indiscrete topology Let X be an infinite set and let $\mathcal T $ be a topology on X.
2) prove that if $\mathcal T $ contains every infinite subset of X, then it is the indiscrete topology.
i think this is untrue, consider $X= \Bbb N $ the indiscrete topology has two sets $\{\emptyset ,\Bbb N\}$  but we have for example $2\Bbb N $ and $\Bbb P $ which are not in the indiscrete topology? 

If every infinite subset of $X$ is declared open, then this topology is the discrete topology, not the indiscrete one.
To see this, take an arbitrary countably infinite subset $Y$ of $X$. Enumerate the elements of $Y$ as $\{y_1,y_2,\ldots\}$ in such a way that all these elements are distinct (that is, $y_n\neq y_m$ whenever $n,m\in\mathbb N$ and $n\neq m$). Now define
\begin{align*}
Y_{\text{o}}\equiv&\;\{y_n\,|\,n\in\mathbb N\text{ is odd}\}=\{y_1,y_3,y_5,\ldots\},\\
Y_{\text{e}}\equiv&\;\{y_n\,|\,n\in\mathbb N\text{ is even}\}=\{y_2,y_4,y_6,\ldots\}.
\end{align*}
Then $Y_{\text{o}}$ and $Y_{\text{e}}$ are both infinite and they are disjoint.
Take any arbitrary $x\in X$. Then, the sets $Y_{\text{o}}\cup\{x\}$ and $Y_{\text{e}}\cup\{x\}$ are both infinite and thus open, and the only common element in them is $x$. The intersection of these two sets, which is $\{x\}$, is therefore open.
It follows that every singleton is an open set, so that the topology must be the discrete one.
