# 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 ProofOfLindelofTheorem 2013-03-22 12:56:31 2013-03-22 12:56:31 Evandar (27) Evandar (27) 5 Evandar (27) Proof msc 54D99