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 $X_1\times \mathrm{\cdots }\times X_n$ 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 $𝒞$ 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 $𝒞$: label the subcover by ${𝒮}^y=\{U_1^y\times V_1^y,\mathrm{\dots },U_{k_y}^y\times V_{k_y}^y\}$. 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 $V^{y_1},\mathrm{\dots },V^{y_l}$. Then ${\bigcup }_{j=1}^l{𝒮}^{y_j}$ 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