# First Countable

1. May 21, 2010

### lttlbbygurl

In a first countable space any point that is adherent to a set S is also the limit of a sequence in S.

In my head, this seems obvious, but I can't seem to get it on paper.. I know that is has to do with inverse functions preserving unions and intersections, but can't seem to write the proof out.

2. May 22, 2010

### qspeechc

We may as well consider the neighbourhood base to consist of open sets; let the open neighbourhood base of the adherence point be $\{ G_n : n\in \mathbb{N} \}$.Put
$B_1 = G_1$
$B_n = G_1\cap \ldots \cap G_n$
Then in each $B_n$ we have a point $s_n$ of S. Then the sequence $(s_n)$ converges to the adherence point.