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Let (X,T) a metric space, if $A\subset X$ is dense, proof that $B=$ { $B_d$(a,$\frac{1}{n}$) :$a\in A$ , $n\in N$ } is base for T. ($B_d$ is a ball).

I don't know how to do it: but here i let you two propositions I think they could be useful (but I don't know how to use them well).

a) B is base if and only if $\forall U \in T$ with U open and $\forall x \in U$ exists $b\in B$: $x\in B \subset U$

b) B is base if we can put every open as an union of elements of the base.

Thank you so much for any help!