Comment by FjolleJagt on How do finite factored sets compare with phase space? · 2022-12-06T21:05:53.150Z · LW · GW

A finite factored set is "just" a set $X$ with a specific choice of decomposition as a product of sets $X_1\times \dots \times X_n$. I'm not sure what definition of phase space you're using, but for a sufficiently general definition of dynamical system (e.g. https://en.wikipedia.org/wiki/Dynamical_system#Formal_definition) I don't think that the phase space necessarily has coordinates in this way. The position / momentum phase space example is a special case, where your phase space happens to look like a product of copies of the real numbers, which is then getting back to the "factored" part of a finite factored set. I'm not convinced that there's a deep connection here, though am not very familiar with either concept so could easily be missing something here!

Comment by FjolleJagt on Finite Factored Sets · 2021-05-28T11:02:16.240Z · LW · GW

Thanks, that makes sense! Could you say a little about why the weak union axiom holds? I've been struggling to prove that from your definitions. I was hoping that $h^F(X|z,w)\subseteq h^F\left(X|z\right)$ would hold, but I don't think that $h^F\left(X|z\right)$ satisfies the second condition in the definition of conditional history for $h^F(X|z,w)$.

Comment by FjolleJagt on Finite Factored Sets · 2021-05-26T12:04:39.616Z · LW · GW

I'm confused by the definition of conditional history, because it doesn't seem to be a generalisation of history. I would expect $h^F\left(X|\varnothing \right)=h^F\left(X\right)$, but both of the conditions in the definition of $h^F\left(X|\varnothing \right)$ are vacuously true if $E=\varnothing$. This is independent of what $H$ is, so $h^F\left(X|\varnothing \right)=\varnothing$. Am I missing something?