A-The addition of the Grothendieck Universe Axiom (for every set x, there exists a set y that is a universe and contains x as member element) to ZFC (ZFC+GU) is considered as giving an almost good solution permitting the insertion of Category theory inside Set theory. It is known that we obtain an equivalent theory (ZFC+TA) by replacing GU with the Axiom A of Tarski (for every set x, there exists a set y that is an A Tarski set and contains x as a member element), or also (ZFC+IN) by replacing GU by the axiom IN (for every ordinal x, there exists a strongly inaccessible ordinal y containing x as a member element.

1-In each three cases, we can prove (using foundation) that "there exists a proper class of (GU sets, TA sets, inaccessible cardinals)" is a theorem. But is it possible to replace the axioms GU, TA and IN by the formulation with proper classes (this seems possible for IN and GU ) ?

B-If we replace ZFC by ZF, the situation seems much more involved. As it is provable that AC is a consequence of TA, ZF+TA is equivalent to ZFC+TA. As it is provable that some GU sets cannot be well-ordered, AC is not a consequence of ZF+GU, that cannot be equivalent to ZFC+GU. For the third case, this is depending of how we define an inaccessible without AC; but if we take the "correct one", Ac is not derivable from ZF+IN'. These question are very thoroughly presented in a message of R. Solovay to FOM "AC and stongly inaccessible cardinals" (29/02/2008). But, in fact, the power set axiom and the Infinity axiom can also be derived from the Tarski a axiom. So that one could think that the theory: Extensionality+Replacement+AT+Union+Foundation is equivalent to ZFC+GU. But when you try to developp such a theory, it seems that you are obliged to also introduce the Pair Axiom before introducing AT that needs a definition of functions.

2-Is it in fact possible to dispense of the axiom of the pair within this theory ?

C-Tarski's A axiom is given inside his paper (auf deutsch) "Über unerreichbare Kardinalzahlen", Fund Math 1938, page 84. 3-On the same page, Tarski gives another axiom, named A'with four conditions (as in the case of A) and writes ""Übrigens sind vershiedene âquivalente Unformung dieses Axioms [A] bekannt. Man kann Z. B. Bedingungen A-1-A4 beziehungsweise durch folgenden Bedingungen [A'1-A'4] ersetzen (und zwar jede Bedingung unhabhängig von denen anderen). Does anyone completely understand what is exactly meant here by Tarski, and how is this proved ?

Gérard LANG

  • $\begingroup$ Could you clarify exactly what you are asking in item A? Also, I would suggest that you might organize your question into paragraphs, in order that it might become easier to read. $\endgroup$ Jun 16, 2010 at 15:07
  • $\begingroup$ In each part A, B, C,of my message I have one question that is 1 (for A),2 (for B) and 3 (for C). My question 1 is: is it possible to have an equivalent theory in each three cases considered in A, when replacing the axiom "For every set x, there exists a set y containing x that is GU/ AT/ IN" by the axiom "There exists a proper class of GU sets/ AT sets/ INcardinals" $\endgroup$ Jun 16, 2010 at 15:30
  • $\begingroup$ Here is a link to Solovay's FOM post: cs.nyu.edu/pipermail/fom/2008-March/012783.html. Note the date is March, not Februrary. $\endgroup$ Jun 16, 2010 at 15:47
  • $\begingroup$ Here is a translation of the passage you quoted from Tarski: "By the way, some equivalent reformulations of this axiom [A] are known. For example, one can replace conditions A1-A4 by the following conditions A1'-A4' (and actually each condition can be replaced independently of the others)." $\endgroup$ Jun 16, 2010 at 16:37

1 Answer 1


The answer to question 1 is affirmative for GU and IN but negative for TA. That is, the proper class formulation of TA is not equivalent to TA, unless both are inconsistent.

