Is the Empty Set a Manifold Is the empty set $\emptyset$ a $C^1$-Manifold?
If this is personal preference, as is $0 \in \mathbb{N}$ for example, then what are the reasons for/against this choice?
 A: Some people demand in their definitions that manifolds be connected. Some people consider the empty set to not be connected (for the same reason $1$ isn't prime). The intersection of these sets of people must therefore consider the empty set not to be a manifold.
If you do allow disconnected manifolds then note that there are two separate ways of dealing with the notion of dimension. Some people say that a manifold must locally look like $\mathbb R^k$ for some $k$, where $k$ can vary over the manifold (it has to be fixed on each connected component, but different components can have different dimensions). Other people say that you have to specify a particular $k$ first, and then every point in the manifold has to locally look like $\mathbb R^k$ for that particular $k$. In the former case $\emptyset$ is a manifold, in the later case there are many manifolds with underlying set $\emptyset$: one for each dimension.
A: If you decided that $\emptyset$ is not a manifold, you'd have to write things like

the boundary of a manifold with boundary is a manifold unless it is empty

or

The preimage of a regular value under a smooth map is a smooth submanifold, unless it is empty

or you'd have to come up with a funny way of stating the usual proof that a volume form $\nu$ over a closed manifold $M$ is not exact, which usually uses Stoke's theorem asserting that $\int_Md\eta=\int_{\partial M}\eta$, where in that case $\partial M$ is empty.
In all, you'd litter everything with lots of «unless it is empty»...
