Let $X$ be a Hausdorff space and let $U$ be a non-empty open locally compact subset of $X$. Since $U$ is non-empty, there exists an $x\in U$. Then there exists an open neighbourhood of $x$, let's call it $F$, such that $F$ is compact in $U$. Can we say that $F$ is compact in $X$?
Local compactness in an open subset
0
$\begingroup$
general-topology
-
0@ege $(0,1)$ is Hausdorff... And it is completely false that cl$F$ closed means $F=$ cl$F$! The closure of $F$ is the smallest closed set containing $F$. – 2016-08-04
1 Answers
1
Let $(X,\mathcal{T})$ a Hausdorff space. A set $F \subseteq X$ is called compact $\Leftrightarrow (F,\mathcal{T}_F)$ is compact where $\mathcal{T}_F := \{O \cap F; O \in \mathcal{T}\}$ denotes the subspace topology on $F$.
In your case: The topology on $U$ is given by $\mathcal{S} := \mathcal{T}_U$. By assumption $F$ is compact in $U$, i.e. $(F,\mathcal{S}_F)$ is compact. Since
$\mathcal{S}_F = \{O' \cap F; O' \in \mathcal{S}=\mathcal{T}_U\} = \{\underbrace{O \cap F \cap U}_{O \cap F}; O \in \mathcal{T}\} = \mathcal{T}_F$
we conclude that $(F,\mathcal{T}_F)$ is compact which means that $F$ is compact in $X$.