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Let $X$ be a first countable topological space and $A \subset X$.

Show that if $x$ is a limit point of A then there exists a sequence of points $(a_n)$ contained in A that converge to $x$.

My Proof: Suppose $x$ is a limit point of $A$. As $X$ is first countable, there exists a nested neighborhood $\{U_n\}$ basis for $x$.That means each neighborhood of $x$ contains $U_n$ for some $n\in \mathbb{N}$. As $x$ is a limit point of $A$, $\forall n\in \mathbb{N}$ $U_n \cap (A-\{x\})$ is non empty. That means for each $n$ we can find points $x_n$ contained in $U_n$ and $A- \{x\}$. Let $U$ be neighborhood of $x$ so $\exists$ $N\in \mathbb{N}$ so that $U_N$ $\subset U$ for $n \geq N$ (Because they are nested). As each $U_N$ contains $x_N$ it follows that $x_n \rightarrow x$.

Is the proof correct?

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  • $\begingroup$ Yes, this seems correct. $\endgroup$ – guidoar Aug 1 '19 at 23:34
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This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proof is correct.

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