Can choice be proved with ZF+Tarski axiom?
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Following the link found in the Wikipedia article about the Tarski–Grothendieck set theory, the required proof (by Tarski himself!) can be found beginning on p.181 of his article "On the well-ordered subsets of any set" published in 1939 in "Fundamenta Mathematicae" (in fact, Tarski shows that his axiom implies the well-ordering theorem/axiom, which is known to be equivalent to the axiom of choice).
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$\begingroup$ Interesting, it is very powerful: "It is Tarski's axiom that distinguishes TG from other axiomatic set theories. Tarski's axiom also implies the axioms of infinity, choice,[1][2] and power set.[3][4] It also implies the existence of inaccessible cardinals, thanks to which the ontology of TG is much richer than that of conventional set theories such as ZFC. " Then the other natural question would be why ZFC is much more popular than the Tarski-Grothendieck. I conjecture GT must have more contra intuitive results than for example the ZFC Banach Tarski paradox. $\endgroup$ Commented Jan 10, 2023 at 21:42
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$\begingroup$ Your reference shows that $\mathsf{A}_2 \to \mathsf{Choice}$. But how does $\mathsf{TA} \to \mathsf{A}_2$? $\endgroup$ Commented Nov 26 at 6:39