# First Countable Spaces and Limit points

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?

• Yes, this seems correct. – guidoar Aug 1 '19 at 23:34