# 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$$distinct topologies. Can someone tell me if my proof is right? • Augh! My eyes! :O Aug 30, 2016 at 22:52 • Generating one extra topology on$A\cup\{t\}$should work, but you need to first show that the$M$topologies you have on$A$lead to$M$distinct topologies on$A\cup\{t\}$, and then that the topology you give here is distinct from these$M$topologies. Aug 30, 2016 at 22:55 • Not if$A$is the empty set, which has one topology on it, as then$A\cup \{t\}=\{t\}$, which has one topology on it. Aug 31, 2016 at 6:55 ## 2 Answers Here’s another approach to the induction step that I find a little easier. Fix an element $$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$$. 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?

• its AU{t} i suppose Aug 30, 2016 at 23:08
• @user359315 Yep. Aug 30, 2016 at 23:08
• And this new topology that you defined is diferent from the other M topologies because it contains the set {t} Aug 30, 2016 at 23:09
• @user359315 Good, yes. Aug 30, 2016 at 23:10
• I see.Thank you for your help! Aug 30, 2016 at 23:11