Family of continuous functions Considering the family of continous functions $\left\lbrace f_i :(X,\tau_x) \rightarrow (X_i,\tau_i) \space / \space i \in I \right\rbrace$; show that for each closed set $B \subseteq X$ and for each $x \notin B$, exists $i \in I$ such that $f_i(x) \notin \overline{f_i(B)}$ if and only if $\left\lbrace f_i^{-1}(V_i) \space / \space i \in I, V_i \in \tau_i \right\rbrace$ is a basis of $\tau_x$.
Given $A \in \tau_x$ then, $X-A$ is a closed set. So $\exists i \in I$ such that $\forall x \notin X-A \space (\forall x \in A); f_i(x) \notin \overline{f_i(X-A)}$. Then, $f_i(x) \in X_i - \overline{f_i(X-A)} = V_i$ because $\overline{f_i(X-A)}$ is closed. And $f_i$ is continuous, so $f_i^{-1}(V_i) \in \tau_x$ and $x \in f^{-1} (V_i)$.
Then, $A = \bigcup_{x \in A} \left\lbrace x \right\rbrace \subseteq \cup_{i \in I} f_i^{-1}(V_i) = f_i^{-1}(X_i - \overline{f_i(X-A)}) \subseteq X - f_i^{-1}(f_i(X-A)) \subseteq X-(X-A) = A.$
Then $A= \cup_{i \in I} f_i^{-1}(V_i); V_i \in \tau_i$.
Please, I need to prove the other way but I don't how to do it. Can you help me please?
 A: "$\Rightarrow$" Since $f_i$ is continous for all $i$, $\left\lbrace f_i^{-1}(V_i) \space / \space i \in I, V_i \in \tau_i \right\rbrace\subset \tau_x$. So just we need to see for any $U\in \tau_x$ and for any $x\in U$ there exists $i\in I$ and $V_i\in X_i$ such that $x\in f_i^{-1}(V_i)\subset U$. 
To see it take any nonempty $U\in \tau_x$ and any $x\in U$. So $x\notin U^c$, since $U^c$ is closed  there is an $i\in I$ such that $f_i(x)\notin \overline{f_i(U^c)}$. It means there exists $V_i\in X_i$ such that $f_i(x)\in V_i$ and $V_i\cap f_i(U^c)=\emptyset$. Take any $y\in f_i^{-1}(V_i)$ so either $y\in U^c$ or $y\in U$. If $y\in U^c$ then $f_i(y)\in f_i(U^c)$. This contradicts with $V_i\cap f_i(U^c)=\emptyset$. So $y\in U$ and $x\in f_i^{-1}(V_i)\subset U$.
"$\Leftarrow$" Say $B\subset X$ closed and $x\notin B$, then $x\in B^c$ and $B^c$ open. So there is an $i\in I$ and $V_i\in \tau_i$ such that $x\in f_i^{-1}(V_i)\subset B^c$. To say  $f_i(x) \notin \overline{f_i(B)}$, it is enaught to see that $V_i \cap f_i(B)=\emptyset$. Suppose not, take any $f_i(y)\in V_i \cap f_i(B)$. So $f_i(y)\in V_i $ and $ y\in B$, then $y\in f_i^{-1}(V_i)$ and $ y\in B$. Therefore $f^{-1}_i (V_i) \cap B\neq \emptyset$ and this is contradiction.
