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Is the set of integers $ \mathbb{Z}$ closed in $\mathbb{R}$ equipped with the usual topology?

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    $\begingroup$ It depends. What's the topology on $\mathbb{R}$? $\endgroup$ – user61527 Mar 17 '14 at 20:54
  • $\begingroup$ @T.Bongers My bad, I should've said that I have the usual topology on $\mathbb{R}$. $\endgroup$ – Rachel Mar 17 '14 at 21:05
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$\mathbb{R}\setminus\mathbb{Z} = \bigcup_{n\in\mathbb{Z}}(n,n+1)$ is a union of open sets and therefore open. Since the complement of $\mathbb{Z}$ is open in $\mathbb{R}$, $\mathbb{Z}$ is closed in $\mathbb{R}$.

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  • $\begingroup$ This is only correct assuming the usual topology on $\mathbb{R}$. $\endgroup$ – user61527 Mar 17 '14 at 20:58
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    $\begingroup$ @T.Bongers The usual topology is what we usually assume we're given! $\endgroup$ – Pedro Tamaroff Mar 17 '14 at 21:01
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    $\begingroup$ I agree with Pedro. If the topology isn't specified, one assumes the usual topology; conversely if the usual topology isn't what is meant, it is the responsibility of the person asking the question to specify the topology. $\endgroup$ – MJD Mar 17 '14 at 21:04

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