# Sequence lemma for general topological spaces

I am struggling to come up with a proof for this. Any ideas?

Prove:If $$X$$ is first countable:If $$x \in \bar A$$, there is a sequence of points $$x_n$$ of $$A$$ converging to $$x$$.

Proof:Let $$x \in \bar A$$. Since $$X$$ is first countable, there is a countable neighborhood basis $$\{B_n\}$$ for $$x$$ such that $$B_{1}\supset B_{2}\supset \cdots$$ and $$B_{n} \cap A \neq \varnothing$$. Define a sequence converging to $$x$$ by setting $$x_{n} \in B_{n}\cap A$$ for each $$n \in \mathbb{N}$$. Then since each open set $$U$$ containing $$x$$ contains some $$B_{n}$$, for any $$U$$ open containing $$x$$, there is a $$N\in \mathbb{N}$$ with $$x \in B_{N} \subset U$$. So if $$n \geq N$$, then $$x_{n} \in U$$ so $$x_{n} \rightarrow x$$.

I realize that this proof is definitely not correct, but I would like to see the correct way to do this, so I took a stab at it. What would be a good way to approach this problem? Also wouldn't this proof be needed as a lemma to prove the sequential characterization of continuity for first countable spaces?

Edit: I realize I should probably not use a sequence of nested basis elements to solve this problem, but nested open sets. But if I do this, how should I define the sequence converging to $$x$$?

• @bof Oh yea let me fix that
– user892057
Apr 3, 2021 at 2:51
• Why do you think that the proof is incorrect? Elements of a local base at $x$ are open sets, and you can always assume that a countable base at $x$ are nested. (If $\{U_n:n\in\Bbb N\}$ is any base at $x$, let $B_n=\bigcap_{k=0}^nU_k$; then $\{B_n:n\in\Bbb N\}$ is a nested base at $x$.) Apr 3, 2021 at 3:31

The proof is completely correct, and if $$x$$ has a countable local base, it has one that is nested, just by taking finite intersections, see Brians's comment. (This is a reason why countable is "nice" or "special": all initial segments are finite and finite intersections of neighbourhoods are still neighbourhoods). For metric spaces the $$B(x,\frac1n)$$ balls are "naturally" nested (by radius), but in general spaces with a countable base, we can mimick it too, and converging sequences (to $$x$$) are then easily constructed, as you showed. You could consider the above a "missing lemma" if you like, I'd rather say it's folklore and well-known (it's used very often without mentioning it).