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Here is the question:

Let $A$ be a subset of a first-countable topological space $(X, \mathcal{T})$. Prove that $x ∈ X$ is in the closure $\bar A$ if and only if there is a sequence $(x_n), x_n ∈ A$, which converges to $x$.

I can't seem to get a handle on proving either direction.

Marcy
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    Concentrate for now on proving that if there is a sequence in $A$ converging to $x$, then $x\in\operatorname{cl}A$; that is true in any topological space, not just first countable ones, and the proof is a very straightforward application of the definitions of limit of a sequence and belongs to the closure of $A$. – Brian M. Scott Apr 04 '20 at 22:27

2 Answers2

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I'll use $\overline{A} = \{x \in X: \forall O \in \mathcal{T}: x \in O \to O \cap A \neq \emptyset\}$, the closure is the set of adherent points.

If such a sequence exists then every open set that contains $x$, must contain a tail of the sequence, and thus contains points of $A$, hence $x \in \overline{A}$. No first countability needed yet.

If OTOH $x \in \overline{A}$, let $(U_n)_{n \in \Bbb N}$ be a countable local base for $x$, using first countability. For each $n$ we pick $x_n \in (\bigcap_{i=1}^n U_i) \cap A$, which can be done as the intersection is open (a finite intersection of open sets), contains $x$ and so intersects $A$ by $x$ being in $\overline{A}$. Now if $O$ is any open neighbourhood of $x$, some $U_N \subseteq O$ (local base!) and then all $x_n$ for $n \ge N$ are in $O$. So $x_n \to x$, and we have the required sequence.

Henno Brandsma
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  • Late late comment or rather a question . I know that if a space is not first countable then this converse you proved is not true. However I am struggling to come up with a simple example disproving it. For example https://math.stackexchange.com/questions/281267/if-a-in-mathrmclos-does-it-follow-that-there-exists-a-sequence-of-points . This post gives an example but uses quotient spaces which I have not yet done. But is there an easy example in something like cofinite topology which is not first countable? – Mr.Gandalf Sauron Oct 23 '21 at 14:41
  • @Mr.GandalfSauron the cocountable topology on $\Bbb R$ that can be used to disprove the question in the lemma. It is indeed not first countable, but it’s not Hausdorff so the quotient space is a nicer example in that sense. – Henno Brandsma Oct 23 '21 at 15:01
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One direction is easy. If $x_n\in A$ with $x_n\to x$, then every nbhd $N$ of $x$ contains an $x_n$, hence $N\cap A\ne\emptyset$. So $x\in \bar A$.

The other direction isn't much harder. Take a countable nbhd base $N_k$ for each $x\in\bar A$. Then for each $n$, there is $x_n\in \bigcap_{k=1}^nN_k\cap A$. Then $x_n\in A$ and $x_n\to x$.