Does there exist a topological space with countably infinite open sets? This is a problem I've been pondering a lot about lately. I haven't come up with a proof, but I feel the answer must be no.
I have a way of show that a sequence of topologies from the trivial to the
discrete topology goes from finite to uncountable infinite. Let $X=\{x_n\}_{n=1}^\infty$ be at least a countably infinite set. The coarsest topology is the trivial topology: $\tau_0=\{\emptyset, X\}$. Suppose I want the next coarsest topology. I can do this by adding $\{x_1\}$ to $\tau_0$ and closing the the collection under union and finite intersection: $\tau_1=\{\emptyset,X,\{x_1\}\}$. Continuing, I get $\tau_2=\{\emptyset,X,\{x_1\},\{x_2\},\{x_1,x_2\}\}$, and then $\tau_3=\{\emptyset,X,\{x_1\},\{x_2\},\{x_3\},\{x_1,x_2\},\{x_1,x_3\},\{x_2,x_3\},\{x_1,x_2,x_3\}\}$. As you can see, $\tau_n=\tau_0 \cup \mathcal{P}\left(\{x_i\}_{i=1}\right)$, where $\mathcal{P}(.)$ is the power set. Clearly, this sequence approaches $\tau_\infty$, the discrete topology, which is uncountably infinite. 
This gives me the idea that there can't be a countably infinite topology, but that's really all I got. Any ideas?
 A: Take $X=\mathbb{N}$, and let $\tau$ consist exactly of:


*

*The empty set.

*The sets of the form $\{0,1,2,3,\ldots,n\}$

*The set $X$.


I claim that $\tau$ is a topology, and that $|\tau|=\aleph_0$, countably infinite. 
Indeed: $\varnothing$ and $X$ are elements of $\tau$. 
To show that $\tau$ is closed under arbitrary unions, consider an arbitrary family $\mathcal{F}$ of elements of $\tau$. Since $\tau$ is totally ordered, either the family  has only finitely many distinct elements, in which case the union is the largest element of the family; or else the family has infinitely many distinct elements, in which case for every $k\in X$ there exists an element $O\in\mathcal{F}$ with $k\in O$, so that $\cup \mathcal{F}=\mathbb{N}\in\tau$. Finally, if $X_1$ and $X_2$ are in $\tau$, then either $X_1\subseteq X_2$ or $X_2\subseteq X_1$, so $X_1\cap X_2\in \mathcal{F}$. Thus, $\tau$ is a topology.
And finally, $\tau$ is clearly countably infinite.

Your error lies in that you did not create the "simplest" topology after $\{\varnothing,\{x_1\}, X\}$ when you added the open set $\{x_2\}$. You can just have a new open set that properly contains $x_1$ and nothing else; somehing like $\{\varnothing,\{x_1\},\{x_1,x_2\},X\}$ (similar to the Sierpinski topology). Etc. 
A: Try $\mathbb N$ with the topology consisting of $\emptyset$ and the sets $\{x: x \ge n\}$ for $n \in \mathbb N$.
