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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?

Thank you.

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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. –  crf Dec 8 '12 at 8:19
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4 Answers 4

up vote 8 down vote accepted

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

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Do you mean ℤ? So, would every finite set be closed and bounded? –  Alti Dec 8 '12 at 0:24
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yes this is because each singleton is closed and the finite union of closed sets is closed again –  Amr Dec 8 '12 at 0:26
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Finally finite implies bounded –  Amr Dec 8 '12 at 0:27
    
Thank you for your help. –  Alti Dec 8 '12 at 0:28
    
@sugataAdhya I was referring to singletons in $\mathbb{R}$ not in any space –  Amr Feb 10 at 16:28
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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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$$\{x\in\mathbb R\mid x\geq 0\}$$

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

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Thanks, can you please see my reply to Amr? –  Alti Dec 8 '12 at 0:25
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Yes, every finite set is closed and definitely bounded. –  Asaf Karagila Dec 8 '12 at 0:26
    
@AsafKaragila: Check out my comment above. –  Sugata Adhya Dec 8 '12 at 8:16
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@Sugata: Real analysis is done in the real numbers with the standard topology. –  Asaf Karagila Dec 8 '12 at 8:45
    
@AsafKaragila: 'General-Topology' is tagged with the question. But as far as your solution is concerned it's all right. –  Sugata Adhya Dec 8 '12 at 9:09
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In $\mathbb R^n$ every non-compact closed set is unbounded.

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