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Asaf Karagila
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This is not really a problem.

If $\kappa$ is a measurable cardinal then $V_\kappa$, or the set of sets which are hereditarily have size smaller than $\kappa$, is a model of $\sf ZFC$. This means that $\sf ZFC$ cannot even prove the consistency of $\sf ZFC+\exists\kappa\text{ measurable}$, because of the incompleteness theorem. Well, at least if $\sf ZFC$ is consistent to begin with.

Furthermore, if $\kappa$ is the least measurable cardinal, and if there is a measurable cardinal then there is a least measurable cardinal, then in $V_\kappa$ there are no measurable cardinal, therefore $V_\kappa\models\sf ZFC+\lnot\exists\kappa\text{ measurable}$.

Regarding the assumption on determinacy, two points are relevant here: $\sf ZF+AD$ proves that the axiom of choice fails. And moreover the consistency strength of $\sf ZF+AD$ is much higher than that of $\sf ZFC$, or even $\sf ZFC+\exists\kappa\text{ measurable}$. This means that if we assume that $\sf ZF+AD$ is consistent then we can prove a lot more than we can from $\sf ZFC$ or $\sf ZFC+\exists\kappa\text{ measurable}$.

So we know that $\sf ZFC$ cannot prove that a measurable cardinal exist, or even the consistency of one, and we know that if $\sf ZFC+\exists\kappa\text{ measurable}$ is consistent, then so is $\sf ZFC+\lnot\exists\kappa\text{ measurable}$.


I am not too familiar with structural set theory or $\sf ETCS$, but Misha Gavrilovich has a characterization of measurable cardinals using his model categorical construction $\rm QtNaamen$. You can find the papers in Misha's homepage, in particular "Exercises de style: A homotopy theory for set theory. Part II" with Assaf Hasson.

This is not really a problem.

If $\kappa$ is a measurable cardinal then $V_\kappa$, or the set of sets which are hereditarily have size smaller than $\kappa$, is a model of $\sf ZFC$. This means that $\sf ZFC$ cannot even prove the consistency of $\sf ZFC+\exists\kappa\text{ measurable}$, because of the incompleteness theorem. Well, at least if $\sf ZFC$ is consistent to begin with.

Furthermore, if $\kappa$ is the least measurable cardinal, and if there is a measurable cardinal then there is a least measurable cardinal, then in $V_\kappa$ there are no measurable cardinal, therefore $V_\kappa\models\sf ZFC+\lnot\exists\kappa\text{ measurable}$.

Regarding the assumption on determinacy, two points are relevant here: $\sf ZF+AD$ proves that the axiom of choice fails. And moreover the consistency strength of $\sf ZF+AD$ is much higher than that of $\sf ZFC$, or even $\sf ZFC+\exists\kappa\text{ measurable}$. This means that if we assume that $\sf ZF+AD$ is consistent then we can prove a lot more than we can from $\sf ZFC$ or $\sf ZFC+\exists\kappa\text{ measurable}$.

So we know that $\sf ZFC$ cannot prove that a measurable cardinal exist, or even the consistency of one, and we know that if $\sf ZFC+\exists\kappa\text{ measurable}$ is consistent, then so is $\sf ZFC+\lnot\exists\kappa\text{ measurable}$.

This is not really a problem.

If $\kappa$ is a measurable cardinal then $V_\kappa$, or the set of sets which are hereditarily have size smaller than $\kappa$, is a model of $\sf ZFC$. This means that $\sf ZFC$ cannot even prove the consistency of $\sf ZFC+\exists\kappa\text{ measurable}$, because of the incompleteness theorem. Well, at least if $\sf ZFC$ is consistent to begin with.

Furthermore, if $\kappa$ is the least measurable cardinal, and if there is a measurable cardinal then there is a least measurable cardinal, then in $V_\kappa$ there are no measurable cardinal, therefore $V_\kappa\models\sf ZFC+\lnot\exists\kappa\text{ measurable}$.

Regarding the assumption on determinacy, two points are relevant here: $\sf ZF+AD$ proves that the axiom of choice fails. And moreover the consistency strength of $\sf ZF+AD$ is much higher than that of $\sf ZFC$, or even $\sf ZFC+\exists\kappa\text{ measurable}$. This means that if we assume that $\sf ZF+AD$ is consistent then we can prove a lot more than we can from $\sf ZFC$ or $\sf ZFC+\exists\kappa\text{ measurable}$.

So we know that $\sf ZFC$ cannot prove that a measurable cardinal exist, or even the consistency of one, and we know that if $\sf ZFC+\exists\kappa\text{ measurable}$ is consistent, then so is $\sf ZFC+\lnot\exists\kappa\text{ measurable}$.


I am not too familiar with structural set theory or $\sf ETCS$, but Misha Gavrilovich has a characterization of measurable cardinals using his model categorical construction $\rm QtNaamen$. You can find the papers in Misha's homepage, in particular "Exercises de style: A homotopy theory for set theory. Part II" with Assaf Hasson.

Source Link
Asaf Karagila
  • 39.9k
  • 8
  • 135
  • 283

This is not really a problem.

If $\kappa$ is a measurable cardinal then $V_\kappa$, or the set of sets which are hereditarily have size smaller than $\kappa$, is a model of $\sf ZFC$. This means that $\sf ZFC$ cannot even prove the consistency of $\sf ZFC+\exists\kappa\text{ measurable}$, because of the incompleteness theorem. Well, at least if $\sf ZFC$ is consistent to begin with.

Furthermore, if $\kappa$ is the least measurable cardinal, and if there is a measurable cardinal then there is a least measurable cardinal, then in $V_\kappa$ there are no measurable cardinal, therefore $V_\kappa\models\sf ZFC+\lnot\exists\kappa\text{ measurable}$.

Regarding the assumption on determinacy, two points are relevant here: $\sf ZF+AD$ proves that the axiom of choice fails. And moreover the consistency strength of $\sf ZF+AD$ is much higher than that of $\sf ZFC$, or even $\sf ZFC+\exists\kappa\text{ measurable}$. This means that if we assume that $\sf ZF+AD$ is consistent then we can prove a lot more than we can from $\sf ZFC$ or $\sf ZFC+\exists\kappa\text{ measurable}$.

So we know that $\sf ZFC$ cannot prove that a measurable cardinal exist, or even the consistency of one, and we know that if $\sf ZFC+\exists\kappa\text{ measurable}$ is consistent, then so is $\sf ZFC+\lnot\exists\kappa\text{ measurable}$.