0
$\begingroup$

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 $x$?

$\endgroup$
2
  • $\begingroup$ @bof Oh yea let me fix that $\endgroup$
    – user892057
    Apr 3, 2021 at 2:51
  • 2
    $\begingroup$ 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 $x$.) $\endgroup$ Apr 3, 2021 at 3:31

1 Answer 1

1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy