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? 


*

*"Homeomorphic by definition"?

*"We'd rather not speak about empty set homeomorphisms…"

*"…"?


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.
