If $C_1$ and $C_2$ are convex sets then $C_1 + C_2$ is a convex set

Question

Definition: $C_1 + C_2 = \{x + y \,\, | \,\,x \in C_1, y \in C_2\}$.

Proposition: Let $C_1$ and $C_2$ be convex sets, then $C_1 + C_2$ is convex.

Proof (sketch): Choose $a = a_1 + a_2$ and $b = b_1 + b_2$ such that $a_1, b_1 \in C_1$ and $a_2, b_2 \in C_2$. Then $a, b \in C$. Suppose there exists $\lambda \in [0,1]$ such that $\lambda a + (1 - \lambda) b \notin C$. Then:

$$\lambda(a_1 + a_2) + (1 - \lambda) (b_1 + b_2) \notin C $$

$$( \lambda a_1 + (1 - \lambda) b_1) + (\lambda a_2 + (1-\lambda) b_2) \notin C $$

Define $A = ( \lambda a_1 + (1 - \lambda) b_1)$ and $B = ( \lambda a_2 + (1 - \lambda) b_2)$, $A \in C_1$ and $B \in C_2$. Therefore $A + B \notin C$, which is an absurd since by definition $A+B$ must belong to $C$.

Is this enough or am I missing something?

Answer 1

Here is another approach, albeit just replacing one problem by another:

Show that $C_1 \times C_2$ is convex first and then show that if $L$ is linear and $C$ convex, then $L(C)$ is convex. Then apply $L(x) = x_1+x_2$ to $C_1 \times C_2$.

Answer 2

This is fine; excellent use of "by definition" and your proof format was really clear. Even better (in my subjective opinion) would be to drop the "proof by contradiction" guise and do it as a direct proof (for all $\lambda$, note that $\lambda a_1+(1-\lambda)b_1\in C_1$ and similar expression for $C_2$, therefore $\lambda a+(1-\lambda)b\in C$).