I am having trouble solving the exercise 4 in chapter 2, section 4 of Introduction to Topology, Gamelin and Greene, 2nd.
Suppose a topological space $X$ satisfies the first axiom of countability, or is first countable, i.e., for each $x\in X$, there exists a sequence of open neighborhoods $\{ U_n\}$ of $x$ such that each neighborhood of $x$ includes one of the $U_n$'s.
Prove the following assertions:
(c) In a first-countable space $X$, any point adherent to a set $S$ is a limit of a sequence in $S$.
Authors suggest that because $S$ meets each $U_n$, just pick any $s_n\in U_n \cap S$ for each n. Then $\{s_n\}$ converges to $x$.
I don't get the last part that "$\{s_n\}$ converges to $x$".
To me it seems reasonable that for any open neighborhood $V$ of $x$, there is some $U_k\subset V$ thus $s_k \in V$. But to say that $\{s_n\}$ converges to $x$, it is necessary to show that there exists some $N$ such that if $n\geq N$ then $s_n \in V$. I have no way to show the existence of such $N$.
Best Answer
If $\{U_n\}_{n\in\omega}$ is an open local base for a point $x$ you always can suppose WLOG that $U_{n+1}\subseteq U_n$ (You only define $V_n:=\cap_{i\leq n}U_n$, then $\{V_n\}_{n\in\omega}$ is also an open local base).
Note that under this assumption you get $s_n\in U_N$ for every $n\geq N$.