first countable implies compactly generated

Proposition 1.

Any first countable topological space is compactly generated.

Proof.

Suppose $X$ is first countable, and $A\subseteq X$ has the property that, if $C$ is any compact set in $X$, the set $A\cap C$ is closed in $C$. We want to show tht $A$ is closed in $X$. Since $X$ is first countable, this is equivalent to showing that any sequence $(x_i)$ in $A$ converging to $x$ implies that $x\in A$. Let $C=\{x_i\mid i=1,2,\mathrm{\dots }\}\cup \{x\}$.

Lemma 1.

$C$ is compact.

Proof.

Let $\{U_j\mid j\in J\}$ be a collection of open sets covering $C$. So $x\in U_j$ for some $j$. Since $U_j$ is open, there is a positive integer $k$ such that $x_i\in U_j$ for all $i\ge k$. Now, each $x_i\in U_{d(i)}$ for $i=1,\mathrm{\dots },k$. So $C$ is covered by $U_{d(1)},\mathrm{\dots },U_{d(k)}$, and $U_j$, showing that $C$ is compact.∎

In addition, as a subspace of $X$, $C$ is also first countable. By assumption, $A\cap C$ is closed in $C$. Since $x_i\in A\cap C$ for all $i\ge 1$, we see that $x\in A\cap C$ as well, since $C$ is first countable. Hence $x\in A$, and $A$ is closed in $X$.∎