How does $0 < 1/j$ and $1/k < 1/N$ imply that $| 1/j - 1/k | \leq 1/N$?

Question

Synopsis

In Tao's Analysis 1, in his proof that the sequence $a_1,a_2, a_3, \dots$ defined by $a_n := 1/n$ is a Cauchy sequence, there is an inequality that I don't really feel comfortable with. I'll highlight this inequality below the proof and give some insight later.

Proof

We have to show that for every $\epsilon > 0$, the sequence $a_1, a_2, \dots$ is eventually $\epsilon$-steady. So let $\epsilon >0$ be arbitrary. We now have to find a number $N \geq 1$ such that the sequence $a_N, a_{N+1}, \dots$ is $\epsilon$-steady. This means that $d(a_j, a_k) \leq \epsilon$ for every $j,k \geq N$, i.e. $$|1/j - 1/k| \leq \epsilon \text{ for every $j,k \geq N$}.$$ Now since $j,k \geq N$, we know that $0<1/j$, $1/k < 1/N$, so that $|1/j - 1/k| \leq 1/N$. So it is sufficient for $N$ to be greater than $1/ \epsilon$.

The Inequality

The inequality that if $0<1/j$ and $1/k \leq 1/N$ is true, then $|1/j - 1/k| \leq 1/N$ is true just doesn't click for me. Probably it's my lack of sleep, but the intuition isn't there. I have, however, been able to prove it partially (I think, though there's probably something wrong since I'm mentally so incapacitated), and I'll highlight my partial proof below.

Partial Proof of Inequality That Is Probably Obvious to Most People Who Are Not Me

Suppose $j \geq k$. Then $1/j \leq 1/k$. So $1/N \geq 1/k > 1/k - 1/j = |1/j - 1/k|$ (since $1/j - 1/k < 0$). Now suppose $k <j$. Then by a similar argument, $1/N \geq 1/j \geq 1/j - 1/k = |1/j - 1/k|$ (isn't $1/j \leq 1/N$ since both $j,k \geq N$? I hope so.). Therefore, $|1/j - 1/k | < 1/N$.

Finale

As you can see, $|1/j - 1/k | < 1/N$ is slightly different from $|1/j - 1/k | \leq 1/N$. What's wrong with my "proof"? Why is my brain so bewildered by this inequality that probably isn't even that important? What is some intuition for this?

EDIT

I now realize that I simply suck at reading textbooks and that Tao was implying that $0<1/j<1/N$ and $0<1/k<1/N$ were both true at the same time wow I am actually a mess right now. This is now so painfully obvious and I cannot believe I spent like 15 minutes just bewildered. Thank you so much for your kind answers.

Answer 1

Perhaps what you're missing is this: when Tao writes

we know that $0 < 1/j, 1/k < 1/N$ is true

he means that there are two inequalities that are true, namely $$ 0 < 1/j < 1/N \qquad \text{and} \qquad 0 < 1/k < 1/N. $$ So, now you have two numbers, $1/j$ and $1/k$, sandwiched strictly between $0$ and $1/N$. So, the distance between these two "inner numbers" is definitely smaller than the distance between the two "outer numbers", that is, $$ \lvert 1/j - 1/k \rvert < 1/N - 0 = 1/N. $$

Answer 2

There, $0<1/j, 1/k < 1/N$ has the following meaning:

$0<1/j < 1/N$ and $0<1/k < 1/N$.

Answer 3

The way I would think about this fact is that that since $j,k\ge N$ we know that $\frac{1}{k},\frac{1}{j}\le\frac{1}{N}$ and so since we are looking at two numbers smaller than $\frac{1}{N}$ (and strictly bigger than 0), they cannot possibly be further apart than $\frac{1}{N}$; the distinction between $<$ and $\le$ in the proof seems to not matter (certainly the former implies the latter).