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Connected components of a manifold

  1. Sep 18, 2008 #1

    quasar987

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    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. jcsd
  3. Sep 19, 2008 #2
    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.
     
  4. Sep 19, 2008 #3

    HallsofIvy

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    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.
     
  5. Sep 19, 2008 #4

    quasar987

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    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 [itex]W_x \cap M\C \neq \emptyset[/itex]. 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 [itex]C\cup W_x[/itex] is path connected, hence connected, which is a contradiction with the fact that C is not properly contained in any connected subset of M.
     
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