The number of topologies on a finite set increases if the number of elements increases 
Suppose we are given $a$ set $A$ with $n$ distinct elements, where
  $n\in$ $\mathbb{N}$.
  Prove that if $n$ increases then the number of topologies on $A$ also increases.

For proving it I did:
We proceed by induction. 
Let a set $A$ with $n$ elements that has $M$ distinct topologies. Then $A\cup\{t\}$ has $n+1$ elements and we assume that $t \ne a$,    $\forall a\in A$.
By hypothesis $A$ has $M$ distinct topologies. If we define $\mathcal{T}=\{A,\emptyset,\{t\}\}$ we see notice that this new topology is different from the other topologies in $A$. So $A\cup\{t\}$ has at least $M+1$$distinct topologies.
Can someone tell me if my proof is right?
 A: Here’s another approach to the induction step that I find a little easier. Fix an element $a_0\in A$. If $\tau$ is a topology on $A$, let
$$\tau'=\{U\in\tau:a_0\notin U\}\cup\big\{U\cup\{t\}:a_0\in U\big\}\;.$$
The idea behind $\tau'$ is that we simply throw $t$ into every open set in $\tau$ that contains $a_0$, so that $a_0$ and $t$ are in exactly the same open sets in $\tau'$: we make $t$ a sort of twin of $a_0$.
It’s straightforward to verify that $\tau'$ is a topology on $A\cup\{t\}$, and if $\tau_0$ and $\tau_1$ are distinct topologies on $A$, then $\tau_0'$ and $\tau_1'$ are distinct topologies on $A\cup\{t\}$. Thus,
$$\mathscr{T}=\{\tau':\tau\text{ is a topology on }A\}$$ is a family of $M$ topologies on $A\cup\{t\}$. Moreover, if $U\in\tau'\in\mathscr{T}$, then either $\{a_0,t\}\subseteq U$, or $\{a_0,t\}\cap U=\varnothing$: no open set in any of these topologies on $A\cup\{t\}$ contains exactly one of the points $a_0$ and $t$. It follows that the discrete topology on $A\cup\{t\}$ is not in $\mathscr{T}$, because it includes the open set $\{t\}$. In fact, for each topology $\tau$ on $A$ we can let
$$\tau''=\tau\cup\big\{U\cup\{t\}:U\in\tau\big\}$$
and verify that $\{\tau'':\tau\text{ is a topology on }A\}$ is another set of $M$ topologies on $A\cup\{t\}$ that is disjoint from the collection $\mathscr{T}$, since $\{t\}=\varnothing\cup\{t\}\in\tau''$ for every topology $\tau$ on $A$. Thus, the number of topologies on $A\cup\{t\}$ is not just more than $M$: it’s at least $2M$.
A: Your proof has a problem. The problem is that a topology on $A$ is not going to be a topology on $A \cup \{t\}$. Remember that a topology $\mathcal{T}$ on $X$ requires that $X \in \mathcal{T}$.
If we take any topology $\mathcal{T}$ on $A$, it will not be a topology on $A \cup \{t\}$ because $A \cup \{t\} \notin \mathcal{T}$.
To fix your proof, you must first turn the topology $\mathcal{T}$ on $A$ into a topology $\mathcal{T'}$ on $A \cup \{t\}$.
To do so, I would suggest simply throwing in the extra element $A \cup \{t\}$ into the topology. You have to verify that it then satisfies the axioms of a topology.
Then you want to show that there are strictly more topologies on $A \cup \{t\}$. You are almost there with this part, when you consider the set of three elements $\{A, \varnothing, \{t\}\}$. But this set needs one more element to be a topology. What element is that?
