Prove that if $(X,\theta)$ is a $T_{3} $ space and A is a dense subset of $X$, $\chi(a,A)=\chi(a,X),\forall a\in A$, where
$\chi(x,X)=min\{|B(x)|\big|B(x)\text{ is a local base at }x\} $
So, because to every local base $B(x)$ at x in X corresponds a local base at x in A, namely $B^{A}(x)=\{U\cap A\big|U\in B(x)\},$ and every open in A subset is in the form $U\cap A \text{ for some }U\in\theta$, we get that $\chi(a,A)\leq\chi(a,X)$ since it is possible that $U_1\cap A=U_2\cap A$ for $U_1\neq U_2$. Is that correct so far? If so, how do I prove that the latter impossible?