In MK (Kelley-Morse) class theory, if i add an axiom that any cardinal except $On$ has an inaccessible greater than it (ie. essentially a Tarski/Grothendiek universe axiom), does that compel me to admit the existence of any other large cardinals (eg. measurable cardinals)?

Broadly what are the known side-effects of an ever increasing sequence of inaccessibles in MK?

For example this answer Is the axiom of universes 'harmless'? states that universes are used to resolve Fermats Last Theorem (although they can be dispensed with).

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    $\begingroup$ You cannot deduce any additional large cardinals from this. $\endgroup$ – Andrés E. Caicedo Mar 18 '18 at 0:07

This has no noteworthy side-effects that I can think of. In particular, it does not imply the existence of any stronger kinds of large cardinals.

There's a nice list of large cardinal axioms ordered by their (consistency) strength at Cantor's attic. As you can see, the Grothendieck universe axiom is near the very bottom, being the first axiom on the list that is stronger than just the existence of a single inaccessible. The list is formulated for ZFC rather than MK, but MK only very slightly increases the strength (for instance, MK with a proper class of inaccessibles is still weaker than ZFC with a $1$-inaccessible cardinal in consistency strength).

In concrete terms, pretty much all the stronger kinds of large cardinal are automatically at least $1$-inaccessible (that is, they are inaccessible and are a limit of inaccessible cardinals). If $\kappa$ is the least $1$-inaccessible cardinal, then $V_{\kappa+1}$ gives a model of MK in which there is a proper class of inaccessible cardinals but no $1$-inaccessible cardinals.

  • $\begingroup$ To the proposer: This also holds in ZFC. For example if there exists a measurable cardinal and $k$ is the least measurable cardinal and $M=(V_k,\epsilon)$ then $M$ satisfies $ ZFC.$ And $\forall x\in M\;( (x\text { is not measurable })^M)$ while $\forall x\in k\;(x$ is inaccessible $\iff (x$ is inaccessible$)^M).$ $\endgroup$ – DanielWainfleet Mar 19 '18 at 0:09
  • $\begingroup$ Assuming the consistency of a super-compact cardinal, is it consistent that $k$ is the least super-compact and that $\exists x\in k ((x\text { is super-compact })^{V_k})$ ? $\endgroup$ – DanielWainfleet Mar 19 '18 at 0:17

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