# proof of Lindelöf theorem

Let $X$ be a second countable topological space^{}, $A\subseteq X$
any subset and $\mathcal{U}$ an open cover of $A$. Let
$\mathcal{B}$ be a countable basis for $X$; then ${\mathcal{B}}^{\prime}=\{B\cap A:B\in \mathcal{B}\}$ is a countable basis of the
subspace topology on A. Then for each $a\in A$ there is some
${U}_{a}\in \mathcal{U}$ with $a\in {U}_{a}$, and so there is
${B}_{a}\in {\mathcal{B}}^{\prime}$ such that $a\in {B}_{a}\subseteq {U}_{a}$.

Then $\{{B}_{a}\in {\mathcal{B}}^{\prime}:a\in A\}\subseteq \mathcal{B}$ is a countable open cover of $A$. For each ${B}_{a}$, choose ${U}_{{B}_{a}}\in \mathcal{U}$ such that ${B}_{a}\subseteq {U}_{{B}_{a}}$. Then $\{{U}_{{B}_{a}}:a\in A\}$ is a countable subcover of $A$ from $\mathcal{U}$.$\mathrm{\square}$

Title | proof of Lindelöf theorem |
---|---|

Canonical name | ProofOfLindelofTheorem |

Date of creation | 2013-03-22 12:56:31 |

Last modified on | 2013-03-22 12:56:31 |

Owner | Evandar (27) |

Last modified by | Evandar (27) |

Numerical id | 5 |

Author | Evandar (27) |

Entry type | Proof |

Classification | msc 54D99 |