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1-In his article written in German "Über unerreichbare Kardinalzahlen" (On inaccessible cardinals), inside Fund. Math. 1938 (pages 68-89), Alfred Tarski states his axioms A and A' as follows. Axiom A: "For every set x, there exists a set y satisfying the four following conditions:

  • A1: x is a member element of y;
  • A2: If z is a member element of y and t is included inside z, then t is a member element of y;
  • A3: If z is a member element of y and t is the set having as member elements exactly all sets u included inside z, then t is a member element of y;
  • A4: If z is included inside y and if the sets z and y are equipotent, then z is a member element of y."

Axiom A':"For every set x, there exists a set y satisfying the four following conditions:

  • A'1: x is included inside y;
  • A'2: If z is a member element of y and t is a member element of z, then t is a member element of y;
  • A'3: If z is a member element of y, there exists a set w that is a member element of y such that every set t that is included inside z is a member element of w;
  • A'4: If z is included inside y and if no set included inside z is equipotent with y, then z is a member element of y."

Tarski asserts, without giving a proof, that axioms A and A' are equivalent.

QUESTION 1: Does anyone know about an explicit proof of the equivalance of A and A' ?

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Why was this question downvoted so much earlier? And why did people who downvoted it not leave any comment? –  Willie Wong Jun 30 '10 at 11:39
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Presumably because it was poorly formatted and because the author posted 4 or 5 questions on the same subject at once. –  Ben Webster Jun 30 '10 at 12:37
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In fact, all five messages I wrote are linked (and ordered) and are intended to make clarification around the "axiomatic power" of Tarski's axioms A and A', but I could not make a unique message because it was too long, so I divided it and asked one question inside each message. And YES, being a beginner with Mathoverflow, maybe my messages can be considered as poorly formatted ! Gérard LANG –  Gérard Lang Jun 30 '10 at 12:44
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1 Answer 1

up vote 2 down vote accepted

First, I note that you appear to be missing a not in A4, and you should say that "if $z$ and $y$ are not equipollant", for otherwise we could take $z=y$ and thereby deduce $y\in y$, contrary to the Foundation axiom.

With this correction, both your axioms are equivalent in ZFC to the assertion that there is a proper class of inaccessible cardinals.

For the one direction, if there are such cardinals, then for any set $x$ we may find an inaccessible cardinal $\kappa$ such that $x\in V_\kappa$, and take $y=V_\kappa=H_\kappa$ to fulfill either $A$ or $A'$, which is easily verified.

Conversely, assume axiom A. Let $x=\alpha$ be an ordinal and let $y$ arise as in axiom A. Let $\kappa=|y|$. As every subset of $\alpha$ is in $y$, it follows that $\alpha\lt\kappa$. If $\beta\lt\kappa$, then there is subset $z\subset y$ of size $\beta$, and this is an element of $y$ by A4. We also know $P(z)\subset y$ and $P(z)\in Y$, so $P(P(z))\subset y$, so $2^\beta\lt\kappa$. Thus, $\kappa$ is a strong limit. For regularity, suppose that $\kappa$ singular with cofinality $\gamma\lt\kappa$. Thus, $y$ is the union of $\gamma$ many subsets, each of size less than $\kappa$. These subsets are elements of $y$, and all their subsets are also in $y$. But every subset of $y$ is determined by a similar $\gamma$ sequence of elements of $y$, and so $y$ will have $2^\kappa$ many elements, a contradiction. So $\kappa$ is an inaccessible cardinal above $\alpha$, as desired.

For the other converse direction, assume axiom $A'$. Consider $x=\alpha+1$, and get $y$ as in $A'$, and again let $\kappa=|y|$. I claim that $\kappa\subset y$, for otherwise the least ordinal $\beta$ not in $y$ would be less than $\kappa$ in size and have all its subsets size less than $\kappa$, and hence in $y$ by $A4'$. Thus, $\kappa\subset y$. Now, for any $\gamma\lt\kappa$, every subset of $\gamma$ is in $y$ and there is an element of $y$ with at least $2^\gamma$ many subsets, all in $y$, so $2^\gamma\lt\kappa$. So $\kappa$ is a strong limit. Regularity follows as before, and so $\kappa$ is an inaccessible cardinal above $\alpha$.

Thus, since the axioms are both equivalent to the assertion that there is a proper class of inaccessible cardinals, they are also equivalent to each other.

Do you have some historical reason to study Tarski's treatment of inaccessibility? If not, I think you might find the contemporary accounts of large cardinals to be more appealing. You might look at Kanamori's book, The Higher Infinite.

If you wanted the equivalence in ZF rather than ZFC, or in ZF-Foundation, then I wouuld have to think more carefully about it, but I will mention that this issue seems to be remarked on in Solovay's letter mentioned in your previous question. In particular, without AC there are competing inequivalent notions of inaccessibility.

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Dear Mr Hamkins, Thankyou very much for correcting my question, and giving a neat answer under ZFC. And naturally I would like to know what happens under ZF only (I suppose that because A proves choice, a remains equivalent with a proper class of inaccessibles; but if A' does not prove choice under ZF, then the equivalence with A must be questionned, in view of the competing inequivalent definitions of inaccessibility ?). –  Gérard Lang Jul 5 '10 at 14:03
    
In fact, my primary interest is with the remarkable axiomatic strength of the "Tarski-Grothendieck Set Theory" as developped inside Metamath or Mizar, even knowing that "a proper class of inaccessibles" is a rather mild assertion is the "Large Cardinals Scala" presented in the beautiful Kanamori reference book. It is known that, starting with "ZFC + A", and adding the Pair Axiom, you can delete the Power-Set, Infinity and Choice axioms. So, in relation with assertion of Tarski on some variants of A and A', i am wondering if Pair and Union could not also be deleted. –  Gérard Lang Jul 5 '10 at 14:48
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