Logical inductors in multistable situations.

I was reading about logical induction at

https://intelligence.org/files/LogicalInduction.pdf

and understand how it resolves paradoxical self reference, but I’m not sure what the inductor will do in situations where multiple stable solutions exist.

Let $f : [0, 1] \to [0, 1]$

If $f$ is continuous then it must have a fixed point. Even if it has finitely many discontinuities, it must have an “almost fixed” point. An $x$ such that $\forall ϵ > 0 : i n f_{y \in (x - ϵ, x)} f (y) \leq x \leq {sup}_{y \in (x, x + ϵ)} f (y)$

However some $f$ have multiple such points.

f (x) = {\begin{matrix} 0 & x < \frac{1}{2} 1 & x \geq \frac{1}{2} \end{matrix}

Has “almost fixed” points at $0$ , $\frac{1}{2}$ and $1$ .

A similar continuous $f$ is

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 0 & x \leq \frac{1}{3} 3 x - 1 & \frac{1}{3} \leq x \leq \frac{2}{3} 1 & x \geq \frac{2}{3} \end{matrix}

With

f (x) = x

Having every point fixed.

Consider $ϕ_{n} =" f (E_{n} (ϕ_{n})) "$

These functions make $ϕ_{n}$ the logical inductor version of “this statement is true”. Multiple values can be consistently applied to this logically uncertain variable. None of the possible values allow a money pump, so the technique of showing that some behaviour would make the market exploitable that is used repeatedly in the paper don’t work here.

Is the value of $E_{n} (ϕ_{n})$ uniquely defined or does it depend on the implementation details of the logical inductor? Does it tend to a limit as $n \to \infty$ ? Is there a sense in which

f (x) = {\begin{matrix} 0.83 & 0.82 \leq x \leq 0.84 0.1 & e l s e \end{matrix}

causes $E_{n} (ϕ_{n})$ has a stronger attractor to $0.1$ than it does to $0.83$ ?

Can $E_{n} (ϕ_{n})$ be 0.6 where

f (x) = {\begin{matrix} 0.6 & x = 0.6 0.1 & e l s e \end{matrix}

because the smallest variation would force it to be $0.1$ ?

[Question] Logical inductors in multistable situations.