How does one show that $\{ \frac{1}{n} | n \in \mathbb{Z_{>0}}\} $is not compact in the standard topology?

Question

How does one show that $\{ \frac{1}{n} | n \in \mathbb{Z_{>0}}\}$ is not compact in the standard topology of $\mathbb{R}$?

I know this is not compact because if we take small enough intervals around the elements of a, we cannot have a finite subcovering. But i can't seem to think about an open covering that fails. I tried to get something in the form $\{(\frac{1}{n} - \epsilon,\frac{1}{n} + \epsilon)\}$, but one can note it is impossible to choose epsilon s.t

$0 \notin (\frac{1}{n} - \epsilon,\frac{1}{n} + \epsilon)$.

Answer 1

How about $\{(\frac1n,2)\}$? . . .

Answer 2

You can make the intervals smaller and smaller as you get close to zero. Something like a cover of the form $(\frac 1n - \frac 1{2n^2},\frac 1n + \frac 1{2n^2})$

Answer 3

Compact implies closed and bounded but the set is not closed: it does not contain its limit point $0$.