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[parent] proof of Tychonoff's theorem (Proof)

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

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$

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

Lemma 3 - Every net has a universal subnet.

Lemma 4 - A universal net $ (x_{\alpha})_{\alpha \in \mathcal{A}}$ in a compact space $ X$ is convergent. (see this entry)

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 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 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$



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Cross-references: Tychonoff's theorem, every net has a universal subnet, subnet, convergent, compact, topological space, coordinate, converges, nets, proof
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This is version 5 of proof of Tychonoff's theorem, born on 2007-07-26, modified 2007-09-04.
Object id is 9797, canonical name is ProofOfTychonoffsTheorem.
Accessed 1197 times total.

Classification:
AMS MSC54D30 (General topology :: Fairly general properties :: Compactness)

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