I got this book here that mentions en passant that the connected components of a (topological) manifold are open in the manifold.
That's not true in a general topological space, so why does Hausdorff + locally euclidean implies it?
I don't see it.
Doodle Bob
Sep19-08, 05:53 AM
I got this book here that mentions en passant that the connected components of a (topological) manifold are open in the manifold.
That's not true in a general topological space, so why does Hausdorff + locally euclidean implies it?
I don't see it.
Let U be a connected component of the manifold. For each x in U, let V_x be an open nbd of x diffeomorphic to a Euclidean ball such that V_x is contained in U. Then U is the union of all such V_x's and hence is open.
HallsofIvy
Sep19-08, 10:44 AM
I got this book here that mentions en passant that the connected components of a (topological) manifold are open in the manifold.
That's not true in a general topological space, so why does Hausdorff + locally euclidean implies it?
I don't see it.
On the contrary, connectedness components are open in a general topological space. In fact, they are both open and closed! And the same is true of connectedness components of manifolds.
quasar987
Sep19-08, 11:53 AM
On the contrary, connectedness components are open in a general topological space. In fact, they are both open and closed! And the same is true of connectedness components of manifolds.
As a counterexample, consider the connected components of the rational numbers. These are the singetons {r} with r rational. These are not open in Q.
Let U be a connected component of the manifold. For each x in U, let V_x be an open nbd of x diffeomorphic to a Euclidean ball such that V_x is contained in U. Then U is the union of all such V_x's and hence is open.
I see. If such a coordinate nbhd V_x did not exist, then there would be a coordinate nbdh W_x homeomorphic to a euclidean ball, with W_x \cap M\C \neq \emptyset. Since the open ball is path connected, it would mean that W_x is too. And C is locally path connected and connected, so it is path connected. So C\cup W_x is path connected, hence connected, which is a contradiction with the fact that C is not properly contained in any connected subset of M.