Proposition 1.11. Let M be a topological manifold. (a) M is locally path-connected. (b) M is connected if and only if it is
path-connected. (c) The components of M are the same as its path
components. (d) M has countably many components, each of which is an
open subset of M anda connected topological manifold.Proof. Since each coordinate ball is path-connected, (a) follows from
the fact that M has a basis of coordinate balls. Parts (b) and (c) are
immediate consequences of (a) and Proposition A.43. To prove (d), note
that each component is open in M by Proposition A.43, so the
collection of components is an open cover of M . Because M is
second-countable, this cover must have a countable subcover. But since
the components are all disjoint, the cover must have been countable to
begin with, which is to say that M has only countably many components.
Because the components are open, they are connected topological
manifolds in the subspace topology
I have a doubt about the last sentence. For a component to be a manifold I need it to be Hausdorf, second countable and locally euclidean. The first two are properties of the subspace topology. For the third one, How do we use the fact that a component is open in order to prove that it is locally euclidean?
Best Answer
A connected subset of a manifold need not be a topological manifold. For instance, the top left corner point of the Warsaw circle does not have any connected neighbourhood (in the subspace topology).
It follows that the Warsaw circle isn't locally euclidean, although it is a (closed, even compact) subspace of $\Bbb R^2$.
Another counter example: the union of the vertical and the horizontal axes of $\Bbb R^2$ is a closed subset which is not a topological manifold: no neighbourhood of the origin is homeomorphic to any euclidean open subset.
Of course, being open is not a necessary condition. The horizontal line in $\Bbb R^2$ is locally euclidean (it is homeomorphic to $\Bbb R$) although it is not open in $\Bbb R^2$.
It turns out (see the comment section) that OP is confused about this fact: an open subset of a locally euclidean space is locally euclidean. Here is a short proof. Let $U\subset M$ be an open subset, with $M$ locally euclidean. Let $x\in U$. Since $M$ is locally euclidean, there exist $V\subset M$ an open neighbourhood of $x$ in $M$, and a map $\varphi\colon V\to \Bbb R^n$, which is an homeomorphism onto its image. Then $V\cap U$ is an open neighbourhood of $x$ in $U$ which is homeomorphic to $\varphi(U\cap V)$, which is itself an open subset of $\Bbb R^n$ (since $\varphi$ is an homeomorphism onto its image). It follows that $U$ is locally euclidean.