# Subtle question concerning intersection of convex sets

I am attempting to convince myself that if $$\{S_{\alpha}: \alpha \in \mathcal{A}\}$$ is any collection of convex sets, then $$\cap_{\alpha \in \mathcal{A}}S_{\alpha}$$ is convex. This is my proof so far:

Let $C = \cap_{\alpha \in \mathcal{A}}S_{\alpha}$ be an intersection of convex sets. If there is only one point $\textbf{v}$ in $C$ then we immediately know that $C$ is convex. Assume, then, that there are at least two points $\textbf{u}, \textbf{v} \in C$. Then $\textbf{u}, \textbf{v}$ are contained in every convex set in $\{S_{\alpha}: \alpha \in \mathcal{A}\}$. Because each of these sets are convex, it follows that the set $\{t\textbf{u} + (1-t)\textbf{v}, 0 \leq t \leq 1\}$ is contained in each of these sets as well and is therefore contained in $C$. Thus, $C$ is convex.

It is my belief, however, that the above proof is incomplete because we have not accounted for the possibility that $C = \emptyset$. Now, I perused Google and found several sources claiming that the null set is convex. I am having trouble believing them; considering that there are no points in the set with which to take linear combinations with nonnegative components summing to 1. Is it really just as simple as saying that because there are no points in the set that need to satisfy the above property, that is is vacuously true that the null set is convex?

• If a statement starts with "$\forall x,y \in C$", it is trivially satisfied by the empty set. – Prahlad Vaidyanathan Oct 13 '13 at 16:12
• Just write down carefully the definition if a convex set. You will immediately see that empty set is convex. – Moishe Kohan Oct 13 '13 at 16:13
• Okay, thanks. I figured this was the case. – Decave Oct 13 '13 at 16:14