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Andrés E. Caicedo
Andrés E. Caicedo
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Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.)

Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so it implies that, provably in $\mathsf{ZFC}$, the relevant cardinals are weakly inaccessible). If $\kappa$ is the least ordinal such cardinalthat $L_\kappa $ is a model of $\mathsf{ZFC}$, then $L_\kappa$ and $V$ agree on all arithmetic statements, since they share the same natural numbers (in fact, they agree on all $\Sigma^1_2$ statements).

Now argue in $L$: Eitherbut there are no such cardinals in $L_\kappa$, and we are done. Notice that there is such an ordinal $\kappa$, or repeat the previous paragraph withsince any weakly inaccessible cardinal $L_\kappa$ in place$\tau$ of $V$ and $L_{\kappa'}$(such as the large cardinal in place ofquestion) is strongly inaccessible in $L_\kappa$, where$L$ and therefore $\kappa'$ is$L_\tau\models\mathsf{ZFC}$. On the least such cardinal fromother hand, the point of viewminimality of $\kappa$ implies that no ordinal smaller than $\kappa$ is weakly (and therefore strongly) inaccessible in $L_\kappa$.

We are done if $L_\kappa'$($\kappa$ itself is a model where there are no such cardinals, otherwise, repeat the previous argument. This process necessarily stops since the sequence of ordinals we are producing"tiny": it is decreasingcountable in $L$.)

Now, if instead of weak inaccessibility you assume that consistency statements or the like qualify as large cardinal axioms, then "of course" the answer is no, as $\lnot A$ would be a false arithmetic statement.

[Note that H. Friedman's results, mentioned in another answerproviding examples of combinatorial statements of low-level arithmetic complexity whose validity depends on large cardinals, assume more than the negation of the large cardinal in question in the "bad" case, where the nice combinatorial statement fails. The models of set theory where his statements failthis happens are in fact not $\omega$-models.]

Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.)

Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so it implies that the relevant cardinals are weakly inaccessible). If $\kappa$ is the least such cardinal, then $L_\kappa$ and $V$ agree on all arithmetic statements, since they share the same natural numbers (in fact, they agree on all $\Sigma^1_2$ statements).

Now argue in $L$: Either there are no such cardinals in $L_\kappa$, and we are done, or repeat the previous paragraph with $L_\kappa$ in place of $V$ and $L_{\kappa'}$ in place of $L_\kappa$, where $\kappa'$ is the least such cardinal from the point of view of $L_\kappa$.

We are done if $L_\kappa'$ is a model where there are no such cardinals, otherwise, repeat the previous argument. This process necessarily stops since the sequence of ordinals we are producing is decreasing.

Now, if instead of weak inaccessibility you assume that consistency statements or the like qualify as large cardinal axioms, then "of course" the answer is no, as $\lnot A$ would be a false arithmetic statement.

[Note that Friedman's results, mentioned in another answer, assume more than the negation of the large cardinal in question in the "bad" case. The models of set theory where his statements fail are not $\omega$-models.]

Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.)

Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so, provably in $\mathsf{ZFC}$, the relevant cardinals are weakly inaccessible). If $\kappa$ is the least ordinal such that $L_\kappa $ is a model of $\mathsf{ZFC}$, then $L_\kappa$ and $V$ agree on all arithmetic statements, since they share the same natural numbers, but there are no such cardinals in $L_\kappa$. Notice that there is such an ordinal $\kappa$, since any weakly inaccessible cardinal $\tau$ of $V$ (such as the large cardinal in question) is strongly inaccessible in $L$ and therefore $L_\tau\models\mathsf{ZFC}$. On the other hand, the minimality of $\kappa$ implies that no ordinal smaller than $\kappa$ is weakly (and therefore strongly) inaccessible in $L_\kappa$. ($\kappa$ itself is "tiny": it is countable in $L$.)

Now, if instead of weak inaccessibility you assume that consistency statements or the like qualify as large cardinal axioms, then "of course" the answer is no, as $\lnot A$ would be a false arithmetic statement.

[Note that H. Friedman's results, providing examples of combinatorial statements of low-level arithmetic complexity whose validity depends on large cardinals, assume more than the negation of the large cardinal in question in the "bad" case, where the nice combinatorial statement fails. The models of set theory where this happens are in fact not $\omega$-models.]

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Andrés E. Caicedo
Andrés E. Caicedo
  • 32.5k
  • 5
  • 133
  • 240

Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.)

Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so it implies that the relevant cardinals are weakly inaccessible). If $\kappa$ is the least such cardinal, then $L_\kappa$ and $V$ agree on all arithmetic statements, since they share the same natural numbers (in fact, they agree on all $\Sigma^1_2$ statements).

Now argue in $L$: Either there are no such cardinals in $L_\kappa$, and we are done, or repeat the previous paragraph with $L_\kappa$ in place of $V$ and $L_{\kappa'}$ in place of $L_\kappa$, where $\kappa'$ is the least such cardinal from the point of view of $L_\kappa$.

We are done if $L_\kappa'$ is a model where there are no such cardinals, otherwise, repeat the previous argument. This process necessarily stops since the sequence of ordinals we are producing is decreasing.

Now, if instead of weak inaccessibility you assume that consistency statements or the like qualify as large cardinal axioms, then "of course" the answer is no, as $\lnot A$ would be a false arithmetic statement.

[Note that Friedman's results, mentioned in another answer, assume more than the negation of the large cardinal in question in the "bad" case. The models of set theory where his statements fail are not $\omega$-models.]