Let’s back up here and clarify definitions before invoking any theorems. In the language of set theory, we have a countably infinite set of finite statements. Some statements imply other statements. A subset A of these statements is said to be consistent if they can all be assigned to true such that, when following the basic rules of logic, one does not arrive at a contradiction.
The compactness theorem is helpful when A is an infinite set.ZFC is a finite set of axioms, so let’s ignore everything about finite subsets of A and the compactness theorem; it’s not relevant. [Edit: as indicated by Amalthea’s reaction, this is wrong; some “axioms” in ZF are actually infinite sets of axioms, such as replacement]
I’ll now rewrite your last sentence as:
ZFC + not Consistent(ZFC) has no model <-> not Consistent(ZFC + not Consistent(ZFC))
This is true but irrelevant. Assuming ZFC is consistent, ZFC will not be able to prove its own consistency so [not Consistent(ZFC)] can be added as an axiom without affecting its consistency. This means that ZFC + [not Consistent(ZFC)] would indeed have a model; I forget how this goes but I think it’s something like “start with a model of ZFC, throw in a c that’s treated as a natural number and corresponds to the contradiction found in ZFC, then close”. I think c is automatically treated as greater than every “actual” natural number (and the way to show that this can be added without issue (I think) involves the compactness theorem).
Okay, I kinda understood where I am wrong spiritually-intuitively, but now I don’t understand where I’m wrong formally. Like which inference in chain
not Consistent(ZFC) → some subsets of ZFC don’t have a model → some subsets of ZFC + not Consistent(ZFC) don’t have a model → not Consistent(ZFC + not Consistent(ZFC))
It’s completely valid. And we can simplify it further to:
not Consistent(ZFC) → not Consistent(ZFC + not Consistent(ZFC))
because if a set of axioms is already inconsistent, then it’s inconsistent with anything added. But you still won’t be able to actually derive a contradiction from this.
Let’s back up here and clarify definitions before invoking any theorems. In the language of set theory, we have a countably infinite set of finite statements. Some statements imply other statements. A subset A of these statements is said to be consistent if they can all be assigned to true such that, when following the basic rules of logic, one does not arrive at a contradiction.
The compactness theorem is helpful when A is an infinite set.ZFC is a finite set of axioms, so let’s ignore everything about finite subsets of A and the compactness theorem; it’s not relevant. [Edit: as indicated by Amalthea’s reaction, this is wrong; some “axioms” in ZF are actually infinite sets of axioms, such as replacement]
I’ll now rewrite your last sentence as:
ZFC + not Consistent(ZFC) has no model <-> not Consistent(ZFC + not Consistent(ZFC))
This is true but irrelevant. Assuming ZFC is consistent, ZFC will not be able to prove its own consistency so [not Consistent(ZFC)] can be added as an axiom without affecting its consistency. This means that ZFC + [not Consistent(ZFC)] would indeed have a model; I forget how this goes but I think it’s something like “start with a model of ZFC, throw in a c that’s treated as a natural number and corresponds to the contradiction found in ZFC, then close”. I think c is automatically treated as greater than every “actual” natural number (and the way to show that this can be added without issue (I think) involves the compactness theorem).
Okay, I kinda understood where I am wrong spiritually-intuitively, but now I don’t understand where I’m wrong formally. Like which inference in chain
not Consistent(ZFC) → some subsets of ZFC don’t have a model → some subsets of ZFC + not Consistent(ZFC) don’t have a model → not Consistent(ZFC + not Consistent(ZFC))
is actually invalid?
It’s completely valid. And we can simplify it further to:
not Consistent(ZFC) → not Consistent(ZFC + not Consistent(ZFC))
because if a set of axioms is already inconsistent, then it’s inconsistent with anything added. But you still won’t be able to actually derive a contradiction from this.
Edit: I think the right thing to do here is look at models for PA + not consistent(PA). I can’t find a nice treatment of this at the moment, but here’s a possibly wrong one by someone who was learning the subject at the time: https://angyansheng.github.io/blog/a-theory-that-proves-its-own-inconsistency