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[parent] a space is compact if and only if the space has the finite intersection property (Theorem)

Theorem. A topological space is compact if and only if the collection of all of its closed sets has the finite intersection property.

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=\emptyset$. Then,

$\displaystyle X$ $\displaystyle =$ $\displaystyle \left(\bigcap_{i\in I} F_i\right)^c$  
  $\displaystyle =$ $\displaystyle \bigcup_{i\in I} F_i^c.$  

(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=\emptyset$, 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\neq \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

$\displaystyle \bigcap_{i\in J} F^c_i$ $\displaystyle =$ $\displaystyle \Big(\bigcup_{i\in J} F_i\Big)^c \neq \emptyset,$  

where the last assertion follows since $ J$ was finite. Then, since $ X$ has the finite intersection property,
$\displaystyle \emptyset$ $\displaystyle \neq$ $\displaystyle \bigcap_{i\in I} F_i^c = \Big(\bigcup_{i\in I} F_i\Big)^c.$  

This contradicts the assumption that $ \{F_i\}_{i\in I}$ is a cover for $ X$. $ \Box$

Bibliography

1
R.E. Edwards, Functional Analysis: Theory and Applications, Dover Publications, 1995.



"a space is compact if and only if the space has the finite intersection property" is owned by CWoo. [ full author list (3) | owner history (2) ]
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Cross-references: contradiction, contains, subset, compactness, open cover, closed, complement, finite, cover, proof, intersections, characterization, unions, open sets, de Morgan's laws, finite intersection property, closed sets, collection, compact, topological space
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This is version 15 of a space is compact if and only if the space has the finite intersection property, born on 2003-04-12, modified 2008-04-30.
Object id is 4181, canonical name is ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty.
Accessed 7354 times total.

Classification:
AMS MSC: 54D30 (General topology :: Fairly general properties :: Compactness)
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