Wikipedia:Reference desk/Archives/Mathematics/2020 June 6

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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:

$\aleph _{\alpha }^{\aleph _{\beta }}=\aleph _{\alpha }$ when β+1 < α and $\aleph _{\beta }<\operatorname {cf} (\aleph _{\alpha })$, 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#Proofs of certain consequences of GCH where I justify the claims in the article. Here I extract the relevant portion:

$\aleph _{\alpha }=\aleph _{\alpha }^{1}\leq \aleph _{\alpha }^{\aleph _{\beta }}\,.$

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

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

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

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

$\aleph _{\alpha }^{\aleph _{\beta }}\leq \aleph _{\alpha }\cdot \aleph _{\alpha }=\aleph _{\alpha }\,$

which means

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

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