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?
- "Homeomorphic by definition"?
- "We'd rather not speak about empty set homeomorphisms…"
- "…"?
Best Answer
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.)