# Is the empty set homeomorphic to itself?

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. "…"?
• It is vacously homeomorphic to itself – Zelos Malum Oct 1 '16 at 1:03
• $\emptyset$ is a function, albeit not very exciting – copper.hat Oct 1 '16 at 1:04
• 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) – peter a g Oct 1 '16 at 1:08
• @peterag Thank you for the recommendation! – Ders Oct 1 '16 at 1:12
• 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 $F=\emptyset.$ – DanielWainfleet 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.)
• 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. – Andreas Blass 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.