# Proof that every infinite set $X$ has an uncountable number of topologies on it.

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.

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 $$Q(X)$$)