# Spaces $X$ in which every subset is either open or closed, and only $\varnothing$ and $X$ are clopen

Let $(X, \tau)$ be a topological space. Then $X, \varnothing \in \tau$ and are both clopen.

But I wonder if it is possible to construct a topological space $X$ in which all subsets are either open or closed, but $X$ and $\varnothing$ are the only clopen subsets.

• Oh, is it? I am new to general topology, so... :)
– Yes
Aug 28, 2015 at 15:35
• My bad, I went from the title and didn't carefully read the second paragraph. Your property implies connectedness, but it is a much stronger property.
– Ian
Aug 28, 2015 at 15:36
• In $\mathbb R$, being open and closed are contradictory (other than $\mathbb R,\varnothing$). The problem is sets that are neither, not both. Aug 28, 2015 at 15:38
• Presumably you want each subset to be open or closed but not both. In the discrete topology, every subset is both open and closed. Aug 28, 2015 at 15:48
• Exactly, want to know more pathological examples.
– Yes
Aug 28, 2015 at 15:50

Let $X$ be an infinite set, and let $\mathcal U$ be a ultrafilter on $X$. Then $\mathcal T = \mathcal U \cup \{ \emptyset \}$ is a topology on $X$ in which every subset of $X$ is either open or closed, and $\emptyset$ and $X$ are the only clopen subsets.

That it is a topology follows from the fact that $\mathcal U$ is a filter, so is closed under finite intersections, and as every superset of an element of $\mathcal U$ is an element of $\mathcal U$, $\mathcal U$ is clearly closed under arbitrary unions.

That $\mathcal U$ is an ultrafilter implies that every subset is either open or closed, and also implies that $\emptyset$ and $X$ are the only clopen subsets (since if $A$ were another clopen subset, then both $A$ and $X \setminus A$ would belong to $\mathcal U$, meaning that $\emptyset = A \cap ( X \setminus A ) \in \mathcal U$, contradicting our assumption that $\mathcal U$ is an ultrafilter).

(If $\mathcal U$ were a principal ultrafilter, we would get the same topology described in N. S.'s answer.)

– Yes
Aug 28, 2015 at 15:53
• Are all solutions ultrafilters, then? Aug 28, 2015 at 16:01
• @GudsonChou Ultrafilters are very useful. They're the basis for nonstandard analysis/topology, for one thing. Also, here's a question: Take all of the integers, draw an edge between every pair of them to make a (complete) graph, and color every edge either red or blue. Is there always an infinite subset $S$ that's monochromatic (the edges between pairs of elements of $S$ are all the same color)? (For example, suppose we color an edge blue if the vertices have the same parity, and red otherwise. Then take the set of even integers.) The answer is, yes, and you can prove this with ultrafilters. Aug 28, 2015 at 16:12
• @columbus8myhw Not quite. You could take the excluded point topology on a set. This is kind of the opposite of the topology from a principal ultrafilter, but is not generated from an ultrafilter in the same way. I don't know if there are any more. Aug 28, 2015 at 16:36
• @hereisnowhy Ah, I see. The fact that $\varnothing$ is in the topology, but not in an ultrafilter (by definition), is annoying… Aug 28, 2015 at 16:39

The spaces where every set is either open or closed are called door spaces. The spaces where the empty set and the whole space are the only clopen sets are called connected spaces. So you are asking about connected door spaces. They are in fact fully classified. There are no other such spaces then already mentioned examples – the empty space, principal ultrafilter spaces i.e. included point topologies, free ultrafilter spaces, principal ultraideal spaces i.e. excluded point topologies.

• Hi, do you have any references for this classification? Jul 28, 2019 at 2:14
• @Luke: S. D. McCartan, Door spaces are identifiable, Proc. Roy. Irish Acad. Sect. A, 87 (1987), pp. 13–16. A door space $(X, τ)$ is either discrete, or has exactly one non-isolated point, or $τ$ is an expansion of an ultrafilter topology by a set of new isolated points (i.e. there is an ultrafilter $\mathcal{U} ⊆ \mathcal{P}(X)$ and a set $A ⊆ X$ such that $U ∈ τ$ if and only if $U ∈ \mathcal{U}$ or $U ⊆ A$). Jul 28, 2019 at 15:58

Yes, for example $X = \{0, 1\}$ with the topology $\tau = \{ \emptyset, \{0\}, \{0, 1\}\}$.

Sure.

$$X=\emptyset, \tau = \{\emptyset\}$$

That's cheating!!!

OK, so what about if $X$ is not empty?

Well, here's another example:

$$X=\{1\}, \tau=\{\emptyset, X\}$$

OK OK, but there is no nontrivial closed set here, that's cheating!

OK then, $$X=\{1,2\}, \tau=\{\emptyset, \{1\}, X\}$$

• Thank you. I think my question rules out this possibility.
– Yes
Aug 28, 2015 at 15:38
• @GudsonChou Why? If $X=\emptyset$ and $\tau=\{\emptyset\}$, then for every set $A$ in $\tau$ such that $A\neq\emptyset$ and $A\neq X$, the set $A$ is either open and closed. There is no contradiction here!
– 5xum
Aug 28, 2015 at 15:39
• @5xum I think this question is probably a simplification of some more complicated question with additional hypotheses.
– Ian
Aug 28, 2015 at 15:42
• @Ian Possibly. But that's why I asked "Why", since I don't see how my possibility can be ruled out...
– 5xum
Aug 28, 2015 at 15:44
• @GudsonChou I added two more possible answers, are all of them ruled out?
– 5xum
Aug 28, 2015 at 15:44

Pick any $X$ and $a \in X$.

Define $Y \subset X$ to be open if and only i f$X= \emptyset$ or $a \in Y$. It follows that a set $Y$ is closed if and only if $Y=X$ or $a \notin Y$.

It is easy to show that this is a topology which has the required properties.

It is likely that there is no separable topology with these properties.

• Err...do you mean on an uncountable set, with your last comment?
– Ian
Aug 28, 2015 at 15:52
• @Ian Do you know any countable such example? Finite is clearly impossible. Aug 28, 2015 at 16:01
• Apparently our definition of separable is different, because to me a separable topological space is one with a countable dense subset, and if the space itself is countable then you can just take this dense subset to be the whole space.
– Ian
Aug 28, 2015 at 16:02
• @Ian Sorry I always forget this, by separable I mean Haussdorff... Aug 28, 2015 at 16:03