# Thread: Cartesian product question

1. ## 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

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I have the answer but I'm completely confused on how it was derived

2. ## 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 $\displaystyle (x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+y_2)$...

3. ## 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

4. ## 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) ?

5. ## 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

6. ## Re: Cartesian product question

Originally Posted by Krahl

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 $\displaystyle S \times T$.

Originally Posted by entrepreneurforum.co.uk
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.