Consider any $\delta \gt c \gt 0$. Prove that there is a $k \in \mathbb N$ and $z \in \mathbb N$ such that $z\cdot \frac{1}{k} \in [c, \delta)$ I needed this lemma for an exercise and wondered if there were more traditional ways of proving it.

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


By the Archimedean Property, there is a $k \in \mathbb N$ such that $\frac{1}{k} \lt |c - \delta|$.
There are three possibilities:
Case 1) $\frac{1}{k} = c$
Case 2) $\frac{1}{k} \gt c$
Case 3) $\frac{1}{k} \lt c$
In Case 1, simply let $z=1$.
In Case 2, recalling that $\frac{1}{k} \lt |c-\delta|=\delta -c$, we have that $\frac{1}{k}+c \lt \delta$. Noting that $c \gt 0$, we then have: $\frac{1}{k} \lt \frac{1}{k}+c \lt \delta$, which implies that $\frac{1}{k} \lt \delta$. Combining this with $\frac{1}{k}\gt c$, it is clear that $\frac{1}{k} \in (c,\delta)$. We can therefore let $z=1$.
In Case 3, we first note that, because $\mathbb N$ is an unbounded set, there is a $z'$ such that $z' \geq ck$. Rearranging, we have that $\frac{1}{k}\cdot z' \geq c$. If $\frac{1}{k}\cdot z'=c$, we are done. So let $\frac{1}{k}\cdot z' \gt c$.
Now, consider the set $S=\left\{z \in \mathbb N: \frac{1}{k}\cdot z \gt c\right\}$. By assumption, we know $\frac{1}{k}\cdot z' \gt c$, so clearly $z' \in S$, which means $S \neq \emptyset$. Therefore, by the Well-Ordering Principle, there is a minimum element $z^* \in S$. Importantly, because $\frac{1}{k} \lt c$, we must have $z^* \geq 2$. Consider the natural number $(z^*-1)$, which we know exists because $(z^*-1) \geq 1$. Clearly, $(z^*-1) \notin S$  because $z^*-1 \lt z^*$ and $z^*$ is the minimum element of $S$. This means that $\frac{1}{k}\cdot (z^*-1) \leq c$. If equal, we are done. So suppose $\frac{1}{k}\cdot (z^*-1) \lt c \quad (*)$.
Recalling that $\frac{1}{k} \lt \delta -c$, we have that $c+\frac{1}{k} \lt \delta$. In combination with $(*)$, we then have:
$$\frac{1}{k}\cdot(z^*-1)+\frac{1}{k} \lt c+\frac{1}{k} \lt \delta$$
Noting that $\frac{1}{k}(z^*-1)+ \frac{1}{k} = \frac{1}{k} \cdot z^*$, we must have $\frac{1}{k}\cdot z^*\lt \delta$. Because $z^* \in S$, we conclude with $c \lt \frac{1}{k} \cdot z^* \lt \delta \iff \frac{1}{k} \cdot z^* \in (c,\delta)$.
 A: 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.
A: Let $\ c,\ \delta\ $ be such that $ \delta > c > 0.$
By the Archimedean Property, there is a $k \in \mathbb N$ such that $\ 0<\frac{1}{k} \lt \delta-c$.
For any $\ x>0,\ $ there is always a maximum non-negative integer $\ m\ $ such that $\ m<x.\quad (*) $
Since $\ ck>0,\ $ there exists a maximum non-negative integer $\ m\ $ such that $\ m < ck,\implies\frac{m}{k} < c.\ $ Then $\ \frac{m+1}{k}\geq c\ $ and  $\ \frac{m}{k} + \frac{1}{k} < c + \frac{1}{k} < c + \delta - c = \delta,\ $ i.e., $\ c\leq\frac{m+1}{k}<\delta,\ $ and the result is proven with $\ z= m+1.$
$\ (*)\ $ is true because $\ 0<x,\ $ and so if $\ (*)\ $ weren't true, then $\ x\ $ is greater than every non-negative integer, and no real number has this property.
