$\mathcal{T}(\mathcal{S})$ the smallest topology of $X$ Given a set $X$ and a family $\mathcal{S}$ of subsets of $X$, prove that there exists a topology $\mathcal{T}(\mathcal{S})$ on $X$ which contains $\mathcal{S}$ and is the smallest with this property (Hint: use the axioms to see what other subsets of $X$, besides the ones from $\mathcal{S}$, must $\mathcal{T}(\mathcal{S})$ contain.)
I checked the axioms of a topology and I think the other subset $\mathcal{T}(\mathcal{S})$ has to contain is $X$ (for the first axiom). Is this correct? and why is this the smallest topology? How do I prove that?
 A: The "smallest" topology $\mathcal{T}(S)$ is the topology such that if $T'$ is any other topology containing the subsets in the collection $S$, then $\mathcal{T}(S) \subseteq T'$.
As the other answerers have pointed out, consider taking every topology on the set $X$ which contains all of the sets in $S$, and then take the intersection of all of those topologies.  We get another topology!  To show this:
Consider $(\mathcal{T}_{i})_{i \in I}$ to be the collection of topologies on $X$ in which each topology $\mathcal{T}_{i}$ contains the sets in $S$.  We are interested in three things:


*

*Is this intersection non-empty?  Yes!  Since the power set of $X$, $\mathcal{P}(X)$ is a topology which contains every set in the collection $S$, and also since the set $X$ is an element of every topology, the intersection of the topologies is non-empty.

*Is the intersection a topology?  Yes!  You should prove this (it's very easy).  Suppose $\mathcal{T}_{i}$ is a topology for each $i \in I$.  Suppose the intersection $\cap_{i \in I} \mathcal{T}_{i}$ is non-empty.  Then this intersection is also a topology.  Prove it.

*Once we establish that for the topologies $\mathcal{T}_{i}$ that contain the collection $S$, $\cap_{i \in I} \mathcal{T}_{i}$ is a topology, then we want to know if this topology contains the collection $S$.  Does it?  Yes! Why?

*Finally, do we have that $\cap_{i \in I} \mathcal{T}_{i} \subseteq T$ for any topology $T$ which contains the collection of sets $S$?  Yes!  Why?
A: The intersection of an arbitrary non-empty family of topologies containing $\mathcal{S}$ is still a topology. In this case the family is non-empty, indeed it contains the power set of $X$.
