0

I've read somewhere that the interior of a cone is once again a cone. By cone I mean a set $S$ with the property that $(\forall x \in S)(\forall \lambda \geq 0)\ \lambda x \in S$.

However, if we consider the cone $C=[0,\infty\rangle$, then its interior is $\mathrm{Int}C=\langle 0,\infty\rangle$, which obiously isn't a cone since for example $2\in\mathrm{Int}C$, but $0 \cdot 2 \notin \mathrm{Int}C$.

Am I missing something?

implicati0n
  • 1,459
  • Have you considered that what you've read may be wrong or different than what you remember? If a cone $C$ is not the entire euclidean space, its interior does not contain $0$. If the interior is nonempty, it is therefore not a cone. However if you take the union of ${0}$ and the interior of a cone, the result is always a cone again. – Anon May 03 '16 at 19:56
  • @McFry I just needed confirmation that my counterexample was correct, thanks. :) – implicati0n May 03 '16 at 19:57
  • Could it be possible that the interior of a cone was defined as the union of ${0}$ and its topological interior? Because in that case the statement is actually true. – Anon May 03 '16 at 19:59
  • @McFry No, we defined the interior as the union of all open sets contained in the given set. – implicati0n May 03 '16 at 20:01

0 Answers0