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  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 22, 2012 at 7:13
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    $\begingroup$ 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. $\endgroup$
    – Stefan Geschke
    Commented Jul 22, 2012 at 8:12
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 22, 2012 at 9:53
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    $\begingroup$ @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 $x \in y$. $\endgroup$
    – Trevor Wilson
    Commented Jul 23, 2012 at 14:25
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    $\begingroup$ @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. $\endgroup$
    – Trevor Wilson
    Commented Jul 23, 2012 at 16:34