Exercise 2.4.4 (First-countable topology)

Proof. Let $x\in X$ and consider the sequence of open balls ${\left\{B\left(x,\frac{1}{n}\right)\right\}}_{n=1}^{\infty }$ around $x$. Then any open set containing $x$ as an interior point must contain some small open ball around $x$ by definition. □

Proof. Let $B={\{U_n\}}_{n=1}^{\infty }$ be the countable basis for $X$. Then any open set (containing $x\in X$) must be a union of the sets in $B$; thus, fulfilling the condition of first-countability. □

Proof. Let $x\in X$ be adherent to $S$ and let ${\{U_n\}}_{n=1}^{\infty }$ be the sequence of open neighbourhoods of $X$ given by the first-countability axiom. The trick is to define $V_n:={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{U}_n$. The sequence ${\{V_n\}}_{n=1}^{\infty }$ then characterises a sequence of open neighbourhoods of $x$ as well. By adherence, any $S\cap V_n$ must be non-empty. By the axiom of choice, we can construct a sequence ${\{s_n\}}_{n=1}^{\infty }$ by taking one $s_n$ out of each $S\cap V_n$. Then ${\{s_n\}}_{n=1}^{\infty }$ converges to $x$. (To see why, pick an arbitrary open neighbourhood $U$ of $x$. Then we can find a $U_i\subseteq U$ and thus $V_n\subseteq U$ for all $n\ge i$, which implies that $s_n\in U$ for all $n\ge i$.) □