Example of inductive sets that are neither least nor greatest fixed point
Here's a non-trivial example:

Suppose we want to define inductively a subset of reals, so we work on the complete lattice $\mathcal{P}(\mathbb{R})$ ordered by inclusion.

Then, consider the rules
$$
\dfrac{\qquad}{0} \qquad \dfrac{x}{x+1}
$$
This induces the (monotonic, Scott-continuous) function $f : \mathcal{P}(\mathbb{R}) \to \mathcal{P}(\mathbb{R})$ given by
$$
f(X) = \{0\} \cup \{ x+1 \ |\ x\in X\}
$$

All of the following are fixed points of $f$:


$\mathbb{N}$ (least)
$\mathbb{Z}$
$\{x/2 \ |\ x\in \mathbb{Z}\}$
$\{x/3 \ |\ x\in \mathbb{Z}\}$
etc.
for any natural $k \geq 1$, the set $\{x/k \ |\ x\in \mathbb{Z} \}$
$\mathbb{Q}$
$\mathbb{R}$ (greatest)


If we want our definition to be well-formed, beyond specifying the rules, we need to single out one the fixed points. This is typically done by taking the least (induction) or the greatest (coinduction).