5
$\begingroup$

Grothendieck's axiom states that any set is a member of a Grothendieck universe (i.e. of a set that is closed under the subset, powerset, pairing and family-union relations), or equivalently, that there is a proper class of inaccessible cardinals. Tarski's lemma states that any set is a member of a set that is closed under the subset and the powerset relations and that contains each of its subsets of smaller cardinality than itself.

So the axioms seem to state similar things, but I can't really work out the precise connection between them. Does one imply the other? If not, is it known which one has stronger consistency strength?

Is Tarski's axiom as useful as the axiom of universes for category theory? Does the axiom of universe, like Tarski's axiom, imply AC (over ZF)?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
12
$\begingroup$

Over ZFC, the theories are equivalent. Both Grothendieck's Universe Axiom and Tarski's Axiom are equivalent to the assertion that there are unboundedly many inaccessible cardinals.

Grothendieck universes are exactly the sets $H_\kappa$ for an inaccessible cardinal $\kappa$, consisting of all sets hereditarily of size less than $\kappa$. They can also be described as $V_\kappa$ for inaccessible $\kappa$, using the cumulative von Neumann hierarchy. The Tarski universes, in contrast, needn't be transitive, and so the notions are not equivalent.

The issue is that over ZF, they lose their equivalence. Solovay explained on the FOM list that this is due to the way that cardinalities are treated in TG, making ZF+TG imply AC, whereas ZF+GU does not imply AC. You can read Solovay's interesting post here.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Who am I to contradict Solovay, but it really comes down to how you define inaccessible cardinals, but arguably the correct definition is "$V_\kappa$ is a model of 2nd order ZF". In that case TG does not imply AC. $\endgroup$
    – Asaf Karagila ♦
    Commented May 7, 2021 at 20:42
5
$\begingroup$

Both axioms are equivalent with ZFC. They are not equivalent with only ZF, because Ac is a cosequence of ZF+Grothendieck, but not of ZF+ tarski A. See the message "AC and strongly inaccessible cardinals" (29/02/2008) of Robert SOLOVAY on the list FOM (Foundation of Mathematics).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .