Proving the infimum of a set? $A : = \left\{ 6 - \frac { 1 } { n } : n \in \mathbb { N } \right\}$
I know 5 is a lower bound for A
I want to use the Archimedean Property to show 5+Epsilon is not a lower bound for all epsilon > 0 
How do I go about doing this?
 A: $\operatorname{inf}(A)=5$ because $5\in A$ and for all $x\in A$ it is true that $x=6-\frac1n\geq 5$ because $\frac1n\leq 1$.
Note that $5$ is an element of the set, so it is also true that $5=\operatorname{min}(A)$; so you can say that this already implies that $5$ is the infimum of the set.
A: 
I want to use the Archimedean Property to show $5+\epsilon$ is not a lower bound for all $\epsilon > 0$

By proving there is a value in $A$ that is less than all $5+\epsilon$. What numbers are less than $5 + \epsilon$?  Which of those are equal to some $6-\frac 1n$?
Answer:

 Well, If $n = 1$ the $6 - \frac 1n = 5$ and $5 < 5 + \epsilon$ for all $\epsilon$.

If somehow those hints arent enough we can be perverse and attempt to solve:
$ 6 - \frac 1n < 5 + \epsilon$  for all $\epsilon >0$.
$-\frac 1n < -1 + \epsilon$ 
$\frac 1n > 1 -\epsilon$
Well if $\epsilon \ge 1$ this is true for all $n$.  And if $\epsilon <1$ then 
$n < \frac 1{1-\epsilon}$.  
Now $\frac 1{1-\epsilon} > 1$ but for any $h > 0$ we can find $1 < \frac 1{1-\epsilon} < 1+h$ by letting $\frac 1{1+h}< 1-\epsilon$ and we can do that by letting $\epsilon < 1-\frac 1{1+ h} < 1$.
And so the only $n$ so that $6 - \frac 1n < 5 + \epsilon$ for all $\epsilon$ is $n =1$ and $6 - \frac 1n = 5$.
