Show that if $0 < \frac{1}{j},\frac{1}{k} \leq \frac{1}{N} \leq 1$, then $|\frac{1}{j}-\frac{1}{k}| \leq \frac{1}{N}$. ($j,k,N \in \mathbb{N}^+$))

Question

I am working through Tao's analysis I book and I am trying to prove that the sequence $a_n = \frac{1}{n}$ is a Cauchy sequence (Proposition 5.1.11). I understand the proof, but I am having trouble showing why it is exactly that

$0 < \frac{1}{j},\frac{1}{k} \leq \frac{1}{N} \leq 1 \rightarrow |\frac{1}{j}-\frac{1}{k}| \leq \frac{1}{N}$.

I have tried manipulaing single inequalitis on their own and combining them in order to reach the final statement, but with no success. I am always left with an extra $\frac{1}{j},\frac{1}{k}$ on the side of the $\frac{1}{N}$ which doesn't allow me to conclude the reasoning. I suspect I am neglecting some simpler inequalities which would allow me to compare $|\frac{1}{j}-\frac{1}{k}|$ with $\frac{1}{N}$, but I don't see it yet. Any clue is more than welcome.

Answer 1

We have $0 < \frac{1}{j} \leq \frac{1}{N}$ and $0 < \frac{1}{k} \leq \frac{1}{N}$. Then we can subtract $\frac{1}{j}$ from the second inequality to get: $$-\frac{1}{j} < \frac{1}{k} - \frac{1}{j} \leq \frac{1}{N} - \frac{1}{j}$$ but we know $\frac{1}{j} > 0$ so $\frac{1}{N} - \frac{1}{j} \leq \frac{1}{N}$ and we know $\frac{1}{j} \leq \frac{1}{N}$ so $-\frac{1}{N} \leq -\frac{1}{j}$ hence $$-\frac{1}{N} \leq -\frac{1}{j} < \frac{1}{k} - \frac{1}{j} \leq \frac{1}{N} - \frac{1}{j} \leq \frac{1}{N}$$ But we can simplify this to get: $$-\frac{1}{N} \leq \frac{1}{k} - \frac{1}{j} \leq \frac{1}{N} \implies |\frac{1}{k} - \frac{1}{j}| \leq \frac{1}{N}$$

Answer 2

You can put this more simply: If $0\le x,y\le z$, then $|x-y|\le z$. And this is obvious, because the interval $[x,y]$ (or $[y,x]$ if $y<x$) is entirely contained in the interval $[0,z]$.