Convergence weak in a topological space Let a set $X$, a topological space $Y$ and a family $\Omega$  of functions $f$ defined from $X$ to $Y$.
I want prove that $x_n \rightarrow x$ in $\tau_{\Omega}$ if and only if $f(x_n) \rightarrow f(x)\quad \forall f \in \Omega$ .
$\tau_ {\Omega}$ is the weak topology.
Can I have a suggestion for starting the proof? 
 A: Because the weak topology makes all functions in $\Omega$ continuous, it is trivial that 
$$(x_n)_n\to x \implies (f(x_n))_n\to f(x), \forall f \in \Omega$$
The other implication requires some work.

A topology makes all functions in $\Omega$ continuous if and only if it includes 
$$S = \{f^{-1}(U) \mid f\in\Omega, U \in \tau_Y\}$$
where $\tau_Y$ is the topology on $Y$. In other words, $\tau_\Omega$ is the weakest topology such that $S \subseteq \tau_\Omega$. This means that $S$ is a subbasis for $\tau_\Omega$. By taking finite intersections we obtain a basis for $\tau_\Omega$:
$$B = \left\{\bigcap_{i=1}^n f_i^{-1}(U_i) \mid \forall i:f_i\in\Omega, \forall i:U_i \in \tau_Y\right\}$$
Now we have that $(x_n)_n\to x$ if and only if for every basis element $G \in B$ containing $x$, the sequence lies eventually in $G$.
What is left is pretty straightforward but can be tedious. You have to verify that everything I said is true, and that $(f(x_n))_n\to f(x), \forall f \in \Omega$ implies that for every basis element $G \in B$ containing $x$, the sequence $(x_n)_n$ lies eventually in $G$.
A: for the $\Leftarrow)$:
let $B=\{\bigcap_{i \in I} f^{-1}(U_i): \forall i \in I, f_i \in \Omega, U_i \in \tau_Y \}$.
Let U a neighborhood of x. Exists $J \subset I$ finished so ,that $\bigcap_{i \in I} f^{-1}(U_i) \subseteq U$ with $U_i$ neighborhood of $f_i(x)$.
Because $f_i(x_n) \rightarrow f_i(x)$ then $\forall i \in J$ exists $N_i \in \mathbb N: f_i(x_n) \in U_i \quad \forall n \ge N_i$.
Let $N=\max_{i \in J} N_i$ then $x_n \in U \quad \forall n \ge N$. Then $x_n \rightarrow x $ in $\tau_{\Omega}$ 
it's right? 
