# Closure of $\frac {1} {n}$ [duplicate]

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?

• You mean the closure of $\{\frac1n:n\in\mathbb N\}$ in $\mathbb R$? Jul 7, 2016 at 1:10
• Jul 7, 2016 at 1:11
• I agree with Jonas, the set you're working with isn't clear here. Jul 7, 2016 at 1:11
• @JonasMeyer edited it seems those questions had a different definition of closure thus why I posted. Jul 7, 2016 at 1:12
• @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. Jul 7, 2016 at 1:14

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?