nets. This is obtained by taking a set $\left\{x_i\right\}_{i\in I}$, where $I$ is not necessarily $\mathbb{N}$ as in the case of sequences, but rather a more general directed set. So for each $x\in \bar{A}$ there might not be a sequence $\left\{x_n\right\}_{n\in \mathbb{N}}\subset A$ such that $x_n\to x$, but there is always a net $\left\{x_i\right\}_{i\in I}\subset A$ such that $x_i\to x$.

Lorenzo Q
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