Given $A=\{(n-5)/4n : n \in \mathbb{N}\}$
I want to show the $\inf(A)=-1.$ I've claimed this and shown that $(n-5)/4n \ge -1$. I now usually use archimedian property to say that there exists an $n>5/(1-4y)$ in order to prove that $y$ can't be the $\inf(A)$ as $y>(n-5)/4n$. However, solving this gives me $y<(n-5)/4n$ instead and working backwards from the correct answer gives me $n<5/(1-4y)$ which isn't shown by the archimedian property? Can anyone help
If $A\subset \Bbb R$ and $A$ has a least member $\min A$ then $\min A =\inf A .$
Because $\min A$ is a lower bound for $A$ and no $y>\min A$ can be a lower bound for $A,$
(... because if $y>\min A$ then $y$ is greater than some member of $A,$ that is, $y>\min A\in A$...)
so $\min A$ must be the largest lower bound for $A$, i.e. $\min A=\inf A .$
For the set in your Q:
$-1 \in A$ because if $n=1$ then $-1=(n-5)/4n \in A.$
If $n\in \Bbb N$ then $(n-5)/4n\ge -1$ because for any $n\in\Bbb N$ we have $$(n-5)/4n\ge-1\iff 4n\cdot (n-5)/4n\ge 4n\cdot (-1)\iff n-5\ge -4n \iff 5n\ge 5.$$
So $\min A=-1 .$