nLab finite intersection property

Idea

The finite intersection property on a set of subsets says that every finite number of them has non-empty intersection (def. below).

This property is typically used in order to re-formulate the condition on topological space to be compact in terms of closed subsets (prop. below).

The formulation of compactness in terms of the finite intersection property is useful for proving the Tychonoff theorem as well as for proving that compact spaces are equivalently those for which every net has a converging subnet.

Statement

Definition

(finite intersection property)

Let $X$ be a set. Then a set of subsets $\{S_i \subset X\}_{i \in I}$ is said to satify the finite intersection property if for every finite subset $J \subset I$ the corresponding intersection is inhabited: $\underset{i \in J \subset I}{\cap} S_i \neq \emptyset$.

Relation to compact topological spaces

Proposition

Let $(X,\tau)$ be a topological space.

Assuming excluded middle, then:

The following are equivalent.

1. $(X, \tau)$ is a compact topological space in the sense that for every open cover $\{U_i \subset X\}_{i \in I}$ there is a finite subset $J \subset I$ such that $\{U_i \subset X\}_{i \in J \subset I}$ is still a cover.

2. Every set of closed subsets $\{C_i \subset X\}_{i \in I}$ which satisfies the finite intersection property (def. ) has even non-empty total intersection $\underset{i \in I}{\cap} C_i \neq \emptyset$.

Proof

In one direction, assume that $(X,\tau)$ is compact, and that $\{C_i \subset X\}_{i \in I}$ satisfies the finite intersection property. We need to show that then $\underset{i \in I}{\cap} C_i \neq \emptyset$.

Assume that this were not the case, hence assume that $\underset{i \in I}{\cap} C_i = \emptyset$. This would imply that the open complements $U_i \coloneqq X \backslash C_i$ were an open cover of $X$, because (using de Morgan's law)

$$\begin{aligned} \underset{i \in I}{\cup} U_i & \coloneqq \underset{i \in I}{\cup} X \backslash C_i \\ & = X \backslash \left( \underset{i \in I}{\cap}C_i \right) \\ & X \backslash \emptyset \\ & = X \end{aligned} \,.$$

But then by compactness of $(X,\tau)$ there were a finite subset $J \subset I$ such that $\{ U_i \subset X\}_{i \in J \subset I}$ were still an open cover, hence that $\underset{i \in J \subset I}{\cup} U_i = X$. Translating this back through the de Morgan's law again this would mean that

$$\begin{aligned} \emptyset & = X \backslash \left( \underset{i \in J \subset I}{\cup} U_i \right) \\ & \coloneqq X \backslash \left( \underset{i \in J \subset I}{\cup} X \backslash C_i \right) \\ & = \underset{i \in J \subset I}{\cap} X \backslash \left( X \backslash C_i\right) \\ & = \underset{i \in J \subset I}{\cap} C_i \,. \end{aligned}$$

This would be in contradiction with the finite intersection property of $\{C_i \subset X\}_{i \in I}$, and hence we have proof by contradiction.

Conversely, assume that every set of closed subsets in $X$ with the finite intersection property has non-empty total intersection. We need to show that the every open cover $\{U_i \subset X\}_{i \in I}$ of $X$ has a finite subcover.

Write $C_i \coloneqq X \backslash U_i$ for the closed complements of these open subsets.

Assume that there were no finite subset $J \subset I$ such that $\underset{i \in J \subset I}{\cup} U_i = X$. By de Morgan's law this means equivalently that there were no finite subset $J$ such that $\underset{i \in J \subset I}{\cap} C_i = \emptyset$, hence it would mean that $\{C_i \subset X\}_{i \in I }$ satisfied the finite intersection property.

But by assumption this would then imply that $\underset{i \in I}{\cap} C_i \neq \emptyset$, which, again by de Morgan, would mean that $\underset{i \in I}{\cup} U_i \neq X$. But this contradicts the assumption that the $\{U_i \subset X\}_{i \in I}$ are a cover. Hence we have a proof by contradiction.