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Joel David Hamkins
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This is not a full answer, but let me mention that your hypothesis has large cardinal strength, because it implies the failure of the singular cardinals hypothesis.

Assume $2^\kappa=\aleph_{\kappa^+}$ for every infinite cardinal $\kappa$. (I am taking your $|X|^+$ to refer to the cardinal successor.) Let $\kappa$ be a singular strong limit cardinal, which is also an $\aleph$-fixed point. So byBy your hypothesis, we have $2^\kappa=\aleph_{\kappa^+}$, whichbut this is strictly bigger than $\aleph_{\kappa+1}$$\kappa^+$, since successor cardinals are never $\aleph$-fixed points, and so this is a counterinstance of the SCH. So SCH fails in every instance.

The failure of SCH has the large cardinal strength of a measurable cardinal $\kappa$ of Mitchell rank $\kappa^{++}$.

Meanwhile, if you assert your property $2^{\kappa}=\aleph_{\kappa^+}$ only for regular cardinals $\kappa$, then this is equiconsistent with ZFC by Easton's theorem.

This is not a full answer, but let me mention that your hypothesis has large cardinal strength, because it implies the failure of the singular cardinals hypothesis.

Assume $2^\kappa=\aleph_{\kappa^+}$ for every infinite cardinal $\kappa$. (I am taking your $|X|^+$ to refer to the cardinal successor.) Let $\kappa$ be a singular strong limit cardinal, which is also an $\aleph$-fixed point. So by your hypothesis, we have $2^\kappa=\aleph_{\kappa^+}$, which is bigger than $\aleph_{\kappa+1}$, and so this is a counterinstance of the SCH. So SCH fails.

The failure of SCH has the large cardinal strength of a measurable cardinal $\kappa$ of Mitchell rank $\kappa^{++}$.

Meanwhile, if you assert your property $2^{\kappa}=\aleph_{\kappa^+}$ only for regular cardinals $\kappa$, then this is equiconsistent with ZFC by Easton's theorem.

This is not a full answer, but let me mention that your hypothesis has large cardinal strength, because it implies the failure of the singular cardinals hypothesis.

Assume $2^\kappa=\aleph_{\kappa^+}$ for every infinite cardinal $\kappa$. (I am taking your $|X|^+$ to refer to the cardinal successor.) Let $\kappa$ be a singular strong limit cardinal. By your hypothesis, we have $2^\kappa=\aleph_{\kappa^+}$, but this is strictly bigger than $\kappa^+$, since successor cardinals are never $\aleph$-fixed points, and so SCH fails in every instance.

The failure of SCH has the large cardinal strength of a measurable cardinal $\kappa$ of Mitchell rank $\kappa^{++}$.

Meanwhile, if you assert your property $2^{\kappa}=\aleph_{\kappa^+}$ only for regular cardinals $\kappa$, then this is equiconsistent with ZFC by Easton's theorem.

Source Link
Joel David Hamkins
  • 236.3k
  • 44
  • 777
  • 1.4k

This is not a full answer, but let me mention that your hypothesis has large cardinal strength, because it implies the failure of the singular cardinals hypothesis.

Assume $2^\kappa=\aleph_{\kappa^+}$ for every infinite cardinal $\kappa$. (I am taking your $|X|^+$ to refer to the cardinal successor.) Let $\kappa$ be a singular strong limit cardinal, which is also an $\aleph$-fixed point. So by your hypothesis, we have $2^\kappa=\aleph_{\kappa^+}$, which is bigger than $\aleph_{\kappa+1}$, and so this is a counterinstance of the SCH. So SCH fails.

The failure of SCH has the large cardinal strength of a measurable cardinal $\kappa$ of Mitchell rank $\kappa^{++}$.

Meanwhile, if you assert your property $2^{\kappa}=\aleph_{\kappa^+}$ only for regular cardinals $\kappa$, then this is equiconsistent with ZFC by Easton's theorem.