Show that $|\frac{1}{2n}-\frac{1}{2m}| < \epsilon$ holds for all $m, n > \frac{1}{\epsilon}.$ In Example 1.5-9 of the book Functional Analysis by Kreyszig it claims that $|\frac{1}{2n}-\frac{1}{2m}| < \epsilon$ holds for all $m, n > \frac{1}{\epsilon}.$ 
My calculations don't lead to that! : 
1st objection: wlog let $m>n$ and $m=n+k$ so $$|\frac{1}{2n}-\frac{1}{2m}| < \epsilon \implies k-2\epsilon(n)(n+k)<0.$$ Differentiation with respect to $k$ and desiring for the $k-2\epsilon(n)(n+k) $ to decrease after its 'right-zero' gives $n > \frac{1}{2\epsilon} $ and thus $m, n > \frac{1}{2\epsilon}.$
2nd objection: $k-2\epsilon(n)(n+k)= k-2\epsilon n^2 -2\epsilon nk <0$ holds for $n> \dfrac{-2\epsilon k + \sqrt{4{\epsilon}^2k^2+8\epsilon k}}{4\epsilon k}$. How change of $k$ doesn't effect $k-2\epsilon n^2 -2\epsilon nk <0$?     
3rd objection: how intersection of two domains for $n,m$ obtained in 1st and 2nd objections is a superset of $n,m > \frac{1}{\epsilon}? $ 

$m,n>{1\over\epsilon}$ implies that ${1\over 2n}<\epsilon/2$. This implies that $|{1\over 2n}-{1\over 2m}|\leq |{1\over 2n}|+|{1\over 2m}|<\epsilon/2+\epsilon/2=\epsilon$.
