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 $\;p\,,\;\;\frac1p&gt;0\;$ .

Second, since for all $\;n\in\Bbb N\;$ there exists a prime $\;p\;$ s.t. $\;p&gt;n\;$ (why?), we then get that

$$\forall\;n\in\Bbb N\;\;\exists\;\text{a prime}\;\;p\;\;s.t. \;\;\frac1p&lt;\frac1n$$

Now, for any $\;\epsilon &gt;0\;$ choose $\;n\in\Bbb N\;$ s.t. $\;\frac1n&lt;\epsilon\;$ (why such a natural exists?), and then choose a prime $\;p\;$ s.t.

$$\frac1p&lt;\frac1n&lt;\epsilon$$

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
First, clearly $\;0\;$ is a lower bound of $\;E\;$ as for any prime $\;p\,,\;\;\frac1p&gt;0\;$ .

Second, since for all $\;n\in\Bbb N\;$ there exists a prime $\;p\;$ s.t. $\;p&gt;n\;$ (why?), we then get that

$$\forall\;n\in\Bbb N\;\;\exists\;\text{a prime}\;\;p\;\;s.t. \;\;\frac1p&lt;\frac1n$$

Now, for any $\;\epsilon &gt;0\;$ choose $\;n\in\Bbb N\;$ s.t. $\;\frac1n&lt;\epsilon\;$ (why such a natural exists?), and then choose a prime $\;p\;$ s.t.

$$\frac1p&lt;\frac1n&lt;\epsilon$$

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