TITLE: how to find the unique smallest topology?
QUESTION [9 upvotes]: I  have come  across  to  the following question : 
Let  $\mathscr{T}_\alpha$  be a family  of  topologies    on  $ X$ .  Show  that   there  is   a  unique  smallest   topology   on  $X$    containing   all   the  collections   $\mathscr{T}_\alpha$ , and  a  unique    largest   topology   contained   in  all  $\mathscr{T}_\alpha$. 
I think  that  the unique smallest  topology  equals  the union  of  all  the $\mathscr{T}_\alpha's$, and the unique largest  topology  contained  in  all  $\mathscr{T}_\alpha$  equals  the  intersection of all  the $\mathscr{T}_\alpha's$ .

REPLY [10 votes]: The basic fact to verify is that if $\cal{T}_i$, $i \in I$ is an an indexed collection of topologies on a set $X$, then their intersection $\cap_{i \in I} T = \left\{ A \subset X: \forall i \in I: A \in\cal{T}_i \right\}$ is a topology on $X$ as well. This is straightforward from the topology axioms. 
Once this is done, then the smallest topology that contains all $\cal{T}_i$ can be defined as the intersection of all topologies $\cal{T}$ that contain $\cup_{i \in I} T_i$ as a subfamily, and there is at least one (the discrete topology) so we take the intersection of a non-empty family of topologies, which is a topology as we saw above. And it is clearly the smallest one that contains all $\cal{T}_i$: if $\cal{T}$ is any such topology it is one of the topologies we take the intersection of, and thus clearly is a subset of $\cal{T}$.
Also, the largest topology contained in all $\cal{T}_i$ is simply their intersection, and this is clearly the largest topology that is contained in all of them.