Is there a connected $T_1$ anticompact topological space?

Question

Now cross-posted to Mathoverflow.

A topological space is a called anticompact if all its compact subsets are finite.

Question 1: Is there a (non-trivial) connected $T_1$ anticompact space?

Obviously the singleton and the empty space satisfy these properties. So let's assume such a space $X$ has at least two points. Here are some observations:

(1) $X$ must be infinite. Because finite $T_1$ spaces are discrete, such a space would not be connected.

(2) $X$ cannot be compact.

(3) $X$ must be totally path disconnected, because anticompact $T_1$ spaces are totally path disconnected. (The image of a continuous map $[0,1]\to X$ is compact, connected and $T_1$, hence finite and $T_1$, so it would be discrete, and connected. Thus it must a single point.)

(4) $X$ cannot be sequential, as a consequence of this. In particular, $X$ cannot be first countable.

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Answer (from Ulli): Yes. Any uncountable set with the cocountable topology is connected and anticompact. It is also $T_1$, but not $T_2$.

So updating the question:

Question 2: Is there a (non-trivial) connected Hausdorff anticompact space?

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I think the following would be an example. Take $X=[0,1]$. Let $\tau_1$ be the usual Euclidean topology on $X$ and take on $X$ the topology $\tau$ generated by $\tau_1$ and the cocountable topology. By using the Lindelof property for $\mathbb{R}$ it's not too difficult to show that the topology is described by $$\tau=\{U\setminus A: U\in\tau_1 \text{ and } A\subseteq X \text { is countable}\}.$$

Every countable subset of $X$ is closed, and is discrete since all its subsets are also countable, hence closed. So if $A$ is an infinite subset of $X$, it cannot be compact; otherwise a countably infinite subset of it would also be compact (as closed in a compact set) and discrete, which is impossible. This shows that $X$ is anticompact.

$X$ is Hausdorff.

To show that $X$ is connected, suppose $X=Y\cup Z$ with $Y$ and $Z$ nonempty disjoint open sets. Take a point $x\in Y$. There is a Euclidean nbhd $U$ of $x$ such that $Y$ contains all the points of $U$ except countably many of them (where countable = finite or countably infinite). Any point $z\in U\setminus Y$ would have to be contained in $Z$. And since $Z$ is open in $X$, $U\cap Z$ would be a countable nbhd of $z$, which is impossible. This shows that both $Y$ and $Z$ are open in the Euclidean topology. Again that is impossible since $[0,1]$ is connected in its usual topology.

I would appreciate if you could confirm that the example above is correct.

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This leads me to the revised question:

Question 3: Is there a countably infinite connected $T_1$ (or $T_2$) anticompact space?

I can't find an example. The Arens-Fort space or the Appert space are anticompact (and not first countable) but not connected. Countably infinite connected examples that I know of, like the irrational slope topology or the Golomb space, are first countable, hence not anticompact.

Answer 1

Here is a partial answer to question 3, namely an example of a connected, anticompact T1-topology on the (positive) integers:

Let $\mathfrak{F}$ be a free ultrafilter on $\mathbb{N}$ and $\tau := \mathfrak{F} \cup \{\emptyset\}$ the ultrafilter topology on $\mathbb{N}$. Since $\mathfrak{F}$ is free, for each $n \in \mathbb{N}$, $\mathbb{N} \setminus \{n\} \in \mathfrak{F}$, hence $\{n\}$ is closed. Thus, the topology is T1.

The intersection of any two non-empty, open sets is non-empty again. Hence the topology is connected.

It is anticompact:
Let $A$ be an infinite subset of $\mathbb{N}$. Then there exists $U \in \mathfrak{F}$ such that $A \setminus U$ is infinite.
[$A$ is the disjoint union of two infinite subsets $S, T$. If $S \in \mathfrak{F}$ or $T \in \mathfrak{F}$, we are done. Otherwise, $\mathbb{N} \setminus S \in \mathfrak{F}$ and $\mathbb{N} \setminus T \in \mathfrak{F}$. Hence $\mathbb{N} \setminus A = \mathbb{N} \setminus (S \cup T) = (\mathbb{N} \setminus S) \cap (\mathbb{N} \setminus S) \in \mathfrak{F}$.]
Now, $A \setminus U$ is discrete.
[For any $x \in A \setminus U$, $\{x\} = (U \cup \{x\}) \cap (A \setminus U)$ is open in $A \setminus U$.]
Furthermore, $A \setminus U \notin \mathfrak{F}$. Hence its complement is open, hence $A \setminus U$ is closed in $\mathbb{N}$ and thus in $A$. Therefore $A$ cannot be compact.

Note: Of course, this topology is far from being Hausdorff.

Answer 2

As answered in the cross-post to mathoverflow, a paper by T. Banakh and Y. Stelmakh provides examples of countably infinite connected Hausdorff anticompact spaces. Their examples are even strongly rigid.