Let $(\mathbb{R},T)$ be the reals with the standard topology. Let $A \subset \mathbb{R}$. Then,

If $A$ is totally disconnected, then $\operatorname{Int}(A) = \emptyset$ where $\operatorname{Int}(A)$ refers to the interior of $A$.

This can be proven by considering the contrapositive. If $\operatorname{Int}(A)$ is not empty then there is a point $x \in A$ that has an open neighborhood, $V$, such that $V \subset A$. As all open sets in $\mathbb{R}$ with the standard topology consist of intervals, that means $A$ contains an interval. As intervals are connected and are not singletons, this means $A$ can not be totally disconnected which completes the proof for the contrapositive.

The issue I'm having is that I don't want to work only with $\mathbb{R}$ with the standard topology. The property that,

If $A$ is totally disconnected, then $\operatorname{Int}(A) = \emptyset$

is not true for general topological spaces. The discrete topological space on $R$ is a counterexample (which also means not all metric spaces satisfy the property). An example of a topological space that isn't discrete, but this property still fails is a finite topological space with 3 elements.

Specifically, let $X = \{a,b,c\}$ and $T_X = \{\emptyset,\{a\}, \{b,c\}, \{a,b,c\}\}$.Then, $A = \{a,b\}$ is totally disconnected, but has an interior of $\{a\}$.

Currently, my intuition is that the property fails only when your topological space has connected components that are singletons from the rest of the space. I'm not sure how to prove that, nor am I entirely confident that's the only way it can fail.

So, the question is how can I characterize the topological spaces where my desired property (totally disconnected implies empty interior) holds?

Side note: The origin of this question comes from me trying to characterize what the pre-image of an interval can look like. Initially, I was trying to do the problem just with the standard topology on $\mathbb{R}$, but after I got partial success when I tried to generalize my argument, I ran into a problem with this property not being true in general. My own relevant math background is the first few chapters of munkres and the first 7 chapters of baby rudin.

Edit: Based on the comments with Matt, the question could also be stated as how can I characterize the topological spaces where an open, totally disconnected set exist? Is it possible for a connected space $(X,T_X)$ to have this property (assuming X has more than one element)?

  • $\begingroup$ A non discrete metric space that utterly fails to have the property is the rationals, or indeed any non discrete totally disconnected metric space, like the Cantor set. $\endgroup$ Feb 26, 2016 at 4:22
  • $\begingroup$ @Matt That would be a set whose components are singletons. The components of any totally disconnected space is just singletons. Even if you relax it to be a space with only a few singleton components you still fail since you can take a set involving those components. My issue is I'm not sure if that's the only route possible for failure. $\endgroup$
    – Mehdi2277
    Feb 26, 2016 at 4:31
  • $\begingroup$ The set only needs to have one totally disconnected open set to fail this. You could take any space with two separated points and declare a particular disconnected two point set to be open, make other sets open as necessary for the axioms, and the space doesn't have the property. $\endgroup$ Feb 26, 2016 at 5:04
  • $\begingroup$ @Matt That would work as necessary and sufficient condition. If the space fails this property than it contains a totally disconnected whose interior is not empty. The interior of a set is a subset of that set and the subset of any totally disconnected set is also totally disconnected. As the interior is an open set that means it contains a totally disconnected open set. If a space contains a totally disconnected set, that set serves to contradict the property. Is there any equivalent way to describe this property (existence of an open totally disconnected set) using more common properties? $\endgroup$
    – Mehdi2277
    Feb 26, 2016 at 5:21
  • $\begingroup$ It's not a global property of the space, so I don't see a better way to characterize it than saying there is a totally disconnected open set. In any case it's past my bed time so I'll leave you with that. Good luck. $\endgroup$ Feb 26, 2016 at 5:23

2 Answers 2


This is too long to be a comment and is currently only a partial answer.

Following the idea that the existence of an open, totally disconnected set should relate to the connectedness of the space, I read about ways a space can be connected. A locally connected space is one where every point in a open set has a connected, open neighborhood and felt most relevant. I wasn't able to show that being locally connected was a sufficient condition for the lack of an open, totally disconnected set. So, I ended up adding connected back as a requirement (and T1 ended up being helpful during the proof) which led to,

Let $(X,\tau)$ be a locally connected, connected, T1 space that is not the topology on a one point set. Then, $(X, \tau)$ is not weakly disconnected.


The proof strategy will be to use proof by contrapositive. Let $(X, \tau)$ be a weakly disconnected space. Let $U$ be an open, non-empty totally disconnected set. Let $(U, \tau_{U})$ be the subspace topology of $U$. As $U$ is totally disconnected and non-empty, its connected components are singletons. Here, we'll split the proof into two cases. Either all of its connected components are open or they there exists a connected component that is not open.

If there exists a connected component that is not open then its connected components are not all open which means it is not locally connected. As locally connected is hereditary for open subspaces, $(X,\tau)$, can not be locally connected.

If there exists a connected component that is open, call that component $A$. As $U$ is totally disconnected, $A$ must be a singleton. Since $U$ is open, any open set in $U$ is also open in $X$. This means $A$ is an open singleton in $X$. As $X$ is a T1 space, singletons are closed making $A$ a clopen set. Since $X$ is not a one point set, $A$ is non-trivial clopen set. This means $(X,\tau)$ is not connected.

In either case, $(X, \tau)$ can not have the three properties of being locally connected, connected, and T1. This completes the proof.

The issue is I'm not sure if those three properties are a necessary condition, so I still lack a nice way of describing all spaces that lack an open, totally disconnected set. While my current proof strategy uses the fact the space is T1, I'm skeptical that there isn't an alternate way to drop that condition.

  • $\begingroup$ It can be slightly refined as follows: On locally connected spaces, the condition is equivalent to not having isolated points. So on locally connected $T_1$ spaces, the condition is equivalent to not having a clopen point, which is strictly weaker than being connected (on a nondegenerated space). Also notice that the property is stable under taking sums, so connectedness is really not needed. $\endgroup$
    – user87690
    Feb 29, 2016 at 20:43
  • $\begingroup$ @user87690 Can you explain what stable under taking sums means? I'm not familiar with that phrase so I don't understand your last sentence. $\endgroup$
    – Mehdi2277
    Mar 1, 2016 at 1:03
  • $\begingroup$ Sum of topological spaces is one of four fundamental constructions (with subspaces, products, and quotients). By stability under sums I mean than whenever you have a family of spaces satisfying your property, then their sum also satisfies the property. $\endgroup$
    – user87690
    Mar 1, 2016 at 8:27

"If $A\subseteq X$ is totally disconnected, then $\text{int}_X (A)=\varnothing$."

The conditions you mentioned in your answer are not necessary in order for this property to hold, because it also holds if $X$ is connected and locally compact (and $|X|>1$):

Proof. Let $\alpha X$ be a compactification of $X$, such as the one-point compactification. Note that $X$ is open in $\alpha X$. Let $A\subseteq X$ with $U:=\text{int}_X (A)\neq\varnothing$. Let $a\in U$. Note that $U$ is open in $\alpha X$, and that $\alpha X$ is compact connected Hausdorff. So we may apply the Boundary Bumping Theorem; the component of $a$ in $U$ meets the boundary (in $\alpha X$) of $U$, and is thus nontrivial. Therefore $A$ is not totally disconnected. $\square$

Of course there are connected locally compact spaces that are not locally connected. In fact you can get one that is nowhere locally connected - take any indecomposable continuum. Furthermore, by taking a composant of an indecomposable continuum which has more than one composant, you get a connected space with your property which is nowhere locally compact AND nowhere locally connected. So I think there is not a simple characterization.


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