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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 $X$?

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    $\begingroup$ Compactness is intrinsic to a topological space and does not depend on the embedding. In other words, $F$ is compact in $U$ if and only if it is compact in $X$. $\endgroup$ – Zhen Lin Dec 31 '12 at 8:37
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    $\begingroup$ You seem confused about the definition of local compactness: there is an open neighbourhood $F$ of $x$ such that $\overline F$ is compact, not $F$. Of course, they're the same if $F$ is closed as well as open, but most of the time that isn't the case. $\endgroup$ – tomasz Dec 31 '12 at 12:57
  • $\begingroup$ @tomasz, No, I did not confuse about the definition. look at $X$ is Hausdorff. By definition cl F compact but, in Hausdorff space every compact space is closed. $\endgroup$ – ege Dec 31 '12 at 16:56
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    $\begingroup$ @ege: cl $F$ is compact, but $F$ isn't, usually.. $\endgroup$ – tomasz Dec 31 '12 at 16:58
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    $\begingroup$ @ege: Is $(0,1)$ locally compact? Does any point have a compact open neighbourhood? $\endgroup$ – tomasz Jan 1 '13 at 11:07
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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 $X$.

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