# Show that if $x$ is a limit point of a subset $A$ of a space $X$ which has a countable basis, then there is a sequence in $A$ converging to $x$

I'd like to show that if $x$ is a limit point of a subset $A$ of a topological space $X$, where $X$ has a countable basis, then there is a sequence of points in $A$ converging to $x$.

I was told that it is possible to show from the hypothesis that there is a nested sequence of open sets $U_1, U_2, U_3, \dots$ ($U_1$ contains $U_2$ and so on) each containing $x$ such that for any open set $U_x$ containing $x$, $U_x$ contains $U_n$ for some natural number $n$ (and obviously contains all subsequent sets in the sequence as well).

Once this is shown, I see that given an open set $U_x$ containing $x$, we can then take a point in $U_n \cap A$ and another in $U_{n+1} \cap A$ (using the fact that $x$ is a limit point of $A$) and so on to obtain a sequence converging to $x$.

However, I don't see how to show the claim in the second paragraph here.

Thanks Alex

• Hi, I added math notation to your post. You can click the edit link to see how the code for this looks like. Also have a look at meta.math.stackexchange.com/questions/5020/… for a tutorial on MathJax. – Stefan Hamcke Feb 26 '15 at 23:54
• italic thanks – Smithey Feb 27 '15 at 2:24

Let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be a countable local base at $x$. For each $n\in\Bbb N$ let
$$U_n=\bigcap_{k\le n}B_k\;,$$
and let $\mathscr{U}=\{U_n:n\in\Bbb N\}$. Clearly $U_n\supseteq U_{n-1}$ for each $n\in\Bbb N$, so all you have to do to finish the argument is show that $\mathscr{U}$ is a local base at $x$.