# 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 $X$.
∎

Title | first countable implies compactly generated |
---|---|

Canonical name | FirstCountableImpliesCompactlyGenerated |

Date of creation | 2013-03-22 19:09:35 |

Last modified on | 2013-03-22 19:09:35 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 4 |

Author | CWoo (3771) |

Entry type | Example |

Classification | msc 54E99 |