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If a set is closed and open, it may be bounded (e.g., the empty set), or it may be unbounded (e.g., the set of real numbers).

But what about a closed set that is not open? Such as:

  • a singleton
  • a set of finite points
  • a closed interval
  • the union of a finite collection of closed intervals
  • the nonempty intersection of an arbitrary (possibly infinite) collection of closed intervals

These are all examples of closed, non-open sets, and they are all bounded.

Prove that if a set is closed and non-open, then it is bounded; or disprove by providing an example of a closed, non-open set that is unbounded.

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    $\begingroup$ $[0, \infty)$? ${}$ $\endgroup$ – Nigel Overmars Aug 20 '15 at 17:15
  • $\begingroup$ @NigelOvermars WolframAlpha says the right end is open. But it does contain all its limit points… $\endgroup$ – chharvey Aug 20 '15 at 17:17
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    $\begingroup$ If you don't like $[0,\infty)$, how about $\mathbb{Z}$, the set of all integers? This set has no limit points, so it's trivially closed. Alternatively, its complement is open since it is the union of all intervals of the form $(n, n+1)$. $\endgroup$ – Bungo Aug 20 '15 at 17:35
  • $\begingroup$ A singleton can perfectly be open, it depends on the topology: if $V$ is a discrete valuation ring, $\operatorname{Spec} V$, endowed with the Zariski topology, has two points: one is closed, the other is open and not closed. Of course it's not a Hausdoff topology. $\endgroup$ – Bernard Aug 21 '15 at 12:41
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The answer is $[0, \infty) \in \mathbb{R}$ as provided by Nigel. It is not open as you can't find an open ball around the point $0$ and it is closed as $(-\infty,0)$ is open.

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  • $\begingroup$ Good answer; the only thing that would make me happier is if you used the definition of closed to prove it is closed. (Not that proving its complement is open isn't a valid proof, but I like a more direct approach.) $\endgroup$ – chharvey Aug 20 '15 at 17:43
  • $\begingroup$ a set is closed iff it contains all its limit points $\endgroup$ – chharvey Aug 20 '15 at 17:58
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    $\begingroup$ @chharvey: The definition of ‘closed’ is that its complement is open — at least if we define a topology starting with open sets. $\endgroup$ – Bernard Aug 20 '15 at 18:40
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    $\begingroup$ Note $[0,\infty) \not \in \mathbb R.$ $\endgroup$ – zhw. Aug 20 '15 at 18:47
  • $\begingroup$ @Bernard that may be true, but I think it depends on the author. My book uses the limit point definition. Incidentally I think it's much more fascinating to define two concepts separately and then prove that they're related. E.g. define $\ln(x)$ as an integral and $\exp(x)$ as a summation, then prove that they're inverse functions (rather than defining one as the inverse of the other). $\endgroup$ – chharvey Aug 21 '15 at 12:18

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