Proof for: $x \in \overline{A} \iff (x_n)_n$ exists in $A$ with $x_n \to x$

Question

Does anyone know the proof for this double implication including the properties of secquences and the topology of sets?

$x \in \overline{A} \iff (x_n)_n$ exists in $A$ with $x_n \to x$

The $\overline{A}$ is the closure of the set $A$.

Answer 1

Right side implies left side for all spaces.
Within a 1st countable space, left side implies right side.

For each point there is a nhood base of a countable descending
nest of open sets. Systematically pick a point from each of
those nhood base sets to construct the needed sequence.

Answer 2

It turns out that the smallest closed set that contains $A$ is equal to the union of $A$ with its limit points. So the closure of $A$, denoted $\overline{A}$, can be taken to be the union of $A$ with its limit points.

1) If we are in a metric space:

If a distance metric $d(x,y)$ exists and the topology of open sets is defined by the $d(\cdot, \cdot)$ function, then $x_n\rightarrow x$ means precisely that $\lim_{n\rightarrow\infty} d(x_n,x)=0$. Also, for every limit point $x$ of $A$, there is a sequence $\{x_n\}$ of points in $A$ that are distinct from $x$ and satisfy $x_n\rightarrow x$.

So in this case, you can prove your desired if and only if statement using the method of my first comment.

[Note that my first comment assumed that a statement such as "$x_n\rightarrow x$" implicitly assumes existence of a distance metric $d(\cdot)$. However, by chasing down links from the William Elliot answer, I find a more general definition of $x_n\rightarrow x$ that does not require a distance metric, see "topological spaces" here: https://en.wikipedia.org/wiki/Limit_of_a_sequence ]

2) What can go wrong otherwise:

Here is a counter-example to show what can go wrong if we have neither a metric space nor a "first countable space" as mentioned in the William Elliot answer.

Let $X= \mathbb{R}$ and let the topology be the cocountable topology, where open sets are those that are either empty or the complement is finite or countable (see link here): https://en.wikipedia.org/wiki/Cocountable_topology

Now take $A = \{x \in \mathbb{R} : x \neq 0\}$.

Then $0$ is a limit point of $A$ since every open set that contains $0$ must contain an infinite number of other points, and hence contains points in $A$. So $\overline{A} = X = \mathbb{R}$.

So $0 \in \overline{A}$. However, take any infinite sequence of points $\{x_n\}_{n=1}^{\infty}$ such that $x_n \in A$ for all $n \in \{1, 2, 3, ...\}$. We want to show that we cannot have $x_n\rightarrow 0$. Consider the open set $U= \{x \in \mathbb{R} : x \notin \{x_n\}_{n=1}^{\infty}\}$. Then $0 \in U$ but, by definition of $U$, there are no points of the sequence $\{x_n\}_{n=1}^{\infty}$ in $U$. So we do not have $x_n\rightarrow 0$. $\Box$

This shows the left-hand-side does not always imply the right-hand-side.