Skip to main content

Maybe a comment:

In the paper ``A large cardinal in the constructible universe'' Silver shows that if $\kappa\to (\alpha)^{<\aleph_0}$ for all countable $\alpha,$ then the same is true for $\kappa$ in $L$.

On the other hand, by a result of Rowbottom, $\kappa \to (\omega_1)^{<\aleph_0} $ contradicts $V=L.$

Silver concludes with the following:

It does not seem extravagant, then, to assert that, for all particularpractical purposes, $\kappa \to (\omega)^{<\aleph_0}$ is the strongest strong axiom of infinity know to be consistent with $V=L,$ and therein lies its chief interest.

Maybe a comment:

In the paper ``A large cardinal in the constructible universe'' Silver shows that if $\kappa\to (\alpha)^{<\aleph_0}$ for all countable $\alpha,$ then the same is true for $\kappa$ in $L$.

On the other hand, by a result of Rowbottom, $\kappa \to (\omega_1)^{<\aleph_0} $ contradicts $V=L.$

Silver concludes with the following:

It does not seem extravagant, then, to assert that, for all particular purposes, $\kappa \to (\omega)^{<\aleph_0}$ is the strongest strong axiom of infinity know to be consistent with $V=L,$ and therein lies its chief interest.

Maybe a comment:

In the paper ``A large cardinal in the constructible universe'' Silver shows that if $\kappa\to (\alpha)^{<\aleph_0}$ for all countable $\alpha,$ then the same is true for $\kappa$ in $L$.

On the other hand, by a result of Rowbottom, $\kappa \to (\omega_1)^{<\aleph_0} $ contradicts $V=L.$

Silver concludes with the following:

It does not seem extravagant, then, to assert that, for all practical purposes, $\kappa \to (\omega)^{<\aleph_0}$ is the strongest strong axiom of infinity know to be consistent with $V=L,$ and therein lies its chief interest.

Source Link
Mohammad Golshani
  • 31.5k
  • 2
  • 97
  • 195

Maybe a comment:

In the paper ``A large cardinal in the constructible universe'' Silver shows that if $\kappa\to (\alpha)^{<\aleph_0}$ for all countable $\alpha,$ then the same is true for $\kappa$ in $L$.

On the other hand, by a result of Rowbottom, $\kappa \to (\omega_1)^{<\aleph_0} $ contradicts $V=L.$

Silver concludes with the following:

It does not seem extravagant, then, to assert that, for all particular purposes, $\kappa \to (\omega)^{<\aleph_0}$ is the strongest strong axiom of infinity know to be consistent with $V=L,$ and therein lies its chief interest.