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June 6

Why does the generalized continuum hypothesis implies the following equality for cardinal exponentiation?

The article about the continuum hypothesis states that the the generalized continuum hypothesis implies the following:

${\mathrm{\aleph }}_{\alpha }^{{\mathrm{\aleph }}_{\beta }}={\mathrm{\aleph }}_{\alpha }$ when β+1 < α and ${\mathrm{\aleph }}_{\beta }<\mathrm{cf}({\mathrm{\aleph }}_{\alpha })$, where cf is the cofinality operation.

How does one prove that?

Thanks!

Dan Gluck (talk) 12:59, 6 June 2020 (UTC)

See Talk:Continuum hypothesis/Archive 1#Proofs of certain consequences of GCH where I justify the claims in the article. Here I extract the relevant portion:

${\mathrm{\aleph }}_{\alpha }={\mathrm{\aleph }}_{\alpha }^1\le {\mathrm{\aleph }}_{\alpha }^{{\mathrm{\aleph }}_{\beta }}.$

Since we are in the case that ${\mathrm{\aleph }}_{\beta }<\mathrm{cf}({\mathrm{\aleph }}_{\alpha })$ where cf is the cofinality operation, then any function from ${\mathrm{\aleph }}_{\beta }$ to ${\mathrm{\aleph }}_{\alpha }$ must be bounded above by some $\gamma <{\mathrm{\aleph }}_{\alpha }.$

And γ has a cardinality $|\gamma |={\mathrm{\aleph }}_{\delta }$ where δ < α. :The cardinality of the set of functions so bounded by γ is

${\mathrm{\aleph }}_{\delta }^{{\mathrm{\aleph }}_{\beta }}\le 2^{({\mathrm{\aleph }}_{\delta }\cdot {\mathrm{\aleph }}_{\beta })}={\mathrm{\aleph }}_{\max (\delta ,\beta )+1}\le {\mathrm{\aleph }}_{\alpha }.$

Adding these together for the ${\mathrm{\aleph }}_{\alpha }$ possible values of γ gives

${\mathrm{\aleph }}_{\alpha }^{{\mathrm{\aleph }}_{\beta }}\le {\mathrm{\aleph }}_{\alpha }\cdot {\mathrm{\aleph }}_{\alpha }={\mathrm{\aleph }}_{\alpha }$

which means

${\mathrm{\aleph }}_{\alpha }^{{\mathrm{\aleph }}_{\beta }}={\mathrm{\aleph }}_{\alpha }.$

OK? JRSpriggs (talk) 04:42, 7 June 2020 (UTC)