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Let $S$ be a set and k a infinite field. The injection $S \to k\mathrm{Alg}(k^S, k)$ (sending a point to the evaluation in it) is a bijection if and only if $S$ is a non-measurable cardinal (see for example Theo Johnson-Freyd's excellent exposition on MO for this classical fact). In this regard, it may be convenient to assume that there are no measurable cardinals (NMC). But it is also often convenient to use the fact that for every cardinal there is a larger inaccessible cardinal. This is also known as the Grothendieck universe axiom (GU).

Does the consistency of ZFC+GU imply the consistency of ZFC+GU+NMC? If not, what is known about the consistency of the second theory?

Perhaps we can take the smallest measurable cardinal in ZFC+GU as the desired model (just as we would do in plain ZFC)?

In fact, Theo Johnson-Freyd mentions my question at the end of his exposition, saying "probably you can still disallow measurable cardinals with impunity". So the question is devoted to refining this guess.

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    $\begingroup$ I'm just curious: Why is the assumption NMC is 'convenient' to assume? As a set theorist, large cardinals are mostly good to assume, and anti-large cardinal axioms for not-too-strong large cardinals look weird. $\endgroup$
    – Hanul Jeon
    Commented Oct 23 at 6:57
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    $\begingroup$ Hanul sometimes it is convenient in general topology. For example, NMC implies "every metric space is realcompact". You will occasionally run into results phrased this way in the literature. $\endgroup$
    – Todd Eisworth
    Commented Oct 23 at 12:53
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    $\begingroup$ The first sentence of the question needs to assume that the field is infinite or that $S$ is finite. Otherwise, the ultrapower by any non-principal ultrafilter on $S$ gives a counterexample. $\endgroup$
    – Andreas Blass
    Commented Oct 23 at 16:35
  • $\begingroup$ @AndreasBlass i really think about infinite fields, thank you $\endgroup$
    – Arshak Aivazian
    Commented Oct 23 at 21:09
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    $\begingroup$ Just take the inner model $L$ which has no measurables but has a proper class of inaccessibles if $V$ does. $\endgroup$
    – Lucenaposition
    Commented Oct 23 at 23:23

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Suppose $M\models\mathsf{ZFC+GU}$. If $M$ also satisfies $\mathsf{NMC}$, then we're done. Otherwise, let $\kappa$ be (what $M$ thinks is) the smallest measurable cardinal. Then since measurable cardinals are limits of inaccessible cardinals, $(V_\kappa)^M\models\mathsf{ZFC+GU+NMC}$ So yes, the consistency of $\mathsf{ZFC+GU}$ implies the consistency of $\mathsf{ZFC+GU+NMC}$.

(This also uses the fact that "cutting off" the cumulative hierarchy at a given point preserves inaccessibility: if $\lambda$ is inaccessible in $M$ and $\lambda<\kappa$ then $\lambda$ is inaccessible in $(V_\kappa)^M$.)

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  • $\begingroup$ Why does $(V_{\kappa})^M$ prove $GU$? That is, let $\lambda < \kappa$. Why can we find an inaccessible cardinal greater than $\lambda$ and strictly less than $\kappa$? $\endgroup$
    – Arshak Aivazian
    Commented Oct 23 at 5:45
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    $\begingroup$ Because measurable cardinals are limits of inaccessible cardinals and, if $\lambda<\lambda'<\kappa$ is inaccessible in $M$, then $\lambda'$ is inaccessible in $(V_\kappa)^M$. $\endgroup$
    – Andrés E. Caicedo
    Commented Oct 23 at 6:16
  • $\begingroup$ Ah, on the second reading I understood what it means "measurable cardinals are the limits of the inaccessible ones", thank you! $\endgroup$
    – Arshak Aivazian
    Commented Oct 23 at 6:55

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