a space is compact iff any family of closed sets having fip has non-empty intersection
Theorem^{}. A topological space^{} is compact^{} if and only if any collection^{} of its closed sets^{} having the finite intersection property has non-empty intersection^{}.
The above theorem is essentially the definition of a compact space rewritten using de Morgan’s laws. The usual definition of a compact space is based on open sets and unions. The above characterization^{}, on the other hand, is written using closed sets and intersections.
Proof. Suppose $X$ is compact, i.e., any collection of open subsets that cover $X$ has a finite collection that also cover $X$. Further, suppose ${\{{F}_{i}\}}_{i\in I}$ is an arbitrary collection of closed subsets with the finite intersection property. We claim that ${\cap}_{i\in I}{F}_{i}$ is non-empty. Suppose otherwise, i.e., suppose ${\cap}_{i\in I}{F}_{i}=\mathrm{\varnothing}$. Then,
$X$ | $=$ | ${\left({\displaystyle \bigcap _{i\in I}}{F}_{i}\right)}^{c}$ | ||
$=$ | $\bigcup _{i\in I}}{F}_{i}^{c}.$ |
(Here, the complement of a set $A$ in $X$ is written as ${A}^{c}$.) Since each ${F}_{i}$ is closed, the collection ${\{{F}_{i}^{c}\}}_{i\in I}$ is an open cover for $X$. By compactness, there is a finite subset $J\subset I$ such that $X={\cup}_{i\in J}{F}_{i}^{c}$. But then $X={({\cap}_{i\in J}{F}_{i})}^{c}$, so ${\cap}_{i\in J}{F}_{i}=\mathrm{\varnothing}$, which contradicts the finite intersection property of ${\{{F}_{i}\}}_{i\in I}$.
The proof in the other direction is analogous. Suppose $X$ has the finite intersection property. To prove that $X$ is compact, let ${\{{F}_{i}\}}_{i\in I}$ be a collection of open sets in $X$ that cover $X$. We claim that this collection contains a finite subcollection of sets that also cover $X$. The proof is by contradiction^{}. Suppose that $X\ne {\cup}_{i\in J}{F}_{i}$ holds for all finite $J\subset I$. Let us first show that the collection of closed subsets ${\{{F}_{i}^{c}\}}_{i\in I}$ has the finite intersection property. If $J$ is a finite subset of $I$, then
$\bigcap _{i\in J}}{F}_{i}^{c$ | $=$ | ${\left({\displaystyle \bigcup _{i\in J}}{F}_{i}\right)}^{c}\ne \mathrm{\varnothing},$ |
where the last assertion follows since $J$ was finite. Then, since $X$ has the finite intersection property,
$\mathrm{\varnothing}$ | $\ne $ | $\bigcap _{i\in I}}{F}_{i}^{c}={\left({\displaystyle \bigcup _{i\in I}}{F}_{i}\right)}^{c}.$ |
This contradicts the assumption^{} that ${\{{F}_{i}\}}_{i\in I}$ is a cover for $X$. $\mathrm{\square}$
References
- 1 R.E. Edwards, Functional Analysis^{}: Theory and Applications, Dover Publications, 1995.
Title | a space is compact iff any family of closed sets having fip has non-empty intersection |
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Canonical name | ASpaceIsCompactIffAnyFamilyOfClosedSetsHavingFipHasNonemptyIntersection |
Date of creation | 2013-03-22 13:34:10 |
Last modified on | 2013-03-22 13:34:10 |
Owner | CWoo (3771) |
Last modified by | CWoo (3771) |
Numerical id | 20 |
Author | CWoo (3771) |
Entry type | Theorem |
Classification | msc 54D30 |