Convex set problem

Link to my solution (config.gif):
http://www.box.net/shared/znvjj1bukl

To John: A is NOT a subset of A∩B (intersection of A and B). Therefore you cannot directly argue αw1 βw2єA∩B because αw1 βw2єA and αw1 βw2єB
My other two parts are similar to yours.

(a)A∩B={wєR²:wєA,wєB}
for all w1,w2єA∩B,
αw1+βw2єA for all α, β >= 0 and α + β = 1 (A is a subset of A∩B)
αw1+βw2єB for all α, β >= 0 and α + β = 1 (B is a subset of A∩B)
therefore αw1+βw2єA∩B for all α, β >= 0 and α + β = 1
thus A∩B is also convex
(b)for all w1,w2єA+B,for all α, β >= 0 and α + β = 1
αw1+βw2
=α(u1+v1)+β(u2+v2) for some u1,u2єA,v1,v2єB
=(αu1+βu2)+(αv1+βv2)
note:αu1+βu2єA (A is convex),αv1+βv2єB(B is convex)
therefore αw1+βw2є(A+B)
thus A+B is also convex
(c)for all w1,w2єγA,for all α, β >= 0 and α + β = 1
αw1+βw2
=αγu1+βγu2 for some u1,u2єγA
=γ(αu1+βu2)
Note:αu1+βu2єA (A is convex)
therefore αw1+βw2єγA
thus γA is also convex

2008-09-01 19:54:23 補充：
αw1+βw2єA for all α, β >= 0 and α + β = 1 (A∩B is a subset of A)
αw1+βw2єB for all α, β >= 0 and α + β = 1 (A∩B is a subset of B)