1. ## Convex set proof

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!

2. You take any two values from set S and show that the formula works for them...

If we take $\displaystyle x,y \in S$
let just take that x > y
that means that

$\displaystyle ax + ( 1 - a )y \in S$
$\displaystyle ( a \in <0,1> )$

what you need to prove is that:

$\displaystyle a( x - y ) + y \in S$

but:

$\displaystyle a( x - y ) \in <0, x-y>$

than:

$\displaystyle a(x-y)+y \in <y,x>$

( this is for x > y , you now prove for y > x )

3. 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?

4. Originally Posted by econlover55
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?
if we take set { 1, 2 }

you need to find only one $\displaystyle a \in [0,1]$

and only one pair x, y $\displaystyle x,y \in S$

where $\displaystyle ax + ( 1 - a )y \notin S$

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.

Originally Posted by econlover55
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?

notice that it can be:

1) x = y
2) x > y
3) x < y

1) if x = y $\displaystyle \in S$

a*x + ( 1 - a )*y = a*x + x - a*x = x $\displaystyle \in S$

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:

$\displaystyle a(x-y)+y \in <y,x>$

or look better only on this part...

$\displaystyle <y,x>$

and we supposed that x > y

so we know for sure that set $\displaystyle <y,x>$ 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...

$\displaystyle <y,x> \subset S$