# Show that dimension of topological manifold is well defined. Dimension of empty manifold?

Definition: A topological manifold of dimension $$m$$ is a topological space which is Hausdorff, second countable and is locally homeomorphic to $$\mathbb{R}^m$$.

Is dimension of a topological manifold well defined? What is the dimension of the empty manifold?

Attempt:

Suppose $$M$$ is a topological manifold of both dimension $$n$$ and $$m$$. Fix $$p \in M$$. By assumption, there are open neighborhoods $$V,W$$ of $$p$$ such that $$\phi: V \to \mathbb{R}^n$$ and $$\psi: V \to \mathbb{R}^m$$ are homeomorphisms onto their images. In particular, $$V \cap W$$ is an open set containing $$p$$ and $$V \cap W$$ is homeomorphic to both $$\phi(V \cap W) \subseteq \mathbb{R}^n$$ and $$\psi(V \cap W)\subseteq \mathbb{R}^m$$. Thus $$\phi(V \cap W)$$ is homeomorphic to $$\psi(V \cap W)$$ and two open subsets of $$\mathbb{R}^m$$ resp. $$\mathbb{R}^n$$ can only be homeomorphic if $$m=n$$ (https://en.wikipedia.org/wiki/Invariance_of_domain). This forces $$m=n$$ and dimension is well defined.

Is this correct? My proof breaks down if $$M = \emptyset$$. It seems that $$\emptyset$$ is a topological manifold of dimension $$m$$ for all $$m \geq 1$$?

You're correct. The empty set qualifies as a topological manifold of every nonnegative dimension (including $$0$$). This is a little awkward, but it's useful enough in certain circumstances (such as manifolds with empty boundary in Stokes's theorem) that it's not worth ruling out the empty manifold. You just have to remember to stipulate that a manifold is nonempty whenever you need that hypothesis.