# Direct sum of convex sets is convex

Let $S_1 \subseteq \mathbb{R}^n$ be a compact convex set and let $S_2 \subseteq \mathbb{R}^n$ be a closed convex set. Prove that then $A=S_1 \oplus S_2$ is convex.

Here is my attempt, where I havent used the fact that the sets are compact and closed:

I must prove that $(s_1+s_2) + (1- \lambda)(s'_1 + s'_2) \in A$ for all $\lambda \in [0,1]$. Well, for me this is straightforward in the sense that the above expression can be written as $(s_1 +(1-\lambda)s'_1)+ (s_2+(1-\lambda)s'_2)$ and since $S_1,S_2$ are convex, the result follows. I dont understand why we need the sets to be compact and closed, what am I missing? I am a little confused here...

• What do you understand under "direct sum" here? Just a sum of two sets? Because the notion of direct sum usually applies to subspaces of a certain vector space. – TZakrevskiy Nov 19 '15 at 21:40
• They are vector spaces, so I am thinking of the direct sum as the set of all elements $s_1+s_2$, and their intersection is trivial (so the elements can be written in a unique way). Am I thinking about it wrong? This is a homework. – Surna Nov 19 '15 at 21:42
• $S_1$ and $S_2$ are not vector spaces, however if you also know that $\operatorname{span} S_1 \cap \operatorname{span} S_2 = {0}$, then it would still make sense to call it a direct sum. By the way, you need $\color{red}\lambda(s_1+s_2) + (1- \lambda)(s'_1 + s'_2)$, etc. – Paul Sinclair Nov 20 '15 at 0:59

However, since you mentioned that this is homework, my guess is that the next question is to prove that your sum $A=S_1+S_2$ is a closed set.
If we talk about usual topology, then there is only one compact linear subspace of $\Bbb R^n$ - the trivial vector space $\{\vec 0\}$.