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# 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 $C=\{x_{i}\mid i=1,2,\ldots\}\cup\{x\}$.

###### 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 $i\geq k$. Now, each $x_{i}\in U_{{d(i)}}$ for $i=1,\ldots,k$. So $C$ is covered by $U_{{d(1)}},\ldots,U_{{d(k)}}$, 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 $i\geq 1$, we see that $x\in A\cap C$ as well, since $C$ is first countable. Hence $x\in A$, and $A$ is closed in $X$. ∎

## Mathematics Subject Classification

54E99*no label found*

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