Existence of a sequence which realizes an element from a closure

Question

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 $$x_n\to\overline{x}\,?$$

Answer 1

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.

Answer 2

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 $x_i\to x$.