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Struggling with arc-connectedness

  1. Oct 26, 2015 #1
    1. The problem statement, all variables and given/known data

    Let ##(E,N)## a finite dimensional normed vector space. Let ##A## be an arc-connected subset of ##E##, and ##P## be a non-empty subset of ##A## that is both open and close in ##A##. Show that ##P=A##

    2. Relevant equations


    3. The attempt at a solution

    By contradiction, I assume ##P\neq A##, which is ##C_A(P) \neq \emptyset##.

    I'd like to find a continuous function from ##A## to a subset of ##S## of ## \mathbb{R}## that is not a segment.
    This would be a contradiction because the image of an arc-connected set by a continuous function is arc-connected, and the arc-connected sets of ##\mathbb{R}## are segments.

    I tried with the caracteristic function of ##P##, but I believe it doesn't work because for any open set ##O## of ##\mathbb{R}##, ##f^{-1}(O)\in\{\emptyset, P,C_A(P), A \}##, and for now, I can't convince myself that ##C_A(P)## is an open set of ##A##.

    Do you have any idea please ?
     
  2. jcsd
  3. Oct 26, 2015 #2

    Samy_A

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    I think you are almost there ...
    Think of all the known properties of ##P##.
     
    Last edited: Oct 26, 2015
  4. Oct 26, 2015 #3
    I think that thanks to you, I have that last point that bothered me :
    Since ##P## is closed in ##A##, there exists a closed set ##C## of ## E## such that ## P = C \cap A##.
    So ##C_A(P) = C_E(P) \cap A = (C_E(C) \cup C_E(A))\cap A = C_E(C) \cap A ##.
    But ##C_E(C)## is an open set of ##E##. Therefore ##C_A(P) ## is open in ##A##.

    So now I am sure that the caracteristic function is continuous from ##A\rightarrow \{0,1\}## which is absurd because ##\{0,1\}## is not arc-connected.

    Do you agree ?
     
  5. Oct 26, 2015 #4

    Samy_A

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    Yes, looks OK to me.
     
  6. Oct 26, 2015 #5
    Thanks!
     
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