diff options
author | Justin Gassner <justin.gassner@mailbox.org> | 2024-02-14 07:24:38 +0100 |
---|---|---|
committer | Justin Gassner <justin.gassner@mailbox.org> | 2024-02-14 07:24:38 +0100 |
commit | 28407333ffceca9b99fae721c30e8ae146a863da (patch) | |
tree | 67fa2b79d5c48b50d4e394858af79c88c1447e51 /pages/general-topology/compactness/index.md | |
parent | 777f9d3fd8caf56e6bc6999a4b05379307d0733f (diff) | |
download | site-28407333ffceca9b99fae721c30e8ae146a863da.tar.zst |
Update
Diffstat (limited to 'pages/general-topology/compactness/index.md')
-rw-r--r-- | pages/general-topology/compactness/index.md | 41 |
1 files changed, 39 insertions, 2 deletions
diff --git a/pages/general-topology/compactness/index.md b/pages/general-topology/compactness/index.md index 60c29a0..37e9b4d 100644 --- a/pages/general-topology/compactness/index.md +++ b/pages/general-topology/compactness/index.md @@ -1,9 +1,46 @@ --- title: Compactness parent: General Topology -nav_order: 1 +nav_order: 5 has_children: true -# cspell:words +has_toc: false --- # {{ page.title }} + +## Compactness in Terms of Closed Sets + +{% theorem %} +A topological space $X$ is compact +if and only if it has the following property: +- Given any collection $\mathcal{C}$ of closed subsets of $X$, + if every finite subcollection of $\mathcal{C}$ has nonempty intersection, + then $\mathcal{C}$ has nonempty intersection. +{% endtheorem %} + +{% proof %} +By definition, a topological space $X$ is compact +if and only if it has the following property: +- Given any collection $\mathcal{O}$ of open subsets of $X$, + if $\mathcal{O}$ covers $X$, + then there exists a finite subcollection of $\mathcal{O}$ that covers $X$. + +If $\mathcal{A}$ is a collection of subsets of $X$, +let $\mathcal{A}^c = \braces{ X \setminus A : A \in \mathcal{A}}$ denote the collection of the complements of its members. +Clearly, $\mathcal{B}$ is a subcollection of $\mathcal{A}$ +if and only if $\mathcal{B}^c$ is a subcollection of $\mathcal{A}^c$. +Moreover, note that $\mathcal{B}$ covers $X$ if and only if +$\mathcal{B}^c$ has empty intersection. +Taking the contrapositive, we reformulate above property: +- Given any collection $\mathcal{O}$ of open subsets of $X$, + if every finite subcollection of $\mathcal{O}^c$ has nonempty intersection, + then $\mathcal{O}^c$ has nonempty intersection. + +To complete the proof, observe that a collection $\mathcal{A}$ consists of open subsets of $X$ +if and only if $\mathcal{A}^c$ consists of closed subsets of $X$. +{% endproof %} + +{% definition Finite Intersection Property%} +TODO +{% enddefinition %} + |