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 compactif the $X_i$ are compact, it suffices (by induction) to prove that $X\times Y$ is compactwhen $X$ and $Y$ are. It also suffices to prove thata 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 subcoverfrom 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 acontinuous embedding of the compact set $X$.Hence $X\times \{y\}$ has a finite subcover in $𝒞$: label the subcoverby ${𝒮}^y=\{U_1^y\times V_1^y,\mathrm{\dots },U_{k_y}^y\times V_{k_y}^y\}$.Do this for each $y\in Y$.

To get the desired subcover of $X\times Y$, we need to pick a finite numberof $y\in Y$. 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 picka finite subcover $V^{y_1},\mathrm{\dots },V^{y_l}$.Then ${\bigcup }_{j=1}^l{𝒮}^{y_j}$ is a finite subcoverof $X\times Y$.∎