Is ZFC with existence of Grothendieck universe (variant: Grothendieck universe containing every given set) provable in ZFC to be equiconsistent with ZFC?

If not, what else it may be equiconsistent with? (And in which formalistic it may be proved?)

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    $\begingroup$ ZFC+Inaccsessible or two. $\endgroup$ – Asaf Karagila Aug 20 '11 at 20:40
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    $\begingroup$ @Asaf Karagila: What is "two" in your comment? $\endgroup$ – porton Aug 20 '11 at 20:55
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    $\begingroup$ One inaccessible cardinal, or two inaccessible cardinals. $\endgroup$ – Asaf Karagila Aug 20 '11 at 21:04

This matter was much discussed on MathOverflow. The links below and the Wikipedia page should answer the question and the one that appears in the comments.





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