Suppose S and T are convex sets in and respectively
,
and
,
with defined as for and and ...
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
Okay, I don't fully understand your answer, heres what I have;
λs1 + (1-λ)s2 S
λt1 + (1-λ)t2 T
and now I can't understand why R^n , R^m makes SxT convex
thanks
Ok so this is what you've written;
λs1 + (1-λ)s2 S
λt1 + (1-λ)t2 T
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 and . This means that the following by definition is true;
λs1 + (1-λ)s2 S
λt1 + (1-λ)t2 T
(If S is a convex set in then λs1 + (1-λ)s2 is in ).
We've just applied the definition of a convex set so far. Now we want to show the set S T (which is of the form x1=(s1,t1)) is also a convex set. So we want to show that;
λx1 + (1-λ)x2 S T
We begin with the equation
λx1 + (1-λ)x2 S T
which becomes (x1 = (s1,t1), x2 = (s2,t2) are elements of )
λ(s1,t1) + (1-λ)(s2,t2) S and T
which means (λ(s1,t1) = (λs1,λt1))
(λs1,λt1) + ((1-λ)s2,(1-λ)t2) S and T
which then becomes ((s1,t1)+(s2,t2) = (s1+s2,t1+t2))
((λs1 + (1-λ)s2),(λt1 + (1-λ)t2)) S and T
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)what can we say about the equations (λs1 + (1-λ)s2) and (λt1 + (1-λ)t2) ?
(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
By definition (written above) (λs1 + (1-λ)s2) is in and (λt1 + (1-λ)t2) is in so that the element ((λs1 + (1-λ)s2),(λt1 + (1-λ)t2)) is in . This means by definition that is a convex set in
What we've done is taken any two elements x1 and x2 of and showed that λx1 + (1-λ)x2 also lies in .
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.