Lets temporarily adopt the (non-standard) definition that for all cardinals $\kappa$ and $\nu$ we have $[\kappa,\nu) = \{\mbox{cardinals } \eta \mid \kappa \leq \eta < \nu\}.$ Note the word cardinals. Furthermore, write $2^\kappa$ for cardinal exponentiation (rather than ordinal). Under these definitions, it is trivial to see that $$\mathrm{ord}[\aleph_\alpha, \aleph_{\alpha+\beta}) = \beta.$$

Now define a function $f$ such that

$$f(\alpha,\beta)=\mathrm{ord}[2^{\aleph_\alpha}, 2^{\aleph_{\alpha+\beta}}).$$

Then GCH proves that $f(\alpha,\beta)=\beta.$ My question is, in the absence of GCH, what can ZFC tell us about $f(\alpha,\beta)$? Is it necessarily independent of $\alpha$? And, does our ability to pin down $f$ fall prey to Easton's theorem, or something like it? (I presume so.)


1 Answer 1


First of all, GCH doesn't quite imply $f(\alpha,\beta)=\beta$. Suppose, for example, that $\alpha=0$ and $\beta=\omega$. Then GCH says $$ f(0,\omega)=\text{ord}[\aleph_1,\aleph_{\omega+1})=\omega+1. $$ Under GCH, this sort of problem is the only exception to your equation $f(\alpha,\beta)=\beta$; when $\beta$ is a limit ordinal you have to add $1$ to get hte right value for $f$.

In the absence of GCH, though, there is practically nothing to be said about $f(\alpha,\beta)$, except for trivialities like "weakly increasing as a function of $\beta$ when $\alpha$ is fixed." In particular, $f(\alpha,\beta)$ need not be independent of $\alpha$. For example, ZFC allows the possibility that GCH holds for the first two alephs but then $2^{\aleph_2}=\aleph_8$. Then $f(0,1)=1$ but $f(1,1)=6$.


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