TITLE: 'Complementary topology'?
QUESTION [6 upvotes]: Let $(X,\mathscr{T})$ be a topological space, and define a new topology $\mathscr{T}'$ on $X$, such that the collection of closed sets $\mathscr{C}$ of the topology $\mathscr{T}$ form a sub-basis for $\mathscr{T}'$. Lets refer to $\mathscr{T}'$ as the 'complementary topology' for convenience.
In fact since the intersections of closed sets are again closed and since $X\in \mathscr{C}$, the sub-basis also covers the space, so the sub-basis $\mathscr{C}$ would be a basis for $\mathscr{T}'$.
I was curious as to the properties of this topology. An initial thought was that this may be the discrete topology. Taking for example the standard euclidean topology $(\mathbb{R},\mathscr{T}_{E})$, if we form the 'complementary topology', we have in our basis $\mathscr{C}_E$, that $\{x\}\in\mathscr{C}_E, \, \forall x\in\mathbb{R}$. Since open sets are formed from unions of these sets, any set is open so $\mathscr{T}_{E}'=\mathscr{T}_{D}$, the discrete topology.
This cannot be true in general since taking for example a three point set $\{1,2,3\}$, with the topology $\mathscr{T}=\{\emptyset, \{1\},\{1,2\},\{1,2,3\}\}$, the complementary topology is given by $\mathscr{T}'=\{\emptyset,\{3\},\{2,3\},\{1,2,3\}\}$, which is certainly not the discrete topology.
However in the second example we do have that $(\mathscr{T}')'=\mathscr{T}$.
So I am curious if there are conditions on the topology such that we have either $(\mathscr{T}')'=\mathscr{T}$ or $(\mathscr{T})'=\mathscr{T}_D$? Is there a simple example where neither hold?
Of course if the topology was discrete in the first place both would hold.

REPLY [2 votes]: First if $(X,\mathscr{T})$ is a finite topology (i.e. $|\mathscr{T}|<\infty$) then any union of closed sets is a finite union of closed sets so that the set of closed sets is closed by union. Whence the topology $\mathscr{T}'$ you get by taking the topology generated by the closed sets is actually the set of closed sets. Whence if $(X,\mathscr{T})$ is a finite topology you always get $(\mathscr{T}')'=\mathscr{T}$. (this is a particular case of Alexandroff's space, see $major^4$'s answer).
Second, if $(X,\mathscr{T})$ is $T_1$ (by definition : all points are closed) then $\mathscr{T}'$ is $\mathscr{T}_D$ the discrete topology (this is Daniel Fischer's comment).
Hence, looking for a counter example, you need $(X,\mathscr{T})$ an infinite topology, not $T_1$. In particular $X$ should be infinite. 
Take $X:=\mathbb{N}$ and define $\mathscr{T}$ by saying that $A$ is open if $A\subseteq \mathcal{P}$ (the prime numbers) and $\mathcal{P}-A$ is finite. This is a topology. Furthermore $2$ is not a closed point. 
Then the set of open sets being not stable by infinite intersection (e.g. we can write the set of prime number equal to $1$ mod $4$ as an intersection of open sets)  it follows that $\mathscr{T}'$ will contain a set which is not closed for $\mathscr{T}$ so that $(\mathscr{T}')'\neq\mathscr{T}$. 
Now we need to show that $\mathscr{T}'$ is not the discrete topology. Since closed sets for $\mathscr{T}$ are included in $\mathcal{P}^c$, it follows that the topology generated by those sets cannot generate a set which contains $2$, in particular $\{2\}\notin \mathscr{T}'$ and the topology cannot be discrete.