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Unit 22: The Tychonoff Theorem
Now D ... D belongs to D, by (a); therefore this intersection is non-empty, by Notes
1 n
hypothesis.
Theorem 1: (Tychonoff theorem): An arbitrary product of compact spaces is compact is the product
topology:
Proof: Let
X X ,
J
where each space X is compact. Let be a collection of subsets of X having the finite intersection
property. We prove that the intersection
A
A
is non-empty. Compactness of X follows:
Applying Lemma 1, choose a collection of subsets of X such that and is maximal with
respect to the finite intersection property. It will suffice to show that the intersection D D is
non-empty.
Given J, let : X X be the projection map, as usual. Consider the collection
{ (D)| D }
of subset of X . This collection has the finite intersection property because does. By compactness
of X , we can for each choose a point x of X such that
x (D).
D
Let x be the point (x ) J of X. We shall show that for x D for every D ; then our proof will
be finished.
First we show that if – 1 ( ) is any sub-basis element (for the product topology on X) containing
–1
x, then ( ) intersects every element of . The set is a neighbourhood of x in X .Since
x (D) by definition, intersects (D) in some point (y), where y D. Then it follows
–1
that y ( ) D.
It follows from (b) of Lemma 2, that every sub-basis element containing x belongs to D. And
then it follows (a) of the same lemma that every basis element containing x belongs to . Since
has the finite intersection property, this means that every basis element containing x intersects
every element of ; hence x D for every D as desired.
22.2 Summary
Let X be a set and a family of subsets of X. Then is said to have the finite intersection
property if for any finite number F , F , ... F of members of . F F ... F .
1 2 n 1 2 n
Let (X, T) be a topology space. Then (X, T) is compact iff every family of closed subsets of
X with the finite intersection property satisfies F .
F
An arbitrary product of compact spaces is compact in the product topology.
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