3
$\begingroup$

[EDIT: The axiom of successor cardinals was found by an answer by Greg Kirmayer, not to be capturing the intended meaning of it, which is simply reflected by its name, i.e. the existence of a successor cardinal for every cardinal. A corrective note had been inserted below that axiom.

Is Z + Rank + Successor cardinals + Ordinal inaccessibility = ZF?

Where:

Ranks: $\forall x \exists \alpha: x \in V_\alpha$

where $V_\alpha$ is the $\alpha^{th}$ stage of Von Neumann's universe (the cumulative hierarchy).

Successor cardinals: if $f$ is a definable function, then: $$\forall \text{ordinal } \alpha \ \exists \text{ ordinal }\beta \, (\neg \forall \gamma < \beta \exists \lambda < \alpha : f(\lambda)=\gamma)$$ By $\text{ordinal}$ it means a von Neumann ordinal defined in the usual manner.

For every ordinal there is an ordinal such that no surjective function (definable in the language of set theory) from the former to the latter can exist.

[NOTE]: The above formulation suffers a flaw, therefore the correct formulation is: $$\forall \text{ordinal } \alpha \, \exists \text{ordinal } \beta: \not \exists f (f: \beta \hookrightarrow \alpha)$$, where $\hookrightarrow$ signify "injective function".

Ordinal inaccessibility: if $f$ is a definable function, then: $$ \forall \gamma \, (\neg \forall \text{ ordinal } \alpha \,\exists \beta < \gamma : \alpha \leq f(\beta))$$ Any function [definable in the language of set theory] coming from an ordinal cannot have every ordinal being smaller than or equal to some ordinal in its range. That is, its range cannot be a cofinal subclass of the class of all ordinals.

The idea is that every uncountable cardinal in ZF is either Regular and therefore a successor cardinal, or otherwise a singular limit cardinal. Here, both of those kinds would be constructed from below, and the axiom of Rank assures that all sets are built successively within those ordinally indexed stages. Ordinal inaccessibility is I think equivalent to Ordinal Replacement which by itself is actually weak, it can only build stages up to $V_{\omega_1}$, and of course successor cardinals can only build the next stages, but together it seems that they can act to build up the whole of Von Neumann's universe. So, I thought that the above would prove full Replacement. Its easy to prove the result with Choice (in the form of every set is bijective to some von Neumann ordinal); but without it the proof is elluding me?

$\endgroup$
7
  • $\begingroup$ Can you specify exactly what you mean by ordinal in this theory? And I think there is a typo in the definition of ordinal inaccessibility; is $\lambda=\gamma$? $\endgroup$
    – Farmer S
    Apr 29 at 21:37
  • $\begingroup$ @FarmerS, thanks for spotting the typo. About ordinals, those are the usual von Neumann ordinals, i.e. transitive sets of transitive sets, that are $\in$-well founded. $\endgroup$ Apr 29 at 21:54
  • 1
    $\begingroup$ Do you mean "transitive set whose elements are strictly linearly ordered by $\in$"? $\endgroup$
    – Farmer S
    Apr 29 at 21:57
  • $\begingroup$ @FarmerS, I mean a transitive set whose elements are strictly well ordered by $\in$. An this is the usual official definition of von Neumann ordinals, and it is also equivalent to the one I gave in my prior comment. $\endgroup$ Apr 29 at 21:59
  • $\begingroup$ Hmm, under ZF (in particular, Foundation), all sets are $\in$-wellfounded, but not in general $\in$-wellordered. $\endgroup$
    – Farmer S
    Apr 29 at 22:05
7
$\begingroup$

This theory doesn't prove Replacement (assuming the consistency of an inaccessible, at least).

Assume ZFC + $\kappa$ is inaccessible and force over $V$ to add $\kappa$-many Cohen reals (i.e. the forcing is finite support product $\Pi_{\alpha<\kappa}\mathbb{C}_\alpha$ where each $\mathbb{C}_\alpha$ is just Cohen forcing). This forcing is ccc, so preserves all cardinals and cofinalities. So $\kappa$ is still a weakly inaccessible cardinal in $V[G]$. Now let $M=V_\kappa^{V[G]}$. Note that $M$ models the axioms you mention (in particular as $\kappa$ is regular in $V[G]$). But it does not satisfy Replacement: Working in $V[G]$, let $X$ be the set of all wellorders of $V_{\omega+1}$ of ordertype ${<\kappa}$. Then $X\in V_\kappa$. Consider the function sending $W\in X$ to the ordertype of $W$. This is definable over $V_\kappa^{V[G]}$ (actually without parameters), and is a surjection $X\to\kappa$ there.

Note that $M$ does satisfy Choice, formulated as "for every function there is a choice function". However, it does not satisfy the statement that "every wellorderable set is bijectable with an ordinal". (The set $X'$ of equivalence classes of wellorders in $X$, where they are equivalent if they have the same ordertype, is such a set.) Although ordinal inaccessibility holds in $M$, $X'$ is wellorderable in ordertype that of the class of ordinals.

