Skip to main content
MathOverflow

Return to Answer

deleted 9 characters in body
Source Link
Ramiro de la Vega
Ramiro de la Vega
  • 11.5k
  • 1
  • 45
  • 56

I don´t think the statement is true as it is. If $J=\omega_1+\omega_1^*$$J=\omega_1+1$ then for any $I \supseteq J$ (of whatever cardinality) you can define a cut as: $x \in A$ iff $x \in I$ and $\exists \alpha \in \omega_1 (x<\alpha) $ and $B=I \setminus A$. This cut can't have cofinality $(\omega, \omega)$.

I don´t think the statement is true as it is. If $J=\omega_1+\omega_1^*$ then for any $I \supseteq J$ (of whatever cardinality) you can define a cut as: $x \in A$ iff $x \in I$ and $\exists \alpha \in \omega_1 (x<\alpha) $ and $B=I \setminus A$. This cut can't have cofinality $(\omega, \omega)$.

I don´t think the statement is true as it is. If $J=\omega_1+1$ then for any $I \supseteq J$ (of whatever cardinality) you can define a cut as: $x \in A$ iff $x \in I$ and $\exists \alpha \in \omega_1 (x<\alpha) $ and $B=I \setminus A$. This cut can't have cofinality $(\omega, \omega)$.

Source Link
Ramiro de la Vega
Ramiro de la Vega
  • 11.5k
  • 1
  • 45
  • 56

I don´t think the statement is true as it is. If $J=\omega_1+\omega_1^*$ then for any $I \supseteq J$ (of whatever cardinality) you can define a cut as: $x \in A$ iff $x \in I$ and $\exists \alpha \in \omega_1 (x<\alpha) $ and $B=I \setminus A$. This cut can't have cofinality $(\omega, \omega)$.