Consider any $\delta \gt c \gt 0$. Prove that there is a $k \in \mathbb N$ and $z \in \mathbb N$ such that $z\cdot \frac{1}{k} \in [c, \delta)$

Solution 1:

I like your proof, but I think this is a shorter proof in the same vein:

$z \cdot \frac{1}{k} \in [c, \delta)$ is equivalent to $c \leq \frac{z}{k} < \delta,$ or in turn $ck \leq z < \delta k.$ As you note, by the Archimedean property we have that there exists some natural $k$ such that $\frac{1}{k} < \delta - c,$ so $1 < \delta k - ck$ and $ck + 1 < \delta k.$

Now let $z$ be the smallest integer greater than or equal to $ck.$ We must have that $z < ck + 1,$ because otherwise if $z \geq ck + 1$ then $z - 1 \geq ck,$ and because $z - 1$ is an integer if $z$ is then $z$ is not the smallest integer greater than $ck,$ causing a contradiction. So, $ck \leq z < ck + 1 < \delta k.$

The key elements here are that we can always make $[ck, \delta k)$ have a length of at least $1,$ and that in any interval of length at least $1$ there must be an integer.