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I am not sure - a manifold is locally connected and has countable basis?

  1. Jul 19, 2012 #1
    I am not sure -- a manifold is locally connected and has countable basis?

    There is an Exercise in a book as following :

    Given a Manifold M , if N is a sub-manifold , an V is open set then V [itex]\cap[/itex] N is a countable collection of connected open sets .

    I am asking why he put this exercise for only the case of sub-manifold , Is n't this an immediate consequence of the fact that a manifold is locally connected and has countable basis ?????

    I am not sure from what I say ?? I think there can't be exercise as easy as I think , I think I am wrong .


    Thanks
     
  2. jcsd
  3. Jul 19, 2012 #2
    Re: I am not sure -- a manifold is locally connected and has countable basis?

    I am sorry , I was having a confusion in the definitions between what is the n-submanifold property and what is the sub-manifold .

    Thanks
     
  4. Jul 20, 2012 #3

    Bacle2

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    Re: I am not sure -- a manifold is locally connected and has countable basis?

    The definition I know has it that S is a

    submanifold o fM if S is embedded in M in the same way as R^k is embedded /sits-in

    R^{k+j} in the "standard way", i.e., where (x_1,..,x_k)-->(x_1,x_2,...,x_k, 0,0,..,0)

    So S is a submanifold of M if there are subspace charts mapping points of S into

    points of the form (x_1,...,x_k, 0,0,..,0).
     
    Last edited: Jul 20, 2012
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