# Definition topological manifold

In the book "An introduction to manifolds" by Tu, a topological manifold is defined to be a topological space $$M$$ that is Hausdorff, second countable and locally Euclidean.

Does this allow things like the disjoint union of a plane and a line? Then we have a component which is locally Euclidean of dimension $$1$$ and one of dimension $$2$$?

Tu allows manifolds having connected components of different dimensions. He explicitly says it in this post. Usually people talk about a space being "locally $$\Bbb R^n$$" or "locally Euclidean of dimension $$n$$" as opposed to just "locally Euclidean", as he does. But it is not hard to show that for each $$n \geq 0$$, the set $$\{ x \in M \mid x \mbox{ has an open neighborhood homeomorphic to }\Bbb R^n \}$$is both open and closed in $$M$$. So this means that the dimension is well defined on each connected component of $$M$$.
• Thanks for the answer. Don't you mean "homeomorphic to an open subset of $\mathbb{R}^n$"? – user745578 Jul 4 '20 at 21:48
• Also, could you briefly sketch why the set you wrote down is closed? This is not immediately clear to me. I tried taking a sequence $x_n \to x$ where $x_n$ is in the set and show that $x$ is in the set as well. – user745578 Jul 4 '20 at 21:55
• @HagenvonEitzen But why? If you take a point $x$ not in that set, then you know that $x$ has no open neighborhood homeomorphic to some open subset of $\Bbb{R}^n$. Then why is this the case for a nbh of $x$ as well? – user745578 Jul 4 '20 at 22:26
• If $x_k\to x$ and each $x_k$ has a neighborhood homemorphic to $\Bbb R^n$, use that $x$ has a neighborhood homeomorphic to some $\Bbb R^m$. This contains one of the $x_k$'s by convergence. The transition between the charts is a homeomorphism between open subsets of $\Bbb R^n$ and $\Bbb R^m$. So $n=m$. – Ivo Terek Jul 4 '20 at 22:29