Consider the empty set $\emptyset$ as a topological space. Since the power set of it is just $\wp(\emptyset)=\{\emptyset\}$, this means that the only topology on $\emptyset$ is $\tau=\wp(\emptyset)$.

Anyway, we can make $\emptyset$ into a topological space and therefore talk about its homeomorphisms. But here, we seem to have an annoying pathology: is $\emptyset$ homeomorphic to itself? In order to this be true, we need to find a homeomorphism $h:\emptyset \to \emptyset$. It would be very unpleasant if such a homeomorphism did not exist.

I was tempted to think that there are no maps from $\emptyset$ into $\emptyset$, but consider the following definition of a map:

Given two sets $A$ and $B$, a map $f:A\to B$ is a subset of the Cartesian product $A\times B$ such that, for each $a\in A$, there exists only one pair $(a,b)\in f\subset A\times B$ (obviously, we denote such unique $b$ by $f(a)$, $A$ is called the domain of the map $f$ and $B$ is called the codomain of the map $f$).

Thinking this way, there is (a unique) map from $\emptyset$ into $\emptyset$! This is just $h=\emptyset\subset \emptyset\times \emptyset$. This is in fact a map, since I can't find any element in $\emptyset$ (domain) which contradicts the definition.

But is $h$ a homeomorphism? What does it mean for $h$ to have an inverse, since the concept of identity map is not clear for $\emptyset$? Nevertheless, $h$ seems to be continuous, since it can't contradict (by emptiness) anything in the continuity definition ("pre-images of open sets are open")…

So is $\emptyset$ homeomorphic to itself? Which is the mathematical consensus about this?

  1. "Homeomorphic by definition"?
  2. "We'd rather not speak about empty set homeomorphisms…"
  3. "…"?
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    $\begingroup$ It is vacously homeomorphic to itself $\endgroup$ Oct 1 '16 at 1:03
  • 4
    $\begingroup$ $\emptyset$ is a function, albeit not very exciting $\endgroup$
    – copper.hat
    Oct 1 '16 at 1:04
  • 5
    $\begingroup$ Ah, the science of voidology ... This reminds me of a story of Harvard's Benedict Gross in his Eilenberg lectures at Columbia.... He recounts how he took a wonderful course, albeit one in which he learnt appropriately nothing, on alg. topology. It was taught by his future supervisor, John Tate, who was trying to understand the topic from the functorial point of view: they spent lectures examining such weighty questions as "Is the empty set simply connected?" (there's more - and I'm probably misquoting - look the course up on youtube - on the langlands program) $\endgroup$
    – peter a g
    Oct 1 '16 at 1:08
  • $\begingroup$ @peterag Thank you for the recommendation! $\endgroup$
    – Ders
    Oct 1 '16 at 1:12
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    $\begingroup$ The set-theoretic def'n of a function $F:X\to Y$ is that $F \subset X\times Y$ such that (1): $\neg \;(\exists a\in X \; (\neg \exists b\in Y \; ((a,b)\in F)))$ , and (2): $\neg (\exists a \in X \; \exists b,c\in Y \;(\;b\ne c \land (a,b)\in F \land (a,c)\in F)).$ In particular if $X=\emptyset$ there is exactly one function $F:X\to Y$, namely $F=\emptyset.$ $\endgroup$ Oct 1 '16 at 4:38

Your map $h$ does exist, and is a homeomorphism. In fact, it's the identity map: for every element $x\in\emptyset$, $h(x)=x$. So since $h\circ h=h=\operatorname{id}$, $h$ is its own inverse. Since both $h$ and $h^{-1}=h$ are continuous, $h$ is a homeomorphism.

(Incidentally, checking that $h$ is continuous isn't entirely vacuous. You have to check that $h^{-1}(U)$ is open for any open subset $U\subseteq\emptyset$. It is not true that there are no choices of $U$: rather, there is exactly one choice of $U$, namely $U=\emptyset$. Of course, $h^{-1}(\emptyset)=\emptyset$ is indeed open.)

  • $\begingroup$ Very good observation! Anyway, it's very relieving to know that the empty set is homeomorphic to itself... Thanks! $\endgroup$
    – Ders
    Oct 1 '16 at 1:10
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    $\begingroup$ If we really wanted to keep things vacuous, we could use the equivalent definition of continuity as "continuous at every point in the domain". Since there are no points in the domain of $h$, it's vacuously continuous. $\endgroup$ Oct 1 '16 at 17:58

fleshing it out a little more, there is the empty function $\emptyset \to \emptyset$ ( that is just $\emptyset$ viewed as a function). This is bijective and bicontinuous.


In every category, every object is isomorphic to itself, via the identity morphism. This follows immediately from the definitions of category theory, the existence of a identity morphism for every object is an axiom. So if you want to call topological spaces a category, and the empty space an object of that category, then you must accept that the identity morphism of that space is a homeomorphism. The same goes for the empty set as object in the category of sets, whose identity map is a bona fide bijection.

By the way "identity" morphism (or arrow) is just a name, and does not have to be an identity map (or a map at all), though it does have to be neutral in the composition with other morphisms (and with itself). But in the categories of sets and topological spaces morphisms are maps, and the identity map is the only map that is neutral in composition with morphisms, so the identity morphism must be the identity map. And in case of an empty set/space, the identity map is the only map to itself around anyway.


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