Certainly ZFC $\vdash$ Con(ZF) $\rightarrow$ Con(ZF+$\lnot$AC), but the usual forcing argument to construct a model of ZF+$\lnot$AC seems to require choice to find a generic filter.
Asked
Active
Viewed 175 times
2
-
2Actually, I believe PRA proves $\text{Con}(\text{ZF}) \to \text{Con}(\text{ZF} + \lnot \text{AC})$. There are two key points. First, that the forcing arguments can be turned into syntactic arguments, and second, that we don't need an actual generic filter, just one that is generic for a certain collection of dense sets. I believe Kunen discusses this at some length, IIRC. I will let an expert write a proper answer. – Carl Mummert Jan 01 '16 at 22:48
-
2Con(ZF) → Con(ZF+¬AC) is an arithmetic statement, and every arithmetic statement provable in ZFC is provable in ZF. – Wojowu Jan 01 '16 at 22:52
-
1Related: http://math.stackexchange.com/questions/378030/does-forcing-need-a-countable-transitive-model – Carl Mummert Jan 01 '16 at 22:53
2 Answers
3
This has a simple, almost cop-out, solution: absoluteness.
Note that the statement is in fact a first-order statement in the language of arithmetic. And such statement is provable from $\sf ZF$ if and only if it is provable from $\sf ZFC$.
Since you know the latter, you actually know the former.
Although in reality, you can just verify the following in $\sf ZF$:
The completeness theorem holds for countable languages without choice. So if $\sf ZFC$ is consistent, it has a model.
Forcing and symmetric extensions work without using the axiom of choice.
So if $\sf ZFC$ is consistent, it has a countable model, where forcing works as usual, symmetric extensions work as usual, and therefore if there is a model of $\sf ZF$ there is one where the axiom of choice fails.
It should also be added that the axiom of choice is not really needed in order to find a generic filter over a countable transitive model. The reason is that the model is in fact countable, so the elements from the forcing poset inside the model make a countable set externally. Using this fact, as well the fact that there are only countably many dense set in the model we can define a generic filter recursively without appealing to the axiom of choice.
Asaf Karagila
- 393,674
-
Dont you need dependent choice, even if everything is countable ? – Rene Schipperus Jan 01 '16 at 23:09
-
1Why would you? Fix an enumeration, and use recursion over that enumeration! – Asaf Karagila Jan 01 '16 at 23:13
-
So you take a well ordering of $\mathbb{P}\cap M$. And at every stage chose the smallest element in the next dense set. – Rene Schipperus Jan 01 '16 at 23:15
-
Yes. Even better, take a well-ordering of $M$, and at each step take the next element in the next dense set, using the same enumeration. – Asaf Karagila Jan 01 '16 at 23:16
-
Is sounds ok to me but isnt DC needed to show Konig infinity lemma ? – Rene Schipperus Jan 01 '16 at 23:19
-
Not exactly. In the general case, Koenig's lemma requires more or less choice from countable families of finite sets; in here we appeal to a bit more (the levels of the tree are generally infinite, which would give us the full power of DC in the general case). However, you can show that if the tree is well-orderable, then exactly no amount of choice is needed. Each step just pick the least point in a fixed enumeration. It's just the recursion theorem, nothing more. – Asaf Karagila Jan 01 '16 at 23:20
-
In KIL the levels are finite. So the tree is countable, but I often see DC mentioned as hypothesis. – Rene Schipperus Jan 01 '16 at 23:24
-
Sure, just makes things easier. DC is the go-to when you want to define something recursively. But again, if the underlying set is countable, or generally well-orderable, then no choice is needed. – Asaf Karagila Jan 01 '16 at 23:24
-
I think DC is needed to show the tree is countable. In any case I agree about the filters, its just the analogy with KIL that gave me hesitation. – Rene Schipperus Jan 01 '16 at 23:32
-
You only need countable choice to show that a tree of height $\omega$ with finite levels is countable (in fact, even less than countable choice!), but it is true that you cannot prove in ZF that every such tree is countable. – Asaf Karagila Jan 01 '16 at 23:33
-
We also need to ensure the existence of a countable model given a possibly-uncountable model, and the downward lowenheim-skolem theorem generally requires choice. Let us work in some universe $V$ of ZF, which contains a model $M$ of $ZF$. I see that if we find the inner model of $M$ in which $V=L$, then that model $N$ has a class-bijection (set-bijection in $V$) from $N$ to $ORD^N$, so has well-ordering... IF $N$ is a standard model. But what if $N$ is a nonstandard model of set theory? How do we, in $V$, then apply downward lowenheim-skolem to find a countable model over which to force? – vhspdfg Jan 01 '16 at 23:36
-
Well I'd be happy to see and exact determination of the strength of KIL. – Rene Schipperus Jan 01 '16 at 23:36
-
Lowenheim-Skolem, in general, is equivalent to DC. You're right about that. But the proof of the completeness theorem gives you a countable model, if I remember correctly. – Asaf Karagila Jan 01 '16 at 23:37
-
As for the Koenig's lemma, it should be countable choice for finite sets. In the one direction, assuming Koenig's lemma, and given such family of sets, we can define a tree whose branch is a choice function; in the other direction countable choice for finite sets give you that countable union of finite sets is countable, so such tree has a branch. – Asaf Karagila Jan 01 '16 at 23:38
-
-
-
1@vhspdfg: It's not an entirely trivial question. Don't feel bad that the answers are short. – Asaf Karagila Jan 01 '16 at 23:44
-
-
2
Even setting aside the question of filters. In ZF you are constructing a forcing language, with truth values in a Boolian algebra, and all the axiom of ZF+$\neg$AC have non zero truth valuations. This suffices for consistence.
Rene Schipperus
- 39,526