What's the difference between ZFC+Grothendieck, ZFC+inaccessible cardinals and Tarski-Grothendieck set theory? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:03:49Z http://mathoverflow.net/feeds/question/102846 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102846/whats-the-difference-between-zfcgrothendieck-zfcinaccessible-cardinals-and-ta What's the difference between ZFC+Grothendieck, ZFC+inaccessible cardinals and Tarski-Grothendieck set theory? Mike Battaglia 2012-07-22T00:38:00Z 2012-08-08T22:32:13Z <p>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 <a href="http://en.wikipedia.org/wiki/Grothendieck_universe" rel="nofollow">this way</a>). 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."</p> <p>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.</p> <p>In addition to the above, there's also Tarski-Grothendieck set theory, which can be found <a href="http://en.wikipedia.org/wiki/Tarski%25E2%2580%2593Grothendieck_set_theory" rel="nofollow">here</a>. The axioms of TG are</p> <ol> <li>The axiom stating everything is a set</li> <li>The axiom of extensionality from ZFC</li> <li>The axiom of regularity from ZFC</li> <li>The axiom of pairing from ZFC</li> <li>The axiom of union from ZFC</li> <li>The axiom schema of replacement from ZFC</li> <li>Tarski's axiom A</li> </ol> <p>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 $y$.</p> <p>These three axioms from ZFC are then implied as theorems of TG:</p> <ol> <li>The axiom of infinity</li> <li>The axiom of power set</li> <li>The axiom schema of specification</li> <li>The axiom of choice</li> </ol> <p>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?</p> <p>In other words, instead of typing ZFC+Grothendieck, can I just type TG and be referencing a different axiomatization of the exact same thing?</p> http://mathoverflow.net/questions/102846/whats-the-difference-between-zfcgrothendieck-zfcinaccessible-cardinals-and-ta/102849#102849 Answer by Trevor Wilson for What's the difference between ZFC+Grothendieck, ZFC+inaccessible cardinals and Tarski-Grothendieck set theory? Trevor Wilson 2012-07-22T01:10:38Z 2012-07-22T02:12:39Z <p>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.</p> <p>To show that the cardinality $\kappa$ of $y$ is a strong limit cardinal, given $\zeta &lt; \kappa$ 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 $2^{\zeta} &lt; \kappa$.</p> <p>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.)</p> http://mathoverflow.net/questions/102846/whats-the-difference-between-zfcgrothendieck-zfcinaccessible-cardinals-and-ta/104310#104310 Answer by Gérard Lang for What's the difference between ZFC+Grothendieck, ZFC+inaccessible cardinals and Tarski-Grothendieck set theory? Gérard Lang 2012-08-08T21:36:08Z 2012-08-08T22:32:13Z <p>This is not an answer, but a repetition of my linked questions already stated in <a href="http://mathoverflow.net/questions/28389" rel="nofollow">http://mathoverflow.net/questions/28389</a>. 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</p>