# proof of Tychonoff’s theorem

This is a proof in of nets. Recall the following facts:

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 $x^{i}\in X_{i}$

2 - A topological space $X$ is compact if and only if every net in $X$ has a convergent subnet.

3 - Every net has a universal subnet.

4 - A universal net (http://planetmath.org/Ultranet) $(x_{\alpha})_{\alpha\in\mathcal{A}}$ in a compact space $X$ is convergent. (see this entry (http://planetmath.org/UniversalNetsInCompactSpacesAreConvergent))

We now prove Tychonoff’s theorem.

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 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 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. $\square$

Title proof of Tychonoff’s theorem ProofOfTychonoffsTheorem 2013-03-22 17:25:24 2013-03-22 17:25:24 asteroid (17536) asteroid (17536) 8 asteroid (17536) Proof msc 54D30