Show that inf(E=1/p)=0 

*Let E be subset of Q (with the usual ordering) consisting of all fractions of the form 1/p, where p is a
prime number. Show that inf(E) = 0.


Here is my attempt to prove this. Suppose there exists a minimum of E min(e). 0

First, clearly $\;0\;$ is a lower bound of $\;E\;$ as for any prime $\;p\,,\;\;\frac1p>0\;$ .
Second, since for all $\;n\in\Bbb N\;$ there exists a prime $\;p\;$ s.t. $\;p>n\;$ (why?), we then get that
$$\forall\;n\in\Bbb N\;\;\exists\;\text{a prime}\;\;p\;\;s.t. \;\;\frac1p<\frac1n$$
Now, for any $\;\epsilon >0\;$ choose $\;n\in\Bbb N\;$ s.t. $\;\frac1n<\epsilon\;$ (why such a natural exists?), and then choose a prime $\;p\;$ s.t.
$$\frac1p<\frac1n<\epsilon$$
and you've shown $\;0\;$ is the highest lower bound, so it is the infimum of $\;E\;$ . It's easy now to see it can't be the minimum
