# universal nets in compact spaces are convergent

Theorem - A universal net ${({x}_{\alpha})}_{\alpha \in \mathcal{A}}$ in a compact space $X$ is convergent.

Proof : Suppose by contradiction^{} that ${({x}_{\alpha})}_{\alpha \in \mathcal{A}}$ was not convergent. Then for every $x\in X$ we would find neighborhoods^{} ${U}_{x}$ such that

$${\forall}_{\alpha \in \mathcal{A}}{\exists}_{\alpha \le {\alpha}_{0}}{x}_{{\alpha}_{0}}\notin {U}_{x}$$ |

The collection^{} of all this neighborhoods cover $X$, and as $X$ is compact^{}, a finite number
${U}_{{x}_{1}},{U}_{{x}_{2}},\mathrm{\dots},{U}_{{x}_{n}}$ also cover $X$.

The net ${({x}_{\alpha})}_{\alpha \in \mathcal{A}}$ is not eventually in ${U}_{{x}_{k}}$ so it must be eventually in $X-{U}_{{x}_{k}}$ (because it is a net). Therefore we can find ${\alpha}_{k}\in \mathcal{A}$ such that

$${\forall}_{{\alpha}_{k}\le \alpha}{x}_{\alpha}\in X-{U}_{{x}_{k}}$$ |

Because we have a finite number ${\alpha}_{1},{\alpha}_{2}\mathrm{\dots},{\alpha}_{n}\in \mathcal{A}$ we can find $\gamma \in \mathcal{A}$ such that ${\alpha}_{k}\le \gamma $ for each $1\le k\le n$.

Then ${x}_{\gamma}\in X-{U}_{{x}_{k}}$ for all $k$, i.e. ${x}_{\gamma}\notin {U}_{{x}_{k}}$ for all $k$. But ${U}_{{x}_{1}},{U}_{{x}_{2}},\mathrm{\dots},{U}_{{x}_{n}}$ cover $X$ and thus we have a contradiction. $\mathrm{\square}$

Title | universal nets in compact spaces are convergent |
---|---|

Canonical name | UniversalNetsInCompactSpacesAreConvergent |

Date of creation | 2013-03-22 17:31:29 |

Last modified on | 2013-03-22 17:31:29 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 4 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 54A20 |