6
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Let open unit ball in $\mathbb{R}^2$. $\endgroup$
    – Nosrati
    Mar 5 '17 at 7:08
  • $\begingroup$ @MyGlasses Why is this a counterexample? It contains no rays and it is bounded, just as for the closed case. $\endgroup$ Mar 5 '17 at 7:12
  • $\begingroup$ @MyGlasses You need an unbounded convex set that contains no rays. $\endgroup$ Mar 5 '17 at 7:12
  • $\begingroup$ Sorry. I thought he wants an example. $\endgroup$
    – Nosrati
    Mar 5 '17 at 7:15
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Mar 6 '17 at 16:32
4
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ Nice work, man. I believe your proof is correct. $\endgroup$
    – ChubbyRuby
    Mar 5 '17 at 23:09
  • $\begingroup$ 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 $X$. $\endgroup$
    – detnvvp
    Mar 6 '17 at 5:43
  • $\begingroup$ @detnvvp Why not...? They are in $C$ hence in $X$. $\endgroup$
    – saz
    Mar 6 '17 at 6:25
  • $\begingroup$ 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. $\endgroup$
    – detnvvp
    Mar 6 '17 at 6:27
  • 1
    $\begingroup$ @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? $\endgroup$
    – saz
    Mar 6 '17 at 6:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.