If $\mathcal T$ and $\mathcal S$ are two topologies, can we defined $\mathcal T\cup \mathcal S$ and $\mathcal T\cap \mathcal S$ on $X$? Let $X$ a topology space, and $\mathcal T$ and $\mathcal S$ two topology on $X$. I would define $\mathcal T\cap \mathcal S$ the thinest topology included in $\mathcal T$ and $\mathcal S$ and $\mathcal T\cup\mathcal S$ being the coarser topology that contain $\mathcal T$ and $\mathcal S$. 
Obviously, each topology is contains in the discrete topology, and all topology contain the coarser topology, but is there any way to construct $\mathcal T\cup \mathcal S$ and $\mathcal T\cap \mathcal S$. 
For $\mathcal T\cup \mathcal S$, I would do set 
$$\left\lbrace\bigcup_{i\in \mathcal I}U_i\cap V_i\mid U_i\in \mathcal T,V_j\in \mathcal S\right\rbrace.$$
This look to be a topology, no ? And for $\mathcal T\cap \mathcal S$, I have no idea... but may be it's not possible ?   
 A: For $\mathcal T \cap \mathcal S$ you can use the set-theoretic intersection. i.e.
$\mathcal T \cap \mathcal S =\{U: U\in \mathcal T \text{and} \ U\in \mathcal S \}$. Then clearly you can see this is the largest topology which lies inside both topologies. 
$\mathcal T \cup \mathcal S$ set $\mathcal B=\{U: U\in \mathcal T \text{or} \ U\in \mathcal S \}$. Then $\mathcal B$ is a subbase for the smallest topology which contains both topologies, and this gives the topology you describe. 
Note that both facts generalise to arbitary sups and infs.
A: Writing $\mathcal{T} \cup \mathcal{S}$ is misleading imo, since a topology is by definition a subset of the powerset, so the union of two such sets is already defined.
Regarding the intersection of two (in fact arbitrary intersections of) topologies (by which I mean the intersection of the subsets of the powerset) note that all the axioms are satisfied, since they are satisfied in all topologies involved.
The coarsest topology containing given two topologies may be defined as the intersection of all topologies containing $\mathcal{T}$ and $\mathcal{S}$. Indeed, one can show that this is explicitly given by your definition.
