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I have as a definition of the closure of a set $E$ in a metric space $(S,d)$ that $E^-$ is the intersection of all closed sets containing E (Elementary analysis the theory of calculus by Kenneth Ross). I have as a hunch that the closure of $E = \{\frac{1}{n}: n \in \Bbb N\} $ is $[0,1]$ since this set seems to contain every element in $E$. How can I prove this to be true?

IntegrateThis
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    You mean the closure of ${\frac1n:n\in\mathbb N}$ in $\mathbb R$? – Jonas Meyer Jul 07 '16 at 01:10
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    See http://math.stackexchange.com/q/1399295/, http://math.stackexchange.com/questions/75162/determine-the-closure-of-the-set-k-frac1n-mid-n-in-mathbb-n-under-eac?rq=1, http://math.stackexchange.com/q/44913/ – Jonas Meyer Jul 07 '16 at 01:11
  • I agree with Jonas, the set you're working with isn't clear here. – Mathemagician1234 Jul 07 '16 at 01:11
  • @JonasMeyer edited it seems those questions had a different definition of closure thus why I posted. – IntegrateThis Jul 07 '16 at 01:12
  • @James: Two of them have different topologies considered, but those include the standard topology. The first one has only the standard topology, and its answer refers to the closure being the intersection of closed sets containing the set. – Jonas Meyer Jul 07 '16 at 01:14
  • The closure of ${ 1/n : n \in \Bbb{N}}$ is not $[0, 1]$. If $x \in E^-$, then that means every closed set containing $E$ contains $x$. – AJY Jul 07 '16 at 01:18

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Wouldn't the non-connected set $[0,0.501] \cup [0.9,1]$ contain every $1/n$ yet still be a proper subset of $[0,1]$, disproving your conjecture?

Trurl
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