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Q&A Defining Bayes’s theorem from scratch in ZFC

1 answer  ·  posted 3mo ago by Julius H.‭  ·  last activity 1mo ago by xamidi‭

#1: Initial revision by user avatar Julius H.‭ · 2024-02-09T22:58:20Z (3 months ago)
Defining Bayes’s theorem from scratch in ZFC
I have tons of interrelated questions which I would like resolved in order to help me answer this [Philosophy SE][1] question about Bayes’s theorem given a certain probability of 0.

I’ll assume I am working in ZFC, which I have tons of questions about. In fact, I am still seeking even a basic understanding of how ZFC works.

ZFC can be formulated as a collection of 9 formation rules written in the language of first order logic. I read somewhere that somehow, FOL implies the existence of at least one “thing”, so you do not need to assume the existence of the empty set as an axiom, for ZFC.

The important thing about first order logic is that it has quantifiers.

The formation rules of ZFC can be written using FOL, as (roughly):

1. For all z (in some universe), if z is in x, and z is in y, then “x = y”. (I am pretty sure that, since FOL includes “relations” as a syntactic feature, “set membership” is actually nothing more than some arbitrary “relation” (and so is “equality”, and so is the implication arrow.) Thus, what we have here is an interaction between three relations?: for all z in some universe, if xR1z R2 yR1z, then xR3y. And this is true for all x and y in this universe, so we also use the universal quantifier over them.)

I’m going to skip ahead since there’s so much to write and learn. The other axioms include well-foundedness (roughly, sets are disjoint with their elements), the axiom (“schema”) of restricted comprehension (that for any formula of FOL, the set of all elements of some set S meeting that formula exists), the axiom of pairing (or, “the ability to put things together”), the capability to take unions and power sets, the existence of any sets under any definable function, and actually I would like to reject the existence of an infinite set if possible, and I read that therefore I do not need the axiom of choice.

A more intuitive way for me to summarize this is basically, “preconditions”: equality and regularity; and ways to generate new sets: power sets, unions, pairings, and “logical conditions” (comprehension). (I would like to know why pairing could not be accommodated by unions, and why the function axiom (“replacement”) is needed, since I thought to specify some function, it would require the existence of the set to begin with.)


Basically, my wish was then to define “probability” and “conditional probability” and then prove Bayes’s theorem. Perhaps someone can assist me with that. Thank you.



I know this question is full of different questions so I will probably break it into individual sub-questions as new posts.




  [1]: https://philosophy.stackexchange.com/a/108474/56485