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 $\mathcal{B}^{\prime}=\{B\cap A\colon B\in\mathcal{B}\}$ 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 $\{B_{a}\in\mathcal{B}^{\prime}\colon a\in A\}\subseteq\mathcal{B}$ 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 $\{U_{B_{a}}\colon a\in A\}$ is a countable subcover of $A$ from $\mathcal{U}$ . $\square$

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