# Limit points of sets in a first-countable space have a sequence converging to them

I have written a rudimentary proof of the title, but I'm not sure just how correct -or incorrect- it is. I'm fairly new to topology, and frankly I always feel out of my element when it comes to sequences, especially constructing them from open sets. I'm hoping that understanding this example thoroughly will help my understanding of similar problems. On to the proof I have:

Let X be a first countable space. Then there is a nested countable local basis $$\mathcal U$$ for every $$x \in X$$ (there's a bit of a leap here, but I think I have that part down already). So let $$x \in X$$ be an arbitrary limit point of a set $$A \subseteq X$$. Then there are nested open sets $$U \in \mathcal U$$ such that $$(U-{x}) \cap A \neq \varnothing$$.

Let these sets $$U$$ form the collection $$\{U_n\}_{n \in \Bbb N}$$ ordered by $$\supseteq$$. Then for some $$N \in \Bbb N$$, all $$U_m \subseteq U_N$$ when $$m>N$$. Then the sequence $$\{x_n | x_n \in U_n$$ and $$x_n \in A\}$$ converges to $$x$$.

• What does $\{x_n | x_n \in U_n$ and $x_n \in A\}$ mean? You claim that it is a sequence. How did you define it? – José Carlos Santos Apr 29 '19 at 22:16

Your idea is correct but you have not expressed it correctly. You should say pick $$x_n \in U_n\cap A \setminus \{x\}$$ for each $$n$$. The $$x_n \to x$$.