How to prove that $H(S_1\cap S_2)\subset H(S_1) \cap H(S_2)$ and $H(S_1 \cup S_2) \supset H(S_1) \cup H(S_2)$

Question

I'm studying convex analysis and my task is to prove the following inclusions:

$S_1, S_2$ are non-empty sets in $\mathbb{R}^n$, and $H(S) $ defined as the convex hull of set $S$.

Show that

$H(S_1\cap S_2)\subset H(S_1) \cap H(S_2)$

and

$H(S_1 \cup S_2) \supset H(S_1) \cup H(S_2)$

Can the inclusions be reversed? Explain why?

How should I proceed with a problem like this? I thought about using contradiction proof? What if I assume for example in the first task that there is a point $\textbf{x}_1 \in H(S_1\cap S_2)$, but $\textbf{x}_1\not\in H(S_1) \cap H(S_2)$ and show why this can not be? Is this the way to go? :)

Thank you for any help! =)

Answer 1

All you need to show is that the convex hull is inclusion-preserving, that is, if $S⊂T$, then $H(S)\subseteq H(T)$. The inclusions in your problem then follow easily.

In order to show that, in general, we don't have equality, consider $S_1=\{0,2\}$ and $S_2=\{1\}$, or $S_1=\{0\}$ and $S_2=\{1\}$.

Answer 2

I think I got myself the first one so I'll add it here:

$$S_1 \cap S_2 \subset S_1$$

$$S_1 \cap S_2 \subset S_2$$

$$=>$$

$$H(S_1 \cap S_2) \subset H(S_1)$$ $$H(S_1 \cap S_2) \subset H(S_2)$$

$$=>$$

$$H(S_1 \cap S_2) \subset H(S_1) \cap H(S_2)$$

I think I can do the same with the second one:

$$H(S_1 \cup S_2) \supset H(S_1) \cup H(S_2)$$

$$<=> H(S_1) \cup H(S_2) \subset H(S_1 \cup S_2)$$ $$S_1 \subset S_1 \cup S_2$$ $$S_2 \subset S_1 \cup S_2$$ $$=>$$

$$H(S_1) \subset H(S_1 \cup S_2)$$ $$H(S_2) \subset H(S_1 \cup S_2)$$

so

$$H(S_1) \cup H(S_2) \subset H(S_1 \cup S_2)$$