I am currently self-studying topology without tears. I am working on the following question:

1.2.7(iii): (open sets) If $X$ is an infinite set of cardinality $\aleph$, prove that there are at least $2^{\aleph}$ distinct topologies on it.

From this the point is to deduce that every infinite set has an uncountable number of distinct topologies on it.

In question $(i)$ we proved that the number of topologies grows when the number of points increases (countably so). We then in (ii) showed that a set on $n \in \mathbb N$ points has $(n-1)!$ distinct topologies. We used induction for this.

I am not sure whether or not I should somehow use these parts to prove this question or that this question is just meant to serve as a continuation of the train of thought. That is: increasing, then find some lower bound and now make the step to cardinal numbers.

If this question is actually a standalone question my best bet would be to make some sort of bijection/injection between the number of distinc topologies and the power set of the set $X$.

This power set has the desired cardinality $2^\aleph$. Am I thinking in the right direction, can somebody drop a small hint to either point me in the right direction or to help me further? If I made any logical errors feel free to point them out also. I'm here to learn.


1 Answer 1


I think you are overcomplicating things a little bit:

Let $X$ be an infinite set. Let also $P(X)$ be its power set. As you already know, $P(X)$ is uncountable. For simplicity, let's define $Q(X):=P(X) - \{\emptyset, X\}$. Obviously, $Q(X)$ is still uncountable

For every $A \in Q(x)$, we can define a topology as $T_A:=\{\emptyset, A, X\}$. Since $A \neq B \implies T_A \neq T_B$, we can conclude that $\mathbb{U}:=\{T_A: A\in Q(X)\}$ is also uncountable (just build a very easy injection from $\mathbb{U}$ to $Q(X)$)

  • $\begingroup$ Ah yes, I was thinking of a way to construct topologies with every single distinct subset but I didn't know how. This is a very simple argument, thank you! Indeed this is the kind of injection I was desiring. $\endgroup$
    – user459879
    Commented Jul 10, 2019 at 11:36
  • $\begingroup$ Thank you!!! much appreciated. $\endgroup$
    – user459879
    Commented Jul 10, 2019 at 11:38

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