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Connectedness and convex

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    It would make one of my proofs easy if it is true that

    " If E is an open connected set then it is convex''.

    I have been spending some time trying to prove this. Is this statement even true?



    2. Relevant equations
    Convex implies that if x is in E and y is in E then
    εx+(1-ε)y is in E where 0<ε<1.

    Connected: E cannot be the union the union of two nonempty separated sets.


    3. The attempt at a solution


    My attempt at the solution looks like this.

    Suppose E is open in ℝn. Since E is open implies that there exists a ball around every x in E s.t. b(x) is in E.

    Let's choose a y in E s.t. b(y) intersected with b(x) is nonempty. Let's choose a point y* from the intersection. Let's also choose a x* in b(x). Since y* is also in b(x) and all balls are convex implies that εx*+(1-ε)y* is in b(x) union with b(y), where 0<ε<1

    Now lets choose another point z s.t. b(z) intersected with b(y) is nonempty. Then similarly,
    εy*+(1-ε)z* is in the union of b(x),b(y), and b(z).

    Now assuming all this is correct... I am stuck here but I feel like if I can show
    εx*+(1-ε)z* is in the union of b(x),b(y), and b(z) then the union of the 3 balls is convex which would imply that E is convex.

    I also didn't mention connectedness but I think that comes into play as follows: I can find balls that have no separation and I can union them allowing me to create a path of balls connecting all points.



    Thank you.
     
  2. jcsd
  3. Jan 31, 2012 #2

    chiro

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    Hey EV33

    This is just a suggestion based on the intuition of convex sets, but if a set were to be concave, then you could easily show an example of a separated set.

    To get an idea of this with regards to the nature of the definition of a convex set, in a non-convex set (concave set) there will be a line that you can draw where one at least one part of that line does not lie in the boundary of the set.

    Using this idea you can show that for a concave set, that you can form sets that are separated and thus don't have the property you have specified above.
     
  4. Jan 31, 2012 #3
    Maybe I am interpreting what you are saying incorrectly but I think that you are talking about the converse of the "theorem" I stated. I want to show that if a set is open and connected then it is convex. I was not trying to say that if a set is convex then it must be connected.
     
  5. Jan 31, 2012 #4

    HallsofIvy

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    Let A= {(x,y)| y= 0, [itex]0\le x\le 1[/itex]}

    Let B= {(x, y)| x= 0, [itex]0\le y\le 1[/itex]}

    Let C= {(x, y)| x= 1, [itex]0\le y\le 1[/itex]}

    Let D be the union of A, B, and C.
     
  6. Jan 31, 2012 #5

    Ray Vickson

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    Suppose [itex] A = \{(x,y) : x^2 + y^2 < 1\} \mbox{ and } B = \{(x,y) : (x-1)^2 + y^2 < 1 \}. [/itex] These are open discs of radius 1 centered at (0,0) and (1,0). If C is their union, is C open? Is C connected? Is C convex?

    RGV
     
  7. Jan 31, 2012 #6
    If I have the two balls A and B and I union them then their union will be open and it will be connected. If I tried to draw a straight line connecting their tops I would fail.

    So this implies that the theorem I was trying to prove is false right?


    Thank you all for the help.
     
  8. Jan 31, 2012 #7

    HallsofIvy

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    You mean, "will NOT be connected".

    Yes, your statement "an open connected set is convex" is false. The U shaped example I gave also shows that.
     
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