proof of Tychonoff’s theorem in finite case

(The finite case of TychonoffPlanetmathPlanetmath’s Theorem is of course a subset of the infiniteMathworldPlanetmath case, but the proof is substantially easier, so that is why it is presented here.)

To prove that X1××Xn is compactPlanetmathPlanetmath if the Xi are compact, it suffices (by inductionMathworldPlanetmath) to prove that X×Y is compact when X and Y are. It also suffices to prove that a finite subcover can be extracted from every open cover of X×Y by only the basis sets of the form U×V, where U is open in X and V is open in Y.


The proof is by the straightforward strategy of composing a finite subcover from a lower-dimensional subcover. Let the open cover 𝒞 of X×Y by basis sets be given.

The set X×{y} is compact, because it is the image of a continuousPlanetmathPlanetmath embeddingMathworldPlanetmathPlanetmath of the compact set X. Hence X×{y} has a finite subcover in 𝒞: label the subcover by 𝒮y={U1y×V1y,,Ukyy×Vkyy}. Do this for each yY.

To get the desired subcover of X×Y, we need to pick a finite number of yY. Consider Vy=i=1kyViy. This is a finite intersectionMathworldPlanetmathPlanetmath of open sets, so Vy is open in Y. The collectionMathworldPlanetmath {Vy:yY} is an open covering of Y, so pick a finite subcover Vy1,,Vyl. Then j=1l𝒮yj is a finite subcover of X×Y. ∎

Title proof of Tychonoff’s theorem in finite case
Canonical name ProofOfTychonoffsTheoremInFiniteCase
Date of creation 2013-03-22 15:26:27
Last modified on 2013-03-22 15:26:27
Owner stevecheng (10074)
Last modified by stevecheng (10074)
Numerical id 4
Author stevecheng (10074)
Entry type Proof
Classification msc 54D30