If a convex set $S \subseteq R^n$ contains no ray, can you show that it's bounded? I read about a slightly different problem:
Show that a closed convex set $S \subseteq R^n$ is bounded, if and only if it contains no ray, as answered here: To show a closed convex set $S \subseteq R^n$ is bounded if and only if $S$ contains no rays.
I'm thinking whether the closedness condition is necessary? It's used in the original proof, but I wasn't able to construct a counterexample with the closedness condition dropped.
 A: We are going to show that any unbounded convex set contains a ray.
Let $C \neq \emptyset$ be an unbounded convex set. Without loss of generality, we may assume that the interior of $C$ is nonempty, i.e. contains some point $x_0$. (Recall that the relative interior of a convex set is non-empty; if the interior of $C$ is empty, we can reduce the dimension of our space and consider $C$ as a subset of the affine hull of $C$ and find that $C$ has a non-empty interior in this space.) Moreover, we can assume that $x_0 =0$; otherwise we consider the shifted set $C-x_0$.
Since, by assumption, $C$ is unbounded there exists a sequence $(x_n)_{n \in \mathbb{N}} \subseteq C$ such that $\|x_n\|>n$. It follows from the Bolzano-Weierstraß theorem that the normalized sequence $(x_n/\|x_n\|)_{n \in \mathbb{N}}$ has a convergent subsequence; without loss of generality we can assume that the sequence itself converges, i.e. $x_n/\|x_n\| \to x$ for some $x$. 
We are going to show that the ray $\{\lambda x; \lambda \geq 0\}$ is contained in $C$. To this end, fix some number $R>0$. Obviously, $\|x_n\| > n \geq R$ for all $n \gg 1$. Since $0 \in C$ and $C$ is convex, we obtain that $y_n := R/\|x_n\| x_n \in C$ for large $n$. Moreover, as $y_n \to Rx$ as $n \to \infty$ and $0 \in \text{int} (C)$, we have $y_n - Rx \in C$ for large $n$. Hence, by convexity,
$$\frac{R}{2} x = \frac{1}{2} (y_n + (Rx-y_n)) \in C.$$
Since $R>0$ is arbitrary, this shows that $C$ contains the ray $\{\lambda x; \lambda \geq 0\}$.
