proof of Tychonoff’s theorem in finite case

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

To prove that $X_{1}\times\cdots\times X_{n}$ is compact if the $X_{i}$ are compact, it suffices (by induction) to prove that $X\times Y$ is compact when $X$ and $Y$ are. It also suffices to prove that a finite subcover can be extracted from every open cover of $X\times Y$ by only the basis sets of the form $U\times V$ , where $U$ is open in $X$ and $V$ is open in $Y$ .

Proof.

The proof is by the straightforward strategy of composing a finite subcover from a lower-dimensional subcover. Let the open cover $\mathcal{C}$ of $X\times Y$ by basis sets be given.

The set $X\times\{y\}$ is compact, because it is the image of a continuous embedding of the compact set $X$ . Hence $X\times\{y\}$ has a finite subcover in $\mathcal{C}$ : label the subcover by $\mathcal{S}^{y}=\{U_{1}^{y}\times V_{1}^{y},\ldots,U_{k_{y}}^{y}\times V_{k_{y% }}^{y}\}$ . Do this for each $y\in Y$ .

To get the desired subcover of $X\times Y$ , we need to pick a finite number of $y\in Y$ . Consider $V^{y}=\bigcap_{i=1}^{k_{y}}V_{i}^{y}$ . This is a finite intersection of open sets, so $V^{y}$ is open in $Y$ . The collection $\{V^{y}:y\in Y\}$ is an open covering of $Y$ , so pick a finite subcover $V^{y_{1}},\ldots,V^{y_{l}}$ . Then $\bigcup_{j=1}^{l}\mathcal{S}^{y_{j}}$ is a finite subcover of $X\times Y$ . ∎

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