MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say that "U" is the axiom that "For each set x, there exists a Grothendieck universe U such that x $\in$ U", where Grothendieck universes are defined in the usual way (or, if that's unclear, in this way). Also, say that "Ca" is the axiom that "For each cardinal κ, there is a strongly inaccessible cardinal $\lambda$ which is strictly larger than κ."

It's known that ZFC+U and ZFC+Ca are completely equivalent and prove the same sentences. A sentence is a theorem of ZFC+U iff it's a theorem of ZFC+Ca.

In addition to the above, there's also Tarski-Grothendieck set theory, which can be found here. The axioms of TG are

  1. The axiom stating everything is a set
  2. The axiom of extensionality from ZFC
  3. The axiom of regularity from ZFC
  4. The axiom of pairing from ZFC
  5. The axiom of union from ZFC
  6. The axiom schema of replacement from ZFC
  7. Tarski's axiom A

Tarski's axiom A states that for any set $x$, there exists a set $y$ containing, $x$ itself, every subset of every member of $y$, the power set of every member of $y$, and every subset of $y$ of cardinality less than $y$.

These three axioms from ZFC are then implied as theorems of TG:

  1. The axiom of infinity
  2. The axiom of power set
  3. The axiom schema of specification
  4. The axiom of choice

My question is as follows: is TG also completely equivalent to ZFC+U and ZFC+Ca, equivalent in the same sense that something is a theorem of TG iff it's a theorem of the other two? Is TG just an axiomatization of ZFC+U/ZFC+Ca which removes redundant axioms and allows them to just be theorems? Or is there some subtle difference between TG and ZFC+U/ZFC+Ca, in that there's some sentence which TG proves that's undecidable in ZFC+U/ZFC+Ca or vice versa?

In other words, instead of typing ZFC+Grothendieck, can I just type TG and be referencing a different axiomatization of the exact same thing?

share|cite|improve this question
"C" already denotes the axiom of choice, so it's not the best abbreviation for another axiom. – John Bentin Jul 22 '12 at 18:58
Fair enough. I'll change it to "Ca". – Mike Battaglia Jul 23 '12 at 6:54
This is not an answer, but a repetition of my linked questions already stated in <>;. 1/ Is it possible to dispense with axiom 4 (the axiom of pairing from ZF(C)) to develop the Tarski-Grothendieck (TG) set theory? 2/ Does anybody know how to prove the 16 equivalences between conditions of axioms A and A' of Tarski ? Gérard Lang – Gérard Lang Aug 8 '12 at 21:36
up vote 13 down vote accepted

Yes. Assume ZFC. If there is a proper class of inaccessible cardinals, then Tarski's Axiom A holds because whenever $\kappa$ is inaccessible, the rank initial segment $V_\kappa$ of $V$ is a Tarski set. Conversely, if Tarski's Axiom A holds then for every set $x$ there is a Tarski set $y$ with $x \in y$. We will show that $|y|$ is an inaccessible cardinal greater than $|x|$, proving the existence of a proper class of inaccessible cardinals.

To show that the cardinality $\kappa$ of $y$ is a strong limit cardinal, given $\zeta < \kappa$ we take a subset $z$ of $y$ of size $\zeta$. We have $z \in y$ because $y$ contains its small subsets. Then we have $\mathcal{P}(z) \in y$ because $y$ is closed under the power set operation. Finally $\mathcal{P}(\mathcal{P}(z)) \subset y$ because $y$ contains all subsets of its elements. This shows that $2^{2^{\zeta}} \le \kappa$ and therefore that $2^{\zeta} < \kappa$.

To show that the cardinality $\kappa$ of $y$ is regular, notice that if $\kappa$ is singular then by the closure of $y$ under small subsets we can get a family of $\kappa^{cof(\kappa)}$ many distinct sets in $y$, contradicting the fact that $\kappa^{cof( \kappa)} > \kappa$ (which is an instance of Koenig's Theorem.)

share|cite|improve this answer
I always thought that it was $2$-inaccessible (which is slightly stronger than a proper class of inaccessibles) that was equivalent to the TG axioms. – Asaf Karagila Jul 22 '12 at 7:13
Is a 2-inaccessible an inaccessible limit of inaccessibles? In this case, if $\kappa$ is 2-inaccessible, then $V_\kappa$ is a model of TG. – Stefan Geschke Jul 22 '12 at 8:12
Stefan, yes. Indeed if $\kappa$ is $2$-inaccessible then $V_\kappa$ is a model of ZFC+proper class of inaccessible. I suppose that I get the minor difference. – Asaf Karagila Jul 22 '12 at 9:53
@Mike: I'll respond to your second comment first, regarding my other attempt at answering this question, which I deleted quickly but it looks like you were still notified of. If there is a single inaccessible cardinal then for every $x$ there is a Tarski set $y$ with $\lbrace x \rbrace \in y$, and this implies that there is a proper class of Tarski sets, but it does not imply Tarski's axiom that for every $x$ there is a Tarski set $y$ with $x \in y$. – Trevor Wilson Jul 23 '12 at 14:25
@Mike: regarding your first comment, modulo ZFC all three axioms are equivalent by what I wrote combined with what you wrote (that ZFC+U and ZFC+Ca are equivalent.) It does seem to be known that Tarski's Axiom A is not equivalent to the other two under ZF for the reason you said. – Trevor Wilson Jul 23 '12 at 16:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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