# What happens if $U\in \textit{P}(X)$ is always either closed or open

Let $$X$$ be a topological space s.t. $$\forall U\in \textit{P}(X)$$, $$U$$ is either open or closed (or both only in the cases $$U=X, U=\varnothing$$).

What can we say about $$X$$?

I found this example of such a space: Let $$X$$ a set. Fix $$u\in X$$, and then say that $$U\in \textit{P}(X)$$ is closed $$\iff$$ $$u\in U$$ or $$U=\varnothing$$.

Note that in the above example I could substitute "open" with "closed" and it would still work.

My conjecture (which is poorly founded) is that all such topological spaces are of the above forms.

• $X=\{0,1\}$ and $T=\{\varnothing,\{0\},\{0,1\}\}$. Then every set except $\{1\}$ is open, so $\{1\}$ is closed, and neither $\{0\}$ nor $\{1\}$ are clopen. – Asaf Karagila Apr 17 '19 at 14:11
• Yes, and that fits my conjecture ($u=1$) – Lucio Tanzini Apr 17 '19 at 14:13
• Consider the set $\{0,1,2\}$ and the topology: $$\{\emptyset, \{0\}, \{0,1\}, \{0,2\}, \{0,1,2\}\}$$ Now $\{1\}$ and $\{2\}$ are both closed, neither contains the other, and every element of the power set is either open or closed. – InterstellarProbe Apr 17 '19 at 14:14
• How do you quantify over $u$? What you wrote doesn't really parse properly. Because if it just means what it says, then $u=0$ doesn't satisfy this. – Asaf Karagila Apr 17 '19 at 14:15
• Isn't this just saying that the opens (without $\emptyset$) form an ultrafilter in $P(X)$ and then your conjecture would probably be something that says that all ultrafilters are principal (and that is where you get to say something, Asaf) – Mark Kamsma Apr 17 '19 at 14:21

This thread is about so-called "door spaces" which is the name for a topological space where every subset is open or closed (like a door). It gives a link to a paper that proves there are three types of connected door spaces: included point topologies (there is a point so that $$O$$ is open iff $$p \in O$$ or $$O$$ is empty), excluded point topologies (the same with closed instead of open, as you name) and one based of a free (non-principal) ultrafilter on $$X$$. These objects only exist under a form of the axiom of choice (they're not "constructive").

Mark Kamsma's comment answers your question. I figured I'd expand on it, since you seem to be ignoring it.

Let $$F$$ be an ultrafilter on a set $$X$$. Recall the definition of an ultrafilter on a set. An ultrafilter on $$X$$ is a nonempty collection $$F$$ of subsets of $$X$$ such that

1. $$\varnothing\not \in F$$,
2. If $$A\in F$$ and $$A\subseteq B$$, then $$B\in F$$.
3. If $$A,B\in F$$, then $$A\cap B\in F$$.
4. For all $$A\subseteq X$$, either $$A\in F$$ or $$A^C\in F$$.

Then the collection $$\tau = F\cup \{\varnothing\}$$ is a topology for $$X$$ satisfying the condition that for every nonempty proper subset $$\varnothing \subsetneq U\subsetneq X$$, exactly one of $$U$$ and $$U^C$$ is in $$\tau$$.

Being a topology follows from properties 2 and 3 of $$F$$ being an ultrafilter, and satisfying the property you want follows from properties 1 and 4.

Thus a nonprincipal ultrafilter, produces a topology which is a counterexample to your conjecture.

• Oh yes, now I see. By the way this example coincides with my example in the finite case. Do you think the conjecture holds for finite spaces? – Lucio Tanzini Apr 17 '19 at 15:51
• @LucioTanzini For finite spaces there are no free ultrafilters, they are all fixed. So then the ultrafilter of closed (or open) sets has a singleton intersection, and the hypothesis holds. – Henno Brandsma Apr 17 '19 at 15:55