-1
$\begingroup$

[EDIT, this posting had been answered to the negative, However it couldn't be deleted, so I've written a salvage for it in the posting titled "What is the consistency strength of Ackermann + the following cardinals to ordinals isomorphism?"]

Working in Morse-Kelley set theory

Add to it the following schema:

Cardinals to Ordinals isomprhism: if $\psi(Y)$ is a formula in which only the symbol $``Y"$ appear free [and only free], and the symbol $``x"$ never occurring, and $\psi(x)$ is the formula obtained from $\psi(Y)$ by merely replacing each occurrence of the symbol $``Y"$ by an occurrence of the symbol $``x"$, then:

$\forall x (x \text{ is an ordinal }\to \exists Y(\psi(Y) \wedge Y \text{ is a cardinal } \wedge x < Y)) \\ \to \{x | x \text{ is a cardinal } \wedge \psi(x) \} \text{ is order isomorphic to } \{x | x \text{ is an ordinal }\} $

is an axiom.

Note: just to avoid confusion: the terms "cardinal" and "ordinal" here are defined in the standard von Neumann definitions, but they range over all classes and not just sets. So an ordinal class is a transitive set of transitive sets, and a cardinal class is an ordinal class is the smallest of all ordinal classes that are bijective to it.

Questions:

  1. Is there a clear inconsistency with this principle?

  2. if this is consistent, then what's the consistency strength of the above theory?

$\endgroup$
  • $\begingroup$ Here is a $\psi$ for you to consider: $\psi(Y):=$"$Y=$ORD". $\endgroup$ – Andrés E. Caicedo Dec 10 '18 at 13:14
  • $\begingroup$ OK you proved this to be inconsistent, this is a correct answer. $\endgroup$ – Zuhair Dec 10 '18 at 13:17
9
$\begingroup$

You should really stop writing soups of symbols and just describe in words what you are doing. It saves a coding headache on your side and a decoding headache on the reader's. Anyway, the formula seems to say that if for every ordinal there is a larger cardinal with property $\psi$, then there is a bijection between the class of ordinals and that of cardinals with property $\psi$. This is trivial under the axiom of choice: any proper class of ordinals is order isomorphic in a unique way to the class of all ordinals. Simply index your class in increasing order, and that is your bijection. There is no difficulty in formalizing this in ZF (if "cardinal" is understood as initial ordinal) or in ZFC.

As written, however, this is inconsistent in MK: Simply take $\psi$ be the formula saying that $Y=$ORD. I think that really what you are after is the formulation of a kind of strong reflection principle. Why don't you ask another question about that? Be explicit about the examples you have in mind, and avoid writing formulas as much as possible.

$\endgroup$
  • 1
    $\begingroup$ Not under the standard use of the terms. I agree that if you use standard words in funny ways, surely there will be counterexamples. $\endgroup$ – Andrés E. Caicedo Dec 10 '18 at 13:00
  • 3
    $\begingroup$ +1 for the “soups of symbols dig”. That’s also a pet peeve of mine. $\endgroup$ – MPW Dec 10 '18 at 13:01
  • 3
    $\begingroup$ I use the word "cardinal" in its standard sense. If you mean something else, be explicit in the body of a new question. $\endgroup$ – Andrés E. Caicedo Dec 10 '18 at 13:04
  • 3
    $\begingroup$ (By the way, it may be reasonable in your end to assume that working set theorists use standard terms in standard ways, and are familiar with standard terms, such as ZF or cardinal.) $\endgroup$ – Andrés E. Caicedo Dec 10 '18 at 13:06
  • 2
    $\begingroup$ OK, Zuhair, I am done here. $\endgroup$ – Andrés E. Caicedo Dec 10 '18 at 13:06

Your Answer

By clicking “Post 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.