Let A and B be convex subsets of $\Bbb R^n$. The join of A and B is the set of all $\vec x$ such that $\vec x$ lies on a line segment with one endpoint in A and the other in B. I am wondering how to show that the join of A and B is a convex set.



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.

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