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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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    $\begingroup$ Yes, this is fine. $\endgroup$ – Brian M. Scott Oct 4 '12 at 21:33
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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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