Is $(X,\mathcal T)$ necessarily an indiscrete space? > Let $X$ be an infinite set with $\mathcal T$ a topology on $X$. If $X$ is the only infinite subset of $X$ that is open, is $(X,\mathcal T)$ necessarily an indiscrete space? My intuitive thinking leads me to think that the answer is NO. We know that, if it is the indiscrete space, then $\mathcal T = \\{ \emptyset, X \\}$. Notice, however, that we are only given tat $X$ is the only INFINITE subset of $\mathcal T$, we have no information regarding the other possible finite open sets of $X$, that is, there may exist a finite subset $A \subset X$ with $A \in \mathcal T$. This is surely not the indiscrete space, yet it still satisfies the criteria given in the question. Is there a better exact example that I can provide that might help show my point in a more rigorous mathematical way?

You have the right idea, but in order to make it convincing, you have to produce an actual counterexample. Here’s a simple one: let $X=\Bbb N$, and let $\tau=\big\\{\varnothing,\\{0\\},\Bbb N\\}$. It’s straightforward to verify that $\tau$ really is a topology on $X$, and that $X$ is the only infinite open subset of $X$.

With a little more work you can make the space $T_0$. Once again let $X=\Bbb N$, for each $n\in\Bbb N$ let $U_n=\\{k\in\Bbb N:k
You cannot make the space $T_1$, however: if $X$ is $T_1$, and $x\in X$, then $X\setminus\\{x\\}$ is open.