Let $S_1$ and $S_2$ be convex sets. Is any of the sets $S _1^c$, $S_1 $ \ $S_2$, $S_1 ∩ S_2$, $S_1 ∪ S_2$ always convex?

Question

I have this question in front of me right now. When I draw examples of these in Venn diagrams I keep finding situations where they are non-convex. So is it right that no alternative is always convex?

And the union need not be convex..not sure why you are drawing Venn diagrams — nemo, Aug 30 '17 at 08:46
@Arthur It is didn't type it correctly. It is supposed to be S1* (Set one, with the * above it) The complement of S1. Ω/S1 — Etak, Aug 30 '17 at 09:02

score 3 · Answer 1 · answered Aug 30 '17 at 08:49

3

$S_1 \cap S_2$ is convex. Try a proof !
Let $S_1=(0,3)$ and $S_2=(1,2)$. Then $S_1 \setminus S_2$ is not convex.
Let $S_1=(0,3)$ and $S_2=(4,5)$. Then $S_1 \cup S_2$ is not convex.

answered Aug 30 '17 at 08:49

Fred

77,394

Don't forget the complement (the complement of a bounded interval is clearly not convex, though). – Arthur Aug 30 '17 at 09:03
Alright so is S1 ∩ S2 then the only alternative that is always convex? – Etak Aug 30 '17 at 09:07
Yes. $S_1 ∩ S_2 $ is always convex, if $S_1$ and $S_2$ are convex. – Fred Aug 30 '17 at 09:17

Let $S_1$ and $S_2$ be convex sets. Is any of the sets $S _1^c$, $S_1 $ \ $S_2$, $S_1 ∩ S_2$, $S_1 ∪ S_2$ always convex?

1 Answers1