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Prove that every basis of a second countable space contains a countable subfamily which is also a basis.

*I try with the intersection of the bases, the open ones of the finite with those of the other, but I can not. Some clue

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  • $\begingroup$ In chapter 1 of General Topology by R.Engelking there is a proof of the generalization of this : If $B$ is a base for the space $X,$ there exists a base$ B'\subset B$ with $|B'|\leq w(X),$.... where $w(X)$ (the weight of $X$) is the least infinite cardinal $k$ such that $X$ has a base $ C$ with $|C|leq k.$.... If $X$ does not have a finite base we can say that $w(X)$ is the least cardinal of a base for $X$, and that if $B$ is a a base for $X$ then there exists a base $B'|subset X$ with $|B'|=w(X).$ $\endgroup$ – DanielWainfleet Jun 20 '17 at 1:59
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Let $\mathcal D=\{D_1,D_2,\dots,D_n,\dots\}$ be a countable base. Let $\mathcal B$ be any base. For each pair $(m,n)\in\mathbb N\times\mathbb N,$ chose a set $B_{m,n}\in\mathbb B$ such that $D_m\subseteq B_{m,n}\subseteq D_n$ if such a set exists. (You need the axiom of choice here.) The collection of sets $B_{m,n}$ so chosen is a countable base for the topology.

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