Let $X$ a set and $\mathcal{A}$ a base for the topology $\tau$. Consider $\{\tau_\alpha\}_{\alpha\in I}$ a family of all topologies over $X$ such that $\mathcal{A}\subset\tau_\alpha$ for any $\alpha\in I$. I want to show that $\bigcap_{\alpha\in I}\tau_\alpha=\tau$.
Proof:
$\tau\subset\bigcap_{\alpha\in I}\tau_\alpha$: Let $u\in\tau$ then, exists $V_\kappa\in\mathcal{A}$ such that $u=\bigcup V_\kappa$ and, as each $V_\kappa$ is a open set for each $\tau_\alpha$ $(\alpha\in I)$, we have that $u\in\tau_{\alpha}$ for $\alpha\in I$. So, we obtain $u\in\bigcap_{\alpha\in I}\tau_{\alpha}$.
I have troubles with the other contention. Can someone give me a hit to solve my problem?
It is only true if $\{\tau_{\alpha}\mid\alpha\in I\}$ is the collection of all topologies that contain $\mathcal A$.
If that is not the case then the other side does not have to be true.
If that is the case then the other side goes like this:
If $U\in\bigcap_{\alpha\in I}\tau_{\alpha}$ then $U\in\tau$ since $\tau=\tau_{\alpha_0}$ for some $\alpha_0\in I$ and $\bigcap_{\alpha\in I}\tau_{\alpha}\subseteq\tau_{\alpha_0}$