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I think I solved a problem from Lee's book and I just like to see if it has any mistake,

Problem Let $X$ be a topological space and let $\mathcal{U}$ be an open cover of $X$. Show that if $\mathcal{U}$ is countable and each $U$ $\in$ $\mathcal{U}$ is second countable, then $X$ is second countable.

Proof:

For each $U$ $\in$ $\mathcal{U}$, let $\mathcal{B}_{U}$ be a countable basis for $U$. We have that the union $\mathcal{B}$ of all such bases is countable and it's a basis for X. Indeed, each of them is open in X and, let $A$ be an arbitraty open set of $X$, we have that $A$ can be written as the union of all the intersections $A$$\cap$$U$, for all $U$ $\in$ $\mathcal{U}$, each such intersection being written as a union of elements of $\mathcal{B}_{U}$. Then $\mathcal{B}$ is a countable basis for $X$.

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Yes, this is fine. –  Brian M. Scott Oct 4 '12 at 21:33
    
Thank you, Brian. =] –  Br09 Oct 4 '12 at 21:40

1 Answer 1

up vote 1 down vote accepted

The proof is fine.

Note that it relies on the axiom of choice to prove that countable unions of countable sets are countable. But then again, topology without choice is a horror story anyway.

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