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Intuitively for me, it seems as if closed sets are bounded, especially considering closed sets contain all limit points. But I know this isn't the case, because $ℝ$ is closed (and open) and is not bounded.

Is this the only case of a closed set not being bounded? Can anyone provide an example that further illustrates the difference between closed and bounded?

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    $\begingroup$ One thing it might be handy to remember is that a closed set is the complement of an open set. So like, the complement of any open ball, for instance, is closed. $\endgroup$
    – crf
    Dec 8, 2012 at 8:19
  • $\begingroup$ An easier way to describe bounded (for me) is that the set lies in a closed ball. $\endgroup$ Jul 24, 2022 at 10:14

4 Answers 4

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We cover each of the four possibilities below.

Closed and bounded: $[0,1]$

Closed and not bounded: $\cup_{n\in Z}[2n,2n+1]$

Bounded and not closed: $(0,1)$

Not closed and not bounded: $\cup_{n\in Z}(2n,2n+1)$

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The integers as a subset of $\Bbb R$ are closed but not bounded.

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    $\begingroup$ Do you mean ℤ? So, would every finite set be closed and bounded? $\endgroup$
    – Alti
    Dec 8, 2012 at 0:24
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    $\begingroup$ yes this is because each singleton is closed and the finite union of closed sets is closed again $\endgroup$
    – Amr
    Dec 8, 2012 at 0:26
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    $\begingroup$ Finally finite implies bounded $\endgroup$
    – Amr
    Dec 8, 2012 at 0:27
  • $\begingroup$ @sugataAdhya I was referring to singletons in $\mathbb{R}$ not in any space $\endgroup$
    – Amr
    Feb 10, 2014 at 16:28
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$$\{x\in\mathbb R\mid x\geq 0\}$$

Also note that there are bounded sets which are not closed, for examples $\mathbb Q\cap[0,1]$.

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  • $\begingroup$ Thanks, can you please see my reply to Amr? $\endgroup$
    – Alti
    Dec 8, 2012 at 0:25
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    $\begingroup$ Yes, every finite set is closed and definitely bounded. $\endgroup$
    – Asaf Karagila
    Dec 8, 2012 at 0:26
  • $\begingroup$ @AsafKaragila: Check out my comment above. $\endgroup$ Dec 8, 2012 at 8:16
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    $\begingroup$ @Sugata: Real analysis is done in the real numbers with the standard topology. $\endgroup$
    – Asaf Karagila
    Dec 8, 2012 at 8:45
  • $\begingroup$ @AsafKaragila: 'General-Topology' is tagged with the question. But as far as your solution is concerned it's all right. $\endgroup$ Dec 8, 2012 at 9:09
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In $\mathbb R^n$ every non-compact closed set is unbounded.

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