You take any two values from set S and show that the formula works for them...
If we take
let just take that x > y
that means that
what you need to prove is that:
but:
than:
( this is for x > y , you now prove for y > x )
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!
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.
if we take set { 1, 2 }
you need to find only one
and only one pair x, y
where
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.
notice that it can be:
1) x = y
2) x > y
3) x < y
1) if x = y
a*x + ( 1 - a )*y = a*x + x - a*x = x
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:
or look better only on this part...
and we supposed that x > y
so we know for sure that set 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...