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

= 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 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 &gamma; has a cardinality $$\vert \gamma \vert = \aleph_{\delta} \,$$ where &delta; < &alpha;. :The cardinality of the set of functions so bounded by &gamma; 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 &gamma; 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)