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?

  • 2
    $\begingroup$ Augh! My eyes! :O $\endgroup$ Aug 30, 2016 at 22:52
  • $\begingroup$ 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. $\endgroup$
    – 211792
    Aug 30, 2016 at 22:55
  • $\begingroup$ 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. $\endgroup$ Aug 31, 2016 at 6:55

2 Answers 2


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


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?

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

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