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 ProofOfTychonoffsTheoremInFiniteCase 2013-03-22 15:26:27 2013-03-22 15:26:27 stevecheng (10074) stevecheng (10074) 4 stevecheng (10074) Proof msc 54D30