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Math Help - Cartesian product question

  1. #1
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    Cartesian product question

    Okay, I have this problem from a maths textbook and its along the same lines as

    If S and T are sets

    the Cartesian product S x T of S and T is defined


    [(s,t) : s ∈ S, t ∈ T}


    Prove if S and T are convex sets in ℝ^n and ℝ^m

    then ℝ^(n+m) is convex also

    ---------

    I have the answer but I'm completely confused on how it was derived
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  2. #2
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    Re: Cartesian product question

    Suppose S and T are convex sets in \Re ^n and \Re ^m respectively
    \Longleftrightarrow \forall x_1,y_1 \in S, \forall t \in [0,1]
    (1-t)x_1 + ty_1 \in \Re ^n
    and
    \forall x_2,y_2 \in T, \forall t \in [0,1]
    (1-t)x_2 + ty_2 \in \Re ^m

    with a(x,y) defined as (ax,ay) for a \in \Re and (x,y)\in S \times T and (x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+y_2)...
    Last edited by Krahl; October 5th 2012 at 04:20 AM.
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  3. #3
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    Re: Cartesian product question

    Okay, I don't fully understand your answer, heres what I have;

    λs1 + (1-λ)s2  forall S  forall R^n

    λt1 + (1-λ)t2  forall T  forall R^m

    and now I can't understand why R^n , R^m makes SxT convex

    thanks
    Last edited by entrepreneurforum.co.uk; October 6th 2012 at 09:08 AM.
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  4. #4
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    Re: Cartesian product question

    Ok so this is what you've written;
    λs1 + (1-λ)s2  \forall S  \forall R^n

    λt1 + (1-λ)t2  \forall T  \forall R^m



    s1 and s2 are elements of set S and t1 and t2 are elements of set T. I wouldn't write the "forall Rn" part since the statements are only true for the convex sets S and T in Rn,Rm. So basically S and T are convex sets in \Re^n and \Re^m. This means that the following by definition is true;
    λs1 + (1-λ)s2 \in \Re^n  \forall s1,s2 \in S

    λt1 + (1-λ)t2 \in \Re^m  \forall t1,t2 \in T

    (If S is a convex set in \Re^n then λs1 + (1-λ)s2 is in \Re^n).

    We've just applied the definition of a convex set so far. Now we want to show the set S \times T (which is of the form x1=(s1,t1)) is also a convex set. So we want to show that;

    λx1 + (1-λ)x2 \in \Re^{n \times m}  \forall x1,x2 \in S  \times T

    We begin with the equation

    λx1 + (1-λ)x2  \forall x1,x2 \in S  \times T

    which becomes (x1 = (s1,t1), x2 = (s2,t2) are elements of S \times T)

    λ(s1,t1) + (1-λ)(s2,t2)  \forall s1,s2 \in S and  \forall t1,t2 \in T

    which means (λ(s1,t1) = (λs1,λt1))

    (λs1,λt1) + ((1-λ)s2,(1-λ)t2)  \forall s1,s2 \in S and  \forall t1,t2 \in T

    which then becomes ((s1,t1)+(s2,t2) = (s1+s2,t1+t2))

    ((λs1 + (1-λ)s2),(λt1 + (1-λ)t2))  \forall s1,s2 \in S and  \forall t1,t2 \in T

    what can we say about the equations (λs1 + (1-λ)s2) and (λt1 + (1-λ)t2) ?
    Last edited by Krahl; October 8th 2012 at 03:40 AM.
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  5. #5
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    Re: Cartesian product question

    what can we say about the equations (λs1 + (1-λ)s2) and (λt1 + (1-λ)t2) ?
    Well am I correct in saying that ((s1,t1)+(s2,t2) = (s1+s2,t1+t2)) therefore S and T are convex IF they both lie in S x T (on a graph)
    (I'm slightly confused)

    I really have no idea


    Edit: If S and T are two convex sets in R^n and R^m then their intersection S∩T is also convex, - I understand that but I don't know how its proved that its convex in R^n+m
    Last edited by entrepreneurforum.co.uk; October 8th 2012 at 02:54 PM.
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  6. #6
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    Re: Cartesian product question

    Quote Originally Posted by Krahl View Post

    what can we say about the equations (λs1 + (1-λ)s2) and (λt1 + (1-λ)t2) ?
    By definition (written above) (λs1 + (1-λ)s2) is in \Re^n and (λt1 + (1-λ)t2) is in \Re^m so that the element ((λs1 + (1-λ)s2),(λt1 + (1-λ)t2)) is in S \times T. This means by definition that S \times T is a convex set in \Re^{n\times m}

    What we've done is taken any two elements x1 and x2 of S \times T and showed that λx1 + (1-λ)x2 also lies in S \times T.

    Quote Originally Posted by entrepreneurforum.co.uk View Post
    Well am I correct in saying that ((s1,t1)+(s2,t2) = (s1+s2,t1+t2)) therefore S and T are convex IF they both lie in S x T (on a graph)
    No the operation (s1,t1)+(s2,t2) = (s1+s2,t1+t2) is just how the elements are added. What we need to show is that for any two elements of S x T the equation λx1 + (1-λ)x2 lies in it as well.
    Last edited by Krahl; October 8th 2012 at 03:21 PM.
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