Compactness

Compactness in Terms of Closed Sets

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.

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 ${\mathcal{A}}^c=\{X\setminus A:A\in \mathcal{A}\}$ 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 $X$. $\square$

Definition (Finite Intersection Property)

TODO