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.

  • $\begingroup$ $X=\{0,1\}$ and $T=\{\varnothing,\{0\},\{0,1\}\}$. Then every set except $\{1\}$ is open, so $\{1\}$ is closed, and neither $\{0\}$ nor $\{1\}$ are clopen. $\endgroup$ – Asaf Karagila Apr 17 '19 at 14:11
  • $\begingroup$ Yes, and that fits my conjecture ($u=1$) $\endgroup$ – Lucio Tanzini Apr 17 '19 at 14:13
  • $\begingroup$ 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. $\endgroup$ – InterstellarProbe Apr 17 '19 at 14:14
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Apr 17 '19 at 14:15
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    $\begingroup$ 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) $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – Lucio Tanzini Apr 17 '19 at 15:51
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Apr 17 '19 at 15:55

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