proof of Tychonoff's theorem in finite case
ProofOfTychonoffsTheoremInFiniteCase
2005-08-03 14:58:45
2005-08-03 14:58:45
Proof
Tychonoff's theorem
0
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(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 \emph{basis sets} of the form $U \times V$, where $U$ is open in $X$ and $V$ is open in $Y$.
\begin{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 $\mathcal{S}^y = \{ U_1^y \times V_1^y, \dotsc, U_{k_y}^y \times V_{k_y}^y \}$.
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}, \dotsc, V^{y_l}$.
Then $\bigcup_{j=1}^l \mathcal{S}^{y_j}$ is a finite subcover
of $X \times Y$.
\end{proof}