Note that "ordinal" can also be defined as (i) the equivalence class of all wellordered sets of a given ordertype. Under ZF, this is equivalent to (ii) the von Neumann ordinals "transitive sets whose elements are strictly linearly ordered by $\in$" (which under ZF is equivalent to (ii') "transitive sets whose elements are well ordered by $\in$"). But in the model $M$, (i) and (ii) do not coincide (but (ii) and (ii') do).

$\endgroup$
3
  • $\begingroup$ so what we need to get replacement is to add to the above the axiom that every well ordered set is bijective to an ordinal. I think we won't need sucessor cardinals by then. $\endgroup$ Apr 30 at 5:45
  • $\begingroup$ I don't know what you mean by "coincide" in your last remark. But it appears to me that should I've used Scott's ordinals instead of von Neumanns in the formulation of the above axioms, then the resulting system would prove full replacement, since every well ordered set would have a Scott ordinal and even an initial segement of Scott ordinals that is order isomorphic to it, then well ordered replacement would follow and this would prove full Replacement over Z + Foundation. $\endgroup$ Apr 30 at 7:23
  • $\begingroup$ If every wellordered set is bijective with a von Neumann ordinal (hence in fact has the ordertype some von Neumann ordinal), then yes, I agree you get replacement. By "coincide" I just mean "are equivalent". If you use Scott's ordinals instead of von Neumann's, I agree you get replacement, but I think to first get well ordered replacement, it seems to require a small argument to observe that there is no Scott ordinal with ordertype that of the class of all von Neumann ordinals... $\endgroup$
    – Farmer S
    May 2 at 19:56
5
$\begingroup$

We are assuming that a formula πœ™(x,y) with 2 free variables defines a function means that for every x there is a unique y such that πœ™(x,y).

We assume that the axiom schema of Successor cardinals and the axiom schema of Ordinal inaccessibility are the result of replacing "f is a definable function" by "πœ™(x,y)

defines a function", replacing "𝑓(πœ†)=𝛾" by πœ™(πœ†,𝛾), and replacing 𝛼≀𝑓(𝛽) by "there is a b such that πœ™(𝛽,b) and 𝛼≀b".

We note that the axiom schema of Successor cardinals, is an immediate consequence of the axiom schema of Ordinal inaccessibility.(Suppose that πœ™(x,y) defines a function and c ia an ordinal.

By Ordinal inaccessibility, there is an ordinal 𝛼 with the property that βˆ€π›½<c:πœ™(𝛽,b)-->("b is not an ordinal" or b<𝛼). Then Β¬βˆ€π›Ύ<(𝛼+1)βˆƒπœ†<c:πœ™(πœ†,𝛾), since Β¬πœ™(πœ†,𝛼) for all πœ†<c.)

Z + Ranks + Ordinal inaccessibility has the same consequences as Z + Ranks + Ordinal Replacement.

Proof: Suppose Ordinal Replacement holds, πœ™(x,y) defines a function, 𝛾 is an ordinal, and for every ordinal 𝛼 (βˆƒπ›½<π›Ύβˆƒb:𝛼≀bβˆ§πœ™(𝛽,b)). Let πœ“(x,y) be the formula ("x is an ordinal" and "y is an ordinal" and πœ™(x,y)) or ("x is an ordinal" and (βˆ€t(πœ™(x,y)-->"t is not an ordinal") and y=0) or ("x is not an ordinal" and y=x). By Ordinal Replacement βˆƒπ΅βˆ€π‘¦(π‘¦βˆˆπ΅β†”βˆƒπ‘₯βˆˆπ›Ύπœ“(π‘₯,𝑦)). But then every ordinal is in UB. Now suppose Ordinal inaccessibility holds, πœ™(x,y) is a formula in two free variables x,y, βˆ€π‘₯[π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘™(π‘₯)β†’βˆƒ!𝑦(π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘™(𝑦)βˆ§πœ™(π‘₯,𝑦))], and A is a set of ordinals. Let πœ“(x,y) be the formula ("x is an ordinal" and "y is an ordinal" and πœ™(x,y)) or ("x is not an ordinal" and x=y). Then πœ“(x,y) defines a function. By Ordinal inaccessibility, there is an ordinal 𝛼 with the property βˆ€π›½<(UA)βˆ€b(πœ“(𝛽,b)-->b<𝛼). Then there is a set B such that

   b∈B<-->bβˆˆπ›Όβˆ§βˆƒπ‘₯∈𝐴(πœ™(x,b)). Therefore Ordinal Replacement holds.

If ZF is consistent then Z + Ranks + Ordinal inaccessibility + Choice does not prove Replacement.

Proof: See the answer of Joel David Hamkins to "Is full Replacement provable in Z + Ordinal Replacement?". (His argument shows that in a model of ZFC, βŸ¨π‘‰πœ”1,∈⟩ satisfies Z + Ranks + Ordinal Replacement + Choice but not all instancess of Replacement.)

$\endgroup$
2
  • $\begingroup$ the form of axiom of choice I meant is that every set is bijective to some ordinal. In that form, it'll prove the result. I'll edit it to that effect. $\endgroup$ May 1 at 7:48
  • $\begingroup$ Thanks for this answer, I see that the ordinal $\beta$ occurs "after" each function, while it should be prior to all of them, which cannot be done easily. I've corrected the answer to the conventional way of expressing it. $\endgroup$ May 1 at 9:11

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.