Asserting that every ordinal is below an inaccessible cardinal is clearly equivalent in ZF to asserting that the inaccessible cardinals form a proper class. And since the Grothendieck universes are exactly the sets $V_\kappa$ for an inaccessible cardinal $\kappa$, the theory GU over ZFC is equivalent to the assertion that there are a proper class of universes. So those cases are relatively straightforward, as you had guessed.

But the case of the Tarski axiom is different, since his universes are not transitive sets. In fact, if there is a single inaccessible cardinal $\kappa$, one can already form a proper class of Tarski sets. To see this, suppose that $\kappa$ is inaccessible and $x$ is any set. Build a Tarski set $U$ as follows:

  • $U_0=\{\{x\}\}$,
  • $U_{\alpha=1}=P(U_\alpha)$ and
  • $U_\lambda=\bigcup_\alpha U_\alpha$ at limit ordinals.

That is, we perform the usual cumulative hiearchy, but starting with object {x} instead of nothing. The resulting set $U=U_\kappa$ contains {x} as an element, has size $\kappa$ and is closed under subsets, power sets and small unions, so it is a Tarski universe. (Note that if $y\in U_{\alpha+1}$, then all subsets of $y$ are also in $U_{\alpha+1}$, and so $P(y)$ is added on the next step. And {x} has only two subsets, which are both in $U$.) But for sufficiently large $x$, the resulting $U$ will not be transitive, since $x$ itself will never be added as an element. Furthermore, since $x$ was arbitrary, from a single inaccessible cardinal we can build a proper class of Tarski universes. And so the proper class formulation of TA will not be equivalent to TA, unless both theories are inconsistent, because if the existence of an inaccessible cardinal is consistent with ZFC, then it is consistent that there is exactly one such cardinal, by chopping the universe off at the second one. The resulting model will satisfy the proper class formulation of TA but not TA itself.

  • $\begingroup$ A-1: Thank you very much for your very clear answer and proof concerning the TA case. Would the answer concerning the GU case be the same within ZF (so without AC ?). C-3: Thank you very much for the translation from german. So, Tarski is asserting that if we take for i=1,2,3,4 Bi to be either Ai or A'i, where Ai-s are the four conditions of his axiom a and a'i are the four conditions of his axiom A', thenevery sixteen axioms obtained as a $\endgroup$ Jun 17, 2010 at 6:52
  • $\begingroup$ then every sixteen axioms obtained as a conjunction of every possible choice of Bi-s are all equivalents. This seems a very interesting result,but the proof does not seem so easy. In particular, if we take B1=A1, B2=A'2, B3=A'3 and B4=A4, then Tarski asserts that the Union axiom is derivable from thus form of axiom B. $\endgroup$ Jun 17, 2010 at 6:56
  • $\begingroup$ Gérard, I think that in ZF then the GU is equivalently formalized in the proper class version, since the universes continue to be $H_\kappa$ for inaccessible $\kappa$. The trick I use above for Tarski sets doesn't work at all for Grothendieck universes, since they must be transitive. (Indeed, it seems somewhat strange to me that Tarski didn't insist on transitivity.) Note also that as Solovay explains (citing Blass), there are various notions of what it means to be inaccessible, which are equivalent in ZFC but not in ZF, and this is part of the explanation for why TA implies AC. $\endgroup$ Jun 17, 2010 at 12:31
  • $\begingroup$ Again, thank you very much for the case FZ=GU. Concerning tra $\endgroup$ Jun 17, 2010 at 14:09
  • $\begingroup$ Again, thank you very much for the case ZF+ GU. Indeed, Solovay's message is building upon Blass's paper that shows four inequivalent possible definitions of an inaccessible ordinal when you do not have AC, that become equivalent in presence of AC. Concerning transitivity and axiom A of Tarski, this is taken care of inside condition A'2, pertaining to Tarski's axiom A' that he offers as equivalent to axiom A, and this is what I am finding intriguing in his assertion of the equivalence of the sixteen axioms B deriving from mixing conditions of axiom A and of axiom A'. $\endgroup$ Jun 17, 2010 at 14:19

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