Editing Compact space (section)

== Sufficient conditions ==

* A closed subset of a compact space is compact.<ref>{{harvnb|Arkhangel'skii|Fedorchuk|1990|loc=Theorem 5.2.3}}</ref>
* A finite [[Union (set theory)|union]] of compact sets is compact.
* A [[continuous function (topology)|continuous]] image of a compact space is compact.<ref>{{harvnb|Arkhangel'skii|Fedorchuk|1990|loc=Theorem 5.2.2}}</ref>
* The intersection of any non-empty collection of compact subsets of a Hausdorff space is compact (and closed);
** If {{mvar|X}} is not Hausdorff then the intersection of two compact subsets may fail to be compact (see footnote for example).{{efn|
Let {{math|1=''X'' = {''a'', ''b''} &cup; <math>\mathbb{N}</math>}}, {{math|1=''U'' = {''a''} &cup; <math>\mathbb{N}</math>}}, and {{math|1=''V'' = {''b''} &cup; <math>\mathbb{N}</math>}}. Endow {{math|X}} with the topology generated by the following basic open sets: every subset of <math>\mathbb{N}</math> is open; the only open sets containing {{mvar|a}} are {{mvar|X}} and {{mvar|U}}; and the only open sets containing {{mvar|b}} are {{mvar|X}} and {{mvar|V}}. Then {{mvar|U}} and {{mvar|V}} are both compact subsets but their intersection, which is <math>\mathbb{N}</math>, is not compact. Note that both {{mvar|U}} and {{mvar|V}} are compact open subsets, neither one of which is closed.
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* The [[product topology|product]] of any collection of compact spaces is compact. (This is [[Tychonoff's theorem]], which is equivalent to the [[axiom of choice]].)
* In a [[metrizable space]], a subset is compact if and only if it is [[sequentially compact]] (assuming [[axiom of countable choice|countable choice]])
* A finite set endowed with any topology is compact.