# Is there a connected $T_1$ anticompact topological space?

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.

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?

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.

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.

• Your uncountable example $X$ is NOT Hausdorff but it IS $T_1.$ Commented Nov 20, 2022 at 6:29
• Just taking the cocountable topology on $\mathbb{R}$ (or $\omega_1$) is also anticompact, connected and T1 (but not Hausdorff).
– Ulli
Commented Nov 20, 2022 at 6:34
• @DanielWainfleet: the example of the OP is finer than the euclidean topology, hence it is Hausdorff. Just taking the cocountable topology wouldn't be Hausdorff.
– Ulli
Commented Nov 20, 2022 at 6:38
• @Ulli I kind of knew that, but thanks for reminding me. I guess the nice thing is it's a Hausdorff example (which should have been a better original question). I'll update the post. So the only thing not obvious is to find a countably infinite example (T_1 or T_2). Commented Nov 20, 2022 at 7:27
• According to space 63, item 9 in Counterexamples your X in Question 2 is connected. I think your proof for anticompact is also correct (item 7 in Counterexamples is exactly this as well, though they didn't track "anticompact") and will contribute it to pi-Base. Commented Nov 22, 2022 at 17:36

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.

• Very nice example for the $T_1$ case. Commented Nov 22, 2022 at 7:43
• Talking about ultrafilters, there is the Single ultrafilter topology = example #114 in "Counterexamples in Topology". It is countable and $T_2$ and it seems the same arguments that you used can show it is anticompact. But it is far from connected, as it's zero-dimensional. Commented Nov 22, 2022 at 7:55
• Gustin's sequence space seems to be a candidate: it's countable, T2, and connected. Anti-compact is open in the pi-Base. EDIT: or not, it's first-countable, that's just not tracked. Commented Nov 23, 2022 at 4:43

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.