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My question is the same as the title. Suppose $P \subset \mathbb{R}^n$ is a given convex polytope, let's say by a system of linear inequalities $Ax \leq b$. Is that a Borel set?

Here they show that not all convex sets are Borel. However polytopes have a special structure and it seems to me they should be Borel.

spacetimewarp
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  • Polytopes are even closed, so a fortiori they're Borel: every closed set is Borel. Note that convexity is irrelevant here: "concave polytopes" (I don't know what the technical term is) are also all closed, hence trivially Borel. – Noah Schweber Jan 17 '22 at 06:09
  • oh, that's true but let's say that some of the inequalities may be strict. – spacetimewarp Jan 17 '22 at 06:10
  • That still lets us represent every such polytope as a closed set minus some finitely many closed sets, which is again Borel. Any finite (or even countable) Boolean combination of open and closed sets is Borel. And again we can go beyond convexity without issue. – Noah Schweber Jan 17 '22 at 06:11
  • yes, I agree! that's helpful, thanks. – spacetimewarp Jan 17 '22 at 06:12

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Turning my comments above into an answer: anything even vaguely polytope flavored is going to be a finite (or at worst countable) Boolean combination of closed sets hence Borel - indeed low-level Borel. To get a non-Borel set one has to do something extremely complicated.

Noah Schweber
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  • yes cool! one follow-up question. If we have a system $Ax \leq b$ it may define some "unbounded polytope" (I don't know how to call that thing). This is no longer closed so I am a bit unclear on how to see it is a Borel set. – spacetimewarp Jan 17 '22 at 06:25
  • @spacetimewarp Actually that is still closed. Closed does not imply bounded! (It isn't compact but that's a different matter.) – Noah Schweber Jan 17 '22 at 06:25
  • I see, yes that makes sense. Thanks again. – spacetimewarp Jan 17 '22 at 06:26