Why the cone generated by a closed convex set containing the origin may not be closed?

I encounter this problem in the proof of Theorem $1.28$, page 21, Skiadas' Asset Pricing Theory.

$X$ is a constrained market which is defined to be a closed convex subset of $\Bbb R^{K+1}$, $C = \{kx : x \in X, k \in \Bbb R_+\}$, why it's possible that $C$ is not closed?

PS: If you take an interest in the proof of equivalence between non-arbitrage and the existence of present value function for a constrained market, you can find them here in the sample chapter.

Added: As David Mitra's example suggests, it's important to point out $X$ contains the origin, which is required in the definition of a constrained market.

• In $\Bbb R^2$, take a disc of radius $1$ centered at $(0,2)$. Its cone does not contain the origin. But the origin is in the closure of the cone. Mar 6, 2014 at 13:59
• @DavidMitra: So in $\mathbb{R}^n$, it seems plausible that the cone (as OP defines it) of a closed set $S$ is closed iff $0\in S$. Does that seem right? I can't see that convexity buys you anything, as long as the original set is already closed (I'm assuming this refers to closure in the topological sense).
– MPW
Mar 6, 2014 at 14:09
• @DavidMitra Got it, thank you. Is there a example of $X$ that contains the origin? I should have added that $X$, as the constrained market, is defined to contain the origin. Mar 6, 2014 at 14:10
• Try the region in $\Bbb R^2$ consisting of the graph of $y=|x^3|$ and the points "above it". The cone is $\{(x,y) : y>0\}\cup\{(0,0)\}$. Mar 6, 2014 at 14:45
• @DavidMitra Tons of thanks! Please consider posting your comment as an answer. I'm more than happy to accept it. Mar 6, 2014 at 14:50

1 Answer

Take, in $\Bbb R^2$, $X$ to be the graph of $y=x^2$ together with the set of points "above it". Then the cone $C$ of $X$ is the non-closed set $\bigl\{\,(x,y)\mid y>0\,\bigr\}\cup\{\,(0,0)\,\}$.

(I think any strictly convex closed body tangent to the origin will do.)