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Apr 28, 2019 at 12:13 comment added James Propp I won’t have to re-ask the question after all. Sam’s comments have answered the question I meant to ask.
Apr 27, 2019 at 3:31 history edited Sam Hopkins CC BY-SA 4.0
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Apr 27, 2019 at 3:26 history edited Sam Hopkins CC BY-SA 4.0
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Apr 27, 2019 at 2:33 history edited Sam Hopkins CC BY-SA 4.0
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Apr 26, 2019 at 23:41 comment added Andreas Blass A quicker (though ultimately equivalent) explanation for why $K$ can't be compact (unless it's $\varnothing$ or $\{0\}$) is that it's closed under multiplication by positive scalars (since you can also multiply the vectors $\mathbf x$ by scalars).
Apr 26, 2019 at 22:27 history edited Sam Hopkins CC BY-SA 4.0
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Apr 26, 2019 at 19:20 history edited Sam Hopkins CC BY-SA 4.0
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Apr 26, 2019 at 19:03 comment added Sam Hopkins If you put an additional constraint like $\sum a_i =1$, then you get the intersection of a polyhedral cone with an affine subspace, which will be a polytope.
Apr 26, 2019 at 19:02 comment added James Propp You're right! I took the coward's way and now I'm paying the price. :-) I'll look at the literature to see which definition of noncompact polytopes is the right one to use and I'll re-ask the question.
Apr 26, 2019 at 18:47 history edited Sam Hopkins CC BY-SA 4.0
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Apr 26, 2019 at 18:40 history answered Sam Hopkins CC BY-SA 4.0