proof of Tychonoff’s theorem

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

1 - A net ${(x_{\alpha })}_{\alpha \in 𝒜}$ in ${\prod }_{i\in I}{X}_i$ converges to $x\in {\prod }_{i\in I}{X}_i$ if and only if each coordinate ${(x_{\alpha }^i)}_{\alpha \in 𝒜}$ 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) ${(x_{\alpha })}_{\alpha \in 𝒜}$ 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 ${(x_{\alpha })}_{\alpha \in 𝒜}$ be a net in ${\prod }_{i\in I}{X}_i$.

Using Lemma 3 we can find a subnet ${(y_{\beta })}_{\beta \in ℬ}$ of ${(x_{\alpha })}_{\alpha \in 𝒜}$.

It is easily seen that each coordinate net ${(y_{\beta }^i)}_{\beta \in ℬ}$ 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 ${(y_{\beta })}_{\beta \in ℬ}$ 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. $\mathrm{\square }$

Title	proof of Tychonoff’s theorem
Canonical name	ProofOfTychonoffsTheorem
Date of creation	2013-03-22 17:25:24
Last modified on	2013-03-22 17:25:24
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	8
Author	asteroid (17536)
Entry type	Proof
Classification	msc 54D30