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A pointed cone $K$ is one such that if any point $x \in K$, then $-x \notin K$. Based on this, are norm cones not pointed?

For example, if $\|x\| \leq t$ then $\|-x\| \leq t$, but this feels wrong. What’s wrong?

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    $\begingroup$ I think you have to consider the cone defined on $(x,t)$ so that if $(x,t) \succeq 0$, then $(-x,-t) \preceq 0$. $\endgroup$
    – jjjjjj
    Feb 16, 2018 at 17:45
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    $\begingroup$ You need to consider the points $(x,t)$ and $(-x,-t)$, not $(-x,t)$. $\endgroup$
    – copper.hat
    Feb 16, 2018 at 17:47
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    $\begingroup$ Also, I presume you mean any non zero $x \in K$? Any closed cone wiull contain $0$ and $-0 = 0$. $\endgroup$
    – copper.hat
    Feb 16, 2018 at 17:48
  • $\begingroup$ Good point, thanks for clarifying the nonzero $x$ $\endgroup$
    – dunno
    Feb 16, 2018 at 17:49

1 Answer 1

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Figured I'd write this up.

As @copper.hat pointed out, if $x \neq 0$, then you must consider the RHS (i.e. $t$) as part of what defines the cone. So, if $x \in K$, you're really talking about $x := (\tilde{x},t) \succeq_K 0$. From this, it's easier to see that if $(\tilde{x},t) \succeq_K 0$, then $(-\tilde{x},-t) \preceq_K 0$. Thus, $y = 0 = (0, 0)$ is the only element such that $y \in K$ and $-y \in K$

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