Answer :
It is proved that the join of A and B is a convex set when the join of A and B is the set of all [tex]$\vec x$[/tex] such that [tex]$\vec x$[/tex] lies on a line segment with one endpoint in A and the other in B.
Let [tex]p_1[/tex] and [tex]p_2[/tex] belong to the join of A and B.
Let q = x[tex]p_1[/tex] + (1 − x) [tex]p_2[/tex] with x ∈ [0,1].
There exist [tex]a_1[/tex] ∈ A and [tex]b_1[/tex] ∈ B and [tex]r_1[/tex] ∈ [0,1] with
[tex]p_1[/tex]= [tex]r_1[/tex] [tex]a_1[/tex] + (1 − [tex]r_1[/tex]) [tex]b_1[/tex]
There exist [tex]a_2[/tex] ∈ A and [tex]b_2[/tex] in B and [tex]r_2[/tex] ∈ [0,1] with
[tex]p_2[/tex] = [tex]r_2[/tex] [tex]a_2[/tex] + (1 − [tex]r_2[/tex]) [tex]b_2[/tex].
Since A and B are convex, we have
A⊃{c [tex]a_1[/tex] + (1−c) [tex]a_2[/tex] : c∈[0,1]}
and B⊃{d [tex]b_1[/tex]+(1−d) [tex]b_2[/tex] : d∈[0,1]}.
So if we can find c, d, e ∈ [0,1] such that
q = e[c [tex]a_1[/tex] + (1−c) [tex]a_2[/tex]] + (1 − e)[d [tex]b_1[/tex] + (1−d) [tex]b_2[/tex] ] -------(1)
then q belongs to the join of A and B.
Now q = x [tex]p_1[/tex] + (1−x) [tex]p_2[/tex] = x[[tex]r_1[/tex] [tex]a_1[/tex] + (1−[tex]r_1[/tex]) [tex]b_1[/tex]] + (1−x) [[tex]r_2[/tex] [tex]a_2[/tex]+(1−[tex]r_2[/tex]) [tex]b_2[/tex]].
Similarly, we can show that there do exist c, d, e∈[0,1] such that
x [tex]r_1[/tex] = ec
(1 − x) [tex]r_2[/tex] = e (1−c)
x(1− [tex]r_1[/tex]) [tex]b_1[/tex] = (1−e)d
(1 − x)(1 − [tex]r_2[/tex]) = (1 − e)(1 − d).
Then equation 1 is satisfied.
Hence, proved that the join of A and B is a convex set.
Read more about the Convex set:
brainly.com/question/29650204
#SPJ4