I guess this sort of “definability does not imply existence” (such as zero-sharp) thing is not uncommon? A simple example is the cardinal c that is immediately after aleph-null but below the continuum. Definable but does not exist in universes that satisfy CH (I wrote universes rather than models with a clear nod to JDH). This leads me to a question. Penelope Maddy in her “Believing the Axioms I” (pp 500) writes about constructing the iterative hierarchy from the ground up. By stage omega+ 2, we have the set of reals and we have a well-ordering of type aleph-one. The question is whether or not a one-to-one correspondence between them is included at the next stage, since it is consistent to do so (my paraphrase). If we do, we get CH, else not (so my cardinal c exists). So, in a sense, that’s the stage where we take a call on CH. In the same way, is it possible to pin down the stage where one has to take a call on whether zero-sharp exists?
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2The question is unreadable. – Asaf Karagila Jan 02 '20 at 11:07
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If you are asking "where does $0^#$ enter the universe?" the answer is that $0^#$ is coded as a real number. So if it exists, it enters the universe very fast. Namely, before the decision about CH needs to be made. – Asaf Karagila Jan 02 '20 at 11:09
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1Not only where the real number enter, but where does it get "qualified" as one with the right properties to make it zero-sharp? – Mangesh Patwardhan Jan 02 '20 at 11:13
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What are these properties? It is a unique real number. It is very definable with very specific properties. – Asaf Karagila Jan 02 '20 at 11:46
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2https://math.stackexchange.com/questions/1888063/why-is-0-sharp-not-definable-in-zfc As JDH points out here, we may have that real in our universe, but in some universes, it does not give us the theory of order-indiscernibles for L. So if zero-sharp exists, it has to be this real, but having this real is not enough. – Mangesh Patwardhan Jan 02 '20 at 12:33
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Again, this is the question of how you're expressing $0^#$. If you define it as a mouse, then this mouse is a real (since it's a countable transitive set and can be coded as a real number). The definition given by Neil in his question, the real is a unique real in any given model of ZFC. And in order to prove that this is $0^#$ and no other, you also need to argue, in some way, that this collection of Gödel codes satisfies some properties, yes. What you seem to be asking is what are these properties, and how much "information" does one need in order to prove these properties hold. – Asaf Karagila Jan 02 '20 at 13:34
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And the point, again, is that there are many ways to formulate $0^#$, and the modern one, using mice, simply requires you to have a countable set, coded by a real number $r$, that can be iterated $\omega_1$ times while still being well-founded. So in order to prove that this you only need to know that for every countable ordinal $\alpha$—which also can be coded by a real—you can iterate along $\alpha$. This means that if you know the real numbers, you know enough to decide whether or not a certain real number is a code for $0^#$. – Asaf Karagila Jan 02 '20 at 13:38
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1@Asaf But you are claiming more in your last paragraph! You are saying you only need countable iterations rather than the $\omega_1$st. – Andrés E. Caicedo Jan 02 '20 at 15:55
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@Andrés: I distinctly recall that $\omega_1$-iterability is equivalent to "$\forall\alpha<\omega_1$, the mouse is $\alpha$-iterable". So this is still within the domain of the real numbers. – Asaf Karagila Jan 02 '20 at 15:57
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1@Asaf. Yes! But none of this is trivial. I am just saying that in the paragraph you assumed this key equivalence without comment. – Andrés E. Caicedo Jan 02 '20 at 16:00
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@Andrés: Yes, that's true. I did assume the equivalence without a comment. :-) – Asaf Karagila Jan 02 '20 at 16:01
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1This is a good question. The thing is that you only need countable linear iterations and well-foundedness of certain countable structures to deal with $0^\sharp=M_0^\sharp$. The analogous object $M_1^\sharp$ (informally, the sharp for an inner model with one Woodin cardinal) is more complicated to pin down in your sense. – Andrés E. Caicedo Jan 03 '20 at 08:13
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Thanks. I picked up the concept of 0-sharp from Jech Chs. 17 & 18. Yet to get to mouse stuff. I'll study that. In case of my definable cardinal c, once we get reals and well-ordering on aleph-1, we decide at the very next stage whether to include a 1-1 mapping between them. If we decide to leave it out, my cardinal c exists in the universe, else not. Similarly, it is possible to say if we take a specific decision to include such and such set at a particular stage, we'll have 0-sharp, else not? So I meant pin down in a more specific sense (at least that's how I see it). – Mangesh Patwardhan Jan 03 '20 at 09:52
1 Answers
The $V$-hierarchy is quite coarse: very early on (e.g. at level $\omega+k$ for very small $k$) we can see the answers to such questions.
