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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 $X_1 \times \dotsm \times X_n$ 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 $Y$ .
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
. 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
. Then $\bigcup_{j=1}^l \mathcal{S}^{y_j}$ is a finite subcover of $X \times Y$ . 