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I'm studying convex analysis and there is a theorem about a function defined on a convex and closed set and the theorem is proven both for the cases when the set is bounded and unbounded. Now, I know examples of closed and unbounded sets, but I failed to come up with an example where the set would also be convex.

H-a-y-K
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2 Answers2

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How about all of $\mathbb{R}^n$?

Sam Freedman
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  • I think it’s convex analysis, but yes same idea! – Sam Freedman Aug 13 '21 at 11:15
  • Is $\mathbb{R}^n$ closed? – H-a-y-K Aug 13 '21 at 11:15
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    There's lots of less trivial examples, but this is the obvious one. The whole space is always open and closed in any topology. If you want something other than the whole space, take the upper half hyperplane including the boundary. (I deleted the complex comment a second after posting it, because I realized the same!) – Alan Aug 13 '21 at 11:16
  • I see now. Thanks! I didn't even know about the existence of both open and closed spaces. – H-a-y-K Aug 13 '21 at 11:20
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Take, for example, $[0,\infty)$. This is closed, unbounded and convex. You can generate a lot more of these sets using this result.

Umesh Shankar
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