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Thread: 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 $\displaystyle \Re ^n$ and $\displaystyle \Re ^m$ respectively
    $\displaystyle \Longleftrightarrow$ $\displaystyle \forall x_1,y_1 \in S$, $\displaystyle \forall t \in [0,1]$
    $\displaystyle (1-t)x_1 + ty_1 \in \Re ^n$
    and
    $\displaystyle \forall x_2,y_2 \in T$, $\displaystyle \forall t \in [0,1]$
    $\displaystyle (1-t)x_2 + ty_2 \in \Re ^m$

    with $\displaystyle a(x,y)$ defined as $\displaystyle (ax,ay)$ for $\displaystyle a \in \Re$ and $\displaystyle (x,y)\in S \times T$ and $\displaystyle (x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+y_2)$...
    Last edited by Krahl; Oct 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 $\displaystyle forall$ S $\displaystyle forall R^n$

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

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

    thanks
    Last edited by entrepreneurforum.co.uk; Oct 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 $\displaystyle \forall$ S $\displaystyle \forall R^n$

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



    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 $\displaystyle \Re^n$ and $\displaystyle \Re^m$. This means that the following by definition is true;
    λs1 + (1-λ)s2 $\displaystyle \in \Re^n$ $\displaystyle \forall s1,s2 \in$ S

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

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

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

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

    We begin with the equation

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

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

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

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

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

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

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

    what can we say about the equations (λs1 + (1-λ)s2) and (λt1 + (1-λ)t2) ?
    Last edited by Krahl; Oct 8th 2012 at 03:40 AM.
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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; Oct 8th 2012 at 02:54 PM.
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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 $\displaystyle \Re^n$ and (λt1 + (1-λ)t2) is in $\displaystyle \Re^m$ so that the element ((λs1 + (1-λ)s2),(λt1 + (1-λ)t2)) is in $\displaystyle S \times T$. This means by definition that $\displaystyle S \times T$ is a convex set in $\displaystyle \Re^{n\times m}$

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

    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; Oct 8th 2012 at 03:21 PM.
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