Show that $|\frac{1}{2n}-\frac{1}{2m}| < \epsilon$ holds for all $m, n > \frac{1}{\epsilon}.$

Question

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}? $

Answer 1

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

Answer 2

You can do a little better.

I write $c$ for $\epsilon$ because lazy.

Since $m, n > \frac1{c} $, $0 \le \frac1{m}, \frac1{n} \le c $.

If $m = n$ then $|\frac1{m}-\frac1{n}| = 0 < c$.

If $m > n$ then $|\frac1{m}-\frac1{n}| =\frac1{n}-\frac1{m} \lt \frac1{n} < c$.

If $m < n$ then $|\frac1{m}-\frac1{n}| =\frac1{m}-\frac1{n} \lt \frac1{m} < c$.

In all cases, $|\frac1{m}-\frac1{n}| < c$ so that $|\frac1{2m}-\frac1{2n}| < \frac{c}{2}$.