Show that a set S in R^n is convex if and only if every convex combination of a finite number of points in S is in S.

Asked Feb 08 '22 at 04:20

Active Feb 08 '22 at 05:04

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I am stuck on the first direction of the if and only if statement:

Let K be an arbitrary number of points in S

S is a convex set $\rightarrow$ $x' = \sum_{i=1}^k c_ix_i\in S$ and $\sum_{i=1}^k c_i = 1$

If I'm understanding this question correctly, how do I go on to prove this?

Update: Induction attempt

Base case: Let k = 2

Show $x' = c_1x_1 + c_2x_2\in S$ and $c_1 + c_2 = 1$

Trivially true by definition of a convex set, if $x_1,x_2 \in S$ so is their line segment.

Hypothesis: Assume $\forall x_i \in S$ such that $x' = \sum_{i = 1}^kc_ix_i \in S$ and $\sum_{i = 1}^kc_i = 1$ holds

Induction step: Show $x' = \sum_{i = 1}^{k + 1}c_ix_i \in S$ and $\sum_{i = 1}^{k+1}c_i = 1$

$x' = \sum_{i = 1}^kc_ix_i + c_{k+1}x_{k+1}$

So for this last line, theres something that makes x' exist in S, but I don't quite know what.

edited Feb 08 '22 at 05:04

asked Feb 08 '22 at 04:20

jxu7

2

Try induction, Try anything. – copper.hat Feb 08 '22 at 04:23
For any element in S, its line segment is also in S @cineel – jxu7 Feb 08 '22 at 04:29
@copper.hat For the base case, if we assume k = 2 for example x' = c1 * x1 + c2 * x2 I assume this is in the convex set but why – jxu7 Feb 08 '22 at 04:30
Ah nevermind, its because x' is a point on the lines segment of two points in S, by definition also in S. Lemme try finishing this proof – jxu7 Feb 08 '22 at 04:38
yeah i'm still stuck on the induction step haha, i'll add it once i make more progress – jxu7 Feb 08 '22 at 04:45
No, because my question is asking about a finite amount of points in S, every convex combination can be infinite. – jxu7 Feb 08 '22 at 05:25
My question is just a weaker version I guess – jxu7 Feb 08 '22 at 05:25
2

convex combination is a finite sum by definition – cineel Feb 08 '22 at 05:32

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