For getting optimal results along these lines, one very useful fact is the existence of flat pairing functions: these are pairing functions $\langle\cdot,\cdot\rangle$ such that whenever $\alpha$ is infinite and $x,y\in V_\alpha$ then $\langle x,y\rangle\in V_\alpha$. Using a flat pairing function we get for example that every binary relation on $V_\alpha$ for $\alpha$ infinite is an element of $V_{\alpha+1}$ (which is clearly the best we could hope for).
This lets us argue as follows:
There is a canonical partition $P$ of $V_{\omega+1}$ into $\omega_1$-many pieces. (That is, $P$ is definable and ZFC proves that it has the above property.)
CH is equivalent to the statement "There is a relation $R\subseteq (V_{\omega+1})^2$ such that $(i)$ for each $x\in V_{\omega+1}$ the set $\{y: R(x,y)\}$ is a $P$-class and $(ii)$ if $x_1\not=x_2$ and $R(x_1,y_1)$ and $R(x_2,y_2)$ hold then $y_1$ and $y_2$ lie in different $P$-classes.
Such $P$ and $R$ - if they exist - live in $V_{\omega+2}$, and moreover their behavior is "verifiable in $V_{\omega+2}$."
Precisely, we've shown:
If $M,N\models ZFC$ with $(V_{\omega+2})^M=(V_{\omega+2})^N$ then we have $M\models CH$ iff $N\models CH$.
(Note that when I say "$(V_{\omega+2})^M=(V_{\omega+2})^N$" I'm requiring agreement of the elementhood relation as well.)
Conversely, this is optimal:
There are models of ZFC with the same $V_{\omega+1}$ which differ about CH.
Namely, if $M\models \neg CH$ then there is a forcing extension $N$ of $M$ with the same reals which satisfies $CH$.
What about $0^\sharp$?
This turns out to be easier: one of the definitions of $0^\sharp$ is projective (indeed, $\Delta^1_3$ - meanwhile $\{0^\sharp\}$ is a $\Pi^1_2$ singleton), and so we know whether it exists as soon as we know what reals we have. Thus, "$0^\sharp$ exists" is decided at level $\omega+1$:
If $M,N\models ZFC$ with $(V_{\omega+1})^M=(V_{\omega+1})^N$, then $M$ and $N$ agree on whether $0^\sharp$ exists.
Note that this demonstrates that there's really no connection between the level at which a principle is decided and the consistency strength of that principle.
OK, what if we replace the $V$-hierarchy with something finer?
One issue here is that finer hierarchies don't (generally) provably exhaust the universe. The [$L$-hierarchy is pretty fine (and we can go even finer in various ways), but $V=L$ (or even $(V_{\omega+1})^L=V_{\omega+1}$) is undecidable in ZFC. So the finer approach only works if we add additional "structurally limiting" axioms to ZFC.
But those axioms tend to be so strong that they decide outright the principles we're interested in.
This leads us unfortunately to:
As far as I can tell, there's no natural version of this question which yields "fine distinctions:" the $V$-hierarchy makes everything happen "as soon as possible," while finer hierarchies tend to require structural limitations which seem to decide all natural principles outright.
EDIT: As Andres says below, that's hiding quite a lot to the point of being at least ethically incorrect. I'll rewrite that part when I have a chance, but the relevant material is sufficiently outside my comfort zone that I can't do that immediately.
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I don't know about the last paragraph. It seems to hide too much. The natural thing to look at would be something like $K^c$, and then $0^\sharp$ does not appear until we reach a stage $\alpha$ such that there is an appropriate measure$U$ to be added to $L_\alpha$ so that the premouse $L_\alpha[U]$ satisfies appropriate requirements. If no such stage is reached, $0^\sharp$ does not exist. If $M\subseteq V$ agree on the ordinals and $0^\sharp\in M$, then $\alpha$, as computed in $M$, should coincide with its computation in $V$. – Andrés E. Caicedo Jan 10 '20 at 19:53
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@AndrésE.Caicedo That's a very good point, and I'm not sufficiently familiar with that material to respond immediately; I've added a caveat to my answer and will change it appropriately when I get the chance. – Noah Schweber Jan 11 '20 at 00:32