# Example of inductive sets that are neither least nor greatest fixed point

Do there exist a set of inductive rules and a fixed point of these rules but is neither the least nor the greatest fixed points?

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).

• A minor nit, but without specifying least/greatest or some other property to uniquely pick out the appropriate subset, this doesn't inductively define a subset of the reals. Mar 22 '17 at 3:35
• @thbl2012 The greatest fixed point is very sensitive to the choice of the complete lattice you work on. Here, I started with $\mathbb{R}$ as the top element of my lattice, but I could have chosen e.g. $\mathbb{Q}$ or $\mathbb{C}$. Another common choice it the set of finite or infinite symbolic applications of the ocnstructors, where, as you say, you do have the largest fixed point being $\mathbb{N} \cup \{\infty\}$ where $\infty$ here represents the successor applied on itself infinitely many times. Hence, your professor is completely correct, (s)he simply chose another complete lattice.
Any set is a fixed point of the empty set of rules or of the trivial rule $x\in X\Rightarrow x\in X$.