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Prove the following statement:

Given a first countable topological space $X$ and let $A$ be a subset of $X$, a point $x$ lies in the closure of $A$ if and only if there exists a sequence $(x_n)_n$ in $A$ which converges to $x$.

Henno Brandsma
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Arsh Gh
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    Have you tried anything yourself? What is the definition of the closure of $A$? Why do you think they specified first countable? – Arthur Jun 11 '16 at 14:08
  • @Arthur i cant prove if x lies in the closure then there is a sequence converging to x , i first consider a ball containing x and choose an element from A like x1 then by first countablity of A i picked another ball around x contained in the last ball and i picked element x2 and so on but i think its not working... – Arsh Gh Jun 11 '16 at 14:14
  • I have discovered a problem with this problem. You need $A$ to be a subset of an ambient topological space $X$, and it is $X$ that needs to be first countable. If there is no abmient $X$, then the closure of $A$ makes little sense (or it is just $A$). – Arthur Jun 11 '16 at 14:19
  • @Arthur you're right... – Arsh Gh Jun 11 '16 at 14:32

1 Answers1

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If $(x_n)_n$ is a sequence from $A$ (so all $x_n \in A$) that converges to $x$, then let $O$ be any open set that contains $x$. Now apply the definition of convergence. This implies (why?) that $A$ intersects $O$. As $O$ was arbitrary, $x \in \overline{A}$. No first countability needed. That's one direction.

If $x \in \overline{A}$, then let $B_n, n \in \mathbb{N}$ be a countable open neighbourhood base for $x$ (here we use first countability).

Define $O_n= \bigcap_{i=1}^n B_i$, which are all open neighbourhoods of $x$ (why?) so they all intersect $A$ (why?) so pick $x_n \in A \cap O_n$ for all $n$.

Now $x_n \rightarrow x$: let $O$ be open containing $x$. There is some $k$ such that $x \in B_k \subseteq O$ (why?). But then $x_n \in O$ for all $n \ge k$ (why?), so we are done.

Henno Brandsma
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