# 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))

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. $\mathrm{\square}$

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 |