Understanding of the topology of pointwise convergence 
Let $X$ be a set, $Y$ be a topological space, and $F$ be a set of functions from $X$ into $Y$. The weak topology on $X$ induced by $F$ is defined as the smallest topology $T$ on $X$ for which each function $f\in F$ is continuous. Show that convergence of nets in this topology is completely characterized by 
  $$
\lim_{\alpha\in A}x_\alpha=x
$$
  iff
  $$
\lim_{\alpha\in A}f(x_\alpha)=f(x)
$$
  for every $f\in F$. 


By using the fact that $T$ is generated by  $\mathscr{B}:=\{f^{-1}(U)\mid f\in F,\ U\ \hbox{open in}\ Y \}$, I can show that one direction is true, namely, if $\lim_{\alpha\in A}x_\alpha=x$, then
$$
\lim_{x\in A}f(x_\alpha)=f(x)
$$
for every $f\in F$. Could anyone give me some ideas about the other direction? I tried to assume a non-convergent net in $X$, but I don't see get some $f\in F$ such that the corresponding net does not converge in $Y$. 
 A: You’re on the right track. I’ll push you a bit further.
Suppose that $\langle f(x_\alpha):\alpha\in A\rangle$ converges to $f(x)$ for each $f\in F$, but $\langle x_\alpha:\alpha\in A\rangle$ does not converge to $x$. Then there must be an open nbhd $U$ of $x$ such that for each $\alpha\in A$ there is a $\beta(\alpha)\in A$ such that $\alpha\preceq\beta(\alpha)$ and $x_{\beta(\alpha)}\notin U$. (Here $\preceq$ is the order on the directed set $A$.)
By the definition of $T$ there must be $f_1,\ldots,f_n\in F$ and for $k=1,\ldots,n$ open nbhds $V_k$ of $f_k(x)$ in $Y$ such that 
$$x\in\bigcap_{k=1}^nf_k^{-1}[V_k]\subseteq U\;.$$
Thus, for each $\alpha\in A$ we have $x_{\beta(\alpha)}\notin\bigcap_{k=1}^nf_k^{-1}[V_k]$.
By hypothesis $\langle f_k(x_\alpha):\alpha\in A\rangle$ converges to $f_k(x)$ for $k=1,\ldots,n$, so for $k=1,\ldots,n$ there is an $\alpha_k\in A$ such that $f_k(x_\alpha)\in V_k$ whenever $\alpha_k\preceq\alpha$. $A$ is directed, so there is an $\alpha_0\in A$ such that $\alpha_k\preceq\alpha_0$ for $k=1,\ldots,n$. Now consider $x_{\beta(\alpha_0)}$.
