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# 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 $\cap_{{i\in I}}F_{i}=\emptyset$. 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 $X$. $\Box$

# References

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

## Mathematics Subject Classification

54D30*no label found*

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## Comments

## Proof

This article is listed in the Unproven list. Perhaps, since this contains its own proof, you could mark it as such.

## Re: Proof

I just did that. Thanks!