nLab finite intersection property




topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory


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topological homotopy theory



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.



(finite intersection property)

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

Relation to compact topological spaces


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

Assuming excluded middle, then:

The following are equivalent.

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

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


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

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

iIU i iIX\C i =X\(iIC i) X\ =X. \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,τ)(X,\tau) there were a finite subset JIJ \subset I such that {U iX} iJI\{ U_i \subset X\}_{i \in J \subset I} were still an open cover, hence that iJIU i=X\underset{i \in J \subset I}{\cup} U_i = X . Translating this back through the de Morgan's law again this would mean that

=X\(iJIU i) X\(iJIX\C i) =iJIX\(X\C i) =iJIC i. \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 iX} iI\{C_i \subset X\}_{i \in I}, and hence we have proof by contradiction.

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

Write C iX\U iC_i \coloneqq X \backslash U_i for the closed complements of these open subsets.

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

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

Last revised on May 10, 2017 at 13:54:08. See the history of this page for a list of all contributions to it.