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.

• Let open unit ball in $\mathbb{R}^2$. Mar 5 '17 at 7:08
• @MyGlasses Why is this a counterexample? It contains no rays and it is bounded, just as for the closed case. Mar 5 '17 at 7:12
• @MyGlasses You need an unbounded convex set that contains no rays. Mar 5 '17 at 7:12
• Sorry. I thought he wants an example. Mar 5 '17 at 7:15
• An example of what? He does want an example but of non-closed unbounded set that contains no rays (or of a non-closed bounded set containing a ray). I unit ball is neither. Mar 6 '17 at 16:32

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 $\{\lambda x; \lambda \geq 0\}$.

• Nice work, man. I believe your proof is correct. Mar 5 '17 at 23:09
• There is a problem with this proof, namely, when you do the reduction to a lower dimension. Suppose for example that the interior of $C$ is empty, and you consider $C\subseteq X$ to apply your argument, where $X$ is a a strict affine subspace of $\mathbb R^d$; then there is no guarantee that the points $x_n$ you mention will be in $X$. Mar 6 '17 at 5:43
• @detnvvp Why not...? They are in $C$ hence in $X$.
– saz
Mar 6 '17 at 6:25
• The problem is the "without loss of generality", since, after reducing the dimension and considering the affine hull, you do not know if $C$ in this affine hull is still unbounded. Mar 6 '17 at 6:27
• @detnvvp If a set $A \subseteq \mathbb{R}^d$ is contained in a linear subspace $X \subseteq \mathbb{R}^d$ then $A \subseteq X$ is unbounded iff $A \subseteq \mathbb{R}^d$ is unbounded; right? Or am I missing something?
– saz
Mar 6 '17 at 6:34