proof of Tychonoff's theorem
ProofOfTychonoffsTheorem
2007-07-26 17:11:56
2007-09-04 18:49:24
Proof
Tychonoff's theorem
0
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This is a proof in \PMlinkescapetext{terms} of nets. Recall the following facts:
{\bf \PMlinkescapetext{Lemma} 1 -} A net $(x_{\alpha})_{\alpha \in \mathcal{A}}$ in $\prod_{i \in I}X_i$ converges to
$x \in \prod_{i \in I}X_i$ if and only if each coordinate $(x_{\alpha}^i)_{\alpha \in \mathcal{A}}$ converges to $x^i \in X_i$
{\bf \PMlinkescapetext{Lemma} 2 -} A topological space $X$ is compact if and only if every net in $X$ has a convergent subnet.
{\bf \PMlinkescapetext{Lemma} 3 -} Every net has a universal subnet.
{\bf \PMlinkescapetext{Lemma} 4 -} A \PMlinkname{universal net}{Ultranet} $(x_{\alpha})_{\alpha \in \mathcal{A}}$ in a compact space $X$ is convergent. (see this \PMlinkname{entry}{UniversalNetsInCompactSpacesAreConvergent})
We now prove Tychonoff's theorem.
{\bf Proof (Tychonoff's theorem) :} Let $(x_{\alpha})_{\alpha \in \mathcal{A}}$ be a net in $\prod_{i \in I}X_i$.
Using Lemma 3 we can find a \PMlinkescapetext{universal} subnet $(y_{\beta})_{\beta \in \mathcal{B}}$ of $(x_{\alpha})_{\alpha \in \mathcal{A}}$.
It is easily seen that each coordinate net $(y_{\beta}^i)_{\beta \in \mathcal{B}}$ is a \PMlinkescapetext{universal} net in $X_i$.
Using Lemma 4 we see that each coordinate net converges, because $X_i$ is compact.
Using Lemma 1 we see that the whole net $(y_{\beta})_{\beta \in \mathcal{B}}$ converges in $\prod_{i \in I}X_i$.
We conclude that every net in $\prod_{i \in I}X_i$ has a convergent subnet, so, by Lemma 2, $\prod_{i \in I}X_i$ must be compact. $\square$