A space $X$ is indiscrete provided its topology is $\{\emptyset,X\}$. With such a restrictive topology, such spaces must be examples/counterexamples for many other topological properties. Then my question is:

Question: What topological properties are trivially/vacuously satisfied by any indiscrete space?


4 Answers 4


Separation properties

Any indiscrete space is perfectly normal (disjoint closed sets can be separated by a continuous real-valued function) vacuously since there don't exist disjoint closed sets. But on the other hand, the only T0 indiscrete spaces are the empty set and the singleton.


Only the empty and singleton indiscrete spaces are metrizable, but every indiscrete space is compatible with the pseudometric $d(x,y)=0$ for all $x,y$.

Covering properties

Any indiscrete space is compact since its only open cover is finite to begin with ($\{X\}$).

Topological size

Every basis for an indiscrete space is finite ($\Rightarrow$ countable), so it is second-countable and therefore separable.


$X$ is the only nonempty clopen set, so indiscrete spaces are connected. They are also:

  • Strongly connected, that is, the only continuous functions $f:X\to\mathbb R$ are constant.
  • Hyperconnected, that is, all nonempty open sets intersect
  • Ultraconnected, that is, all nonempty closed sets intersect
  • Path connected since all maps [from $\mathbb R$] to $X$ are continuous. This strengthens to arc connected if the space has the cardinality of the reals or greater (as arc connected requires injectivity).
  • $\begingroup$ RE: separation properties, I'd argue that what's really going on is that a couple separation properties like perfect normality aren't quite defined correctly: they need the hypothesis of $T_0$ness thrown in as well. And that the *correct* separation properties are never had by any indiscrete topology (with $>1$ points anyways). $\endgroup$ Sep 28, 2020 at 20:17
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    $\begingroup$ @NoahSchweber: I disagree. I prefer to distinguish (perfectly) normal from (perfectly) $T_4$, completely regular from Tikhonov ($T_{3½}$), regular from $T_3$, and so on, where the $T_k$ properties are a genuine hierarchy. $\endgroup$ Sep 28, 2020 at 20:30
  • $\begingroup$ @BrianM.Scott Fair - out of curiosity, why? Are there situations where it's useful to have the non-$T_0$ versions? (I don't think folding in $T_1$ness or higher is a good idea, but I don't see what's lost with requiring $T_0$.) $\endgroup$ Sep 28, 2020 at 20:39
  • $\begingroup$ @NoahSchweber: Given my interests, I generally assumed that all spaces were $T_1$ in anything that I was doing, so for me the problem didn’t arise. I don’t know that routinely including $T_0$ separation would really lose anything useful, but on general principles I like to keep the generality (and the simplicity of saying that disjoint closed sets have disjoint open nbhds). $\endgroup$ Sep 28, 2020 at 20:47
  • $\begingroup$ Non-$T_1$ spaces are considered in asymmetric topology. I think assuming all topologies are $T_0$ wouldn't lose much: every space has a $T_0$ quotient obtained from identifying topologically-indistinguishable points. $\endgroup$ Sep 29, 2020 at 0:59

Indiscrete spaces are trivially second countable and regular. So they witness the necessity of "Hausdorff" in Urysohn's metrization theorem: "Every second countable Hausdorff regular space is metrizable."

  • $\begingroup$ More generally, they're a great example of how separation axioms above $T_0$ are often necessary hypotheses. $\endgroup$ Sep 28, 2020 at 20:40
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    $\begingroup$ Indeed, "Hausdorff" can be replaced with T0 in this theorem. And if you drop T0, I believe metrizable can be swapped with pseudometrizable. (Certainly it's true for indiscrete spaces: let d(x,y)=0 always.) $\endgroup$ Sep 29, 2020 at 0:24

Connectedness (which can be defined as "$O$ clopen implies $O=\emptyset$ or $O=X$", but in the indiscrete case, clopen can be even replaced by just open...).

Path-connectedness (as every map with codomain an indiscrete space is continuous automatically, so any function can be a path..).

Regular and normal as there are no closed sets to separate from disjoint closed sets or points outside..

Being first countable (and second countable) obviously, as well as Lindelöf, compact, countably compact etc. as there is but one open cover which is already finite.


From wikipedia A topological space is termed an indiscrete space if it satisfies the following equivalent conditions:

1/It has an empty subbasis.

2/It has a basis comprising only the whole space.

3/The only open subsets are the whole space and the empty subset.

4/The only closed subsets are the whole space and the empty subset.

5/The space is either an empty space or its Kolmogorov quotient is a one-point space.

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    $\begingroup$ I think the OP is looking for topological properties which are not equivalent to indiscreteness, but are held by indiscrete topologies. $\endgroup$ Sep 28, 2020 at 20:40

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