General topology: $ x \in \bar{A}$ iff there exists $(x_n) \subset A$ that converges to $x$.

Question

Let $(X, \tau )$ be a topological space, let $A \subseteq X$ and $x \in X$. I know that if there exists a sequence $(x_n)$ contained in $A$ that converges to $x$, then $x$ is in the closure of $A$.

I also know that the reverse implication is not true in general. So my question is: under what conditions one can affirm that the reverse implication is also true, ie, that the statement is "if and only if"?

Yes, but I am looking for general properties like Hausdorffness, first/second countable... — Minkowski, Nov 04 '17 at 15:30
This answer can help you https://math.stackexchange.com/a/1822224. The condition for the "if and only if" is that your space $X$ be first countable. — Carlos Jiménez, Nov 04 '17 at 15:40

score 0 · Answer 1 · answered Nov 04 '17 at 15:36

You can have a look here: https://en.wikipedia.org/wiki/T1_space

A topological space $X$ is $T_1$ if and only if every open neighbourhood of a limit point of $S \subseteq X$ contains infinitely many points of $S$. From this you can construct your sequence - you need the axiom of choice.

General topology: $ x \in \bar{A}$ iff there exists $(x_n) \subset A$ that converges to $x$.

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