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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?

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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).

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  • $\begingroup$ 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. $\endgroup$ – Derek Elkins Mar 22 '17 at 3:35
  • $\begingroup$ @DerekElkins Agreed. I added a note about that. $\endgroup$ – chi Mar 22 '17 at 9:04
  • $\begingroup$ @DerekElkins The answer doesn't claim that it defines anything. It says to suppose that we want to define something, gives some rules and says that those rules have a whole bunch of fixed points. $\endgroup$ – David Richerby Mar 22 '17 at 9:06
  • $\begingroup$ @DavidRicherby Indeed, that was my intention. Still, I can see that it is possible to read a bit beyond what I wrote, so I added a final note anyway. $\endgroup$ – chi Mar 22 '17 at 9:11
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    $\begingroup$ @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. $\endgroup$ – chi Mar 22 '17 at 16:11
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Any set is a fixed point of the empty set of rules or of the trivial rule $x\in X\Rightarrow x\in X$.

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