# Connected components of a manifold

1. Sep 18, 2008

### quasar987

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.

2. Sep 19, 2008

### Doodle Bob

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.

3. Sep 19, 2008

### HallsofIvy

Staff Emeritus
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.

4. Sep 19, 2008

### quasar987

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.

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.