Let $C\subseteq \Bbb R^n$ be a nonempty closed, convex set. Prove that $C$ has a recession direction if and only if $C$ is unbounded.
If $C$ has a recession direction then trivially it is unbounded.
Suppose $C$ is unbounded. I have seen proofs that consider a sequence which goes to infinity, and the corresponding sequence of normalized vectors, which then uses the compactness of the unit circle, and so on. It's a nice proof, but it seems like it should also be possible to give a proof by considering the collection of halfspaces $H_\alpha$ such that $C=\bigcap H_\alpha$.
Each half-space is associated with some set of recession directions, call them $rec(H_\alpha)$. In general for any sets $rec(\bigcap A_\alpha)=\bigcap (rec(A_\alpha))$ so that in this case $$ rec(C) = \bigcap rec(H_\alpha) $$ And we have $rec(H_\alpha)=\{\vec d\in\Bbb R^n:\vec p_\alpha\cdot\vec d \le 0\}$ for some corresponding $\vec p_\alpha$. And we know that intersections of closed sets are closed, so the set of recession directions is closed ... I'm not sure what that gets me, but I'm brainstorming possibly useful facts. I tried thinking of a contradiction I could derive from the hypothesis that $rec(C)=\{\vec 0\}$ but couldn't think of anything.
Can this be completed to a proof?