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Math Help - Convex set proof

  1. #1
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    Convex set proof

    Given any set S, say [0,1], how do I prove that this set is convex for all values of x, y, and a, given the following definition of convexity:

    ax+(1-a)y is an element of the set S, for every 'a' such that 0<a<1, and for every x and y such that both x and y are elements of S.

    Thanks!
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  2. #2
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    You take any two values from set S and show that the formula works for them...

    If we take x,y \in S
    let just take that x > y
    that means that

    ax + ( 1 - a )y \in S
       ( a \in <0,1> )

    what you need to prove is that:

    a( x - y ) + y \in S

    but:

    a( x - y ) \in <0, x-y>

    than:

    a(x-y)+y \in <y,x>

    ( this is for x > y , you now prove for y > x )
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  3. #3
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    Sorry, I'm new to proofs so bear with me. 2 questions:

    1) say the set was the finite set {1,2} where does the proof show non-convexity, as opposed to the (0,1) interval?

    2) why do we have to say that x>y, in other words, take out the x>y part, where would the remaining proof be invalid?

    Thanks for your help.
    Last edited by econlover55; October 13th 2009 at 08:36 AM. Reason: more questions
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  4. #4
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    Quote Originally Posted by econlover55 View Post
    Sorry, I'm new to proofs so bear with me. 2 questions:

    1) say the set was the finite set {1,2} where does the proof show non-convexity, as opposed to the (0,1) interval?
    if we take set { 1, 2 }

    you need to find only one a \in [0,1]

    and only one pair x, y x,y \in S

    where ax + ( 1 - a )y \notin S

    for set { 1, 2 } i can take a = 1/2;

    and now if i take x = 1, y = 2 ... (1/2)*1 + (1/2)*2 = 3/2 and this is not in {1, 2} ..... now this is enough, but you can find more examples.


    Quote Originally Posted by econlover55 View Post
    2) why do we have to say that x>y, in other words, take out the x>y part, where would the remaining proof be invalid?

    Thanks for your help.
    notice that it can be:

    1) x = y
    2) x > y
    3) x < y

    1) if x = y \in S

    a*x + ( 1 - a )*y = a*x + x - a*x = x \in S

    so this is ok from definition of convex set, but this will be for any a, not only for a in [0,1], so this case is not very interesting

    2) and 3)

    i can say where the proof would be invalid and hope that you will understand ( its easy )

    look at this part of proof:

    a(x-y)+y \in <y,x>

    or look better only on this part...

    <y,x>

    and we supposed that x > y

    so we know for sure that set <y,x> is really a set
    ( that expression make sense, <x,y> dont make sense in this example )
    and most important is that with x > y we have that...

    <y,x> \subset S
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