Arithmetic on infinite cardinal numbers

Question

I am stuck on the following problem that says:

Assuming the Generalized Continuum Hypothesis (GCH), that is, the statement $2^{\aleph_{\alpha}}$ = $\aleph_{\alpha+1}$ for every ordinal $\alpha$, find the corresponding $\aleph$ numbers of the following computations in cardinal arithmetic:

$a$) $\aleph_0^{\aleph_1^{\aleph_2}}$

$b)$ $(\aleph_{\aleph_0})^{\aleph_{\aleph_1}}$

$c)$ $(\aleph_{\aleph_\omega})^{2^{\aleph_7}}$

$d$) $(\aleph_{\aleph_1})^{\aleph_2}$

My Attempt:

For parts $a)$ and $b)$ I know that

If GCH holds, then if $x,k$ are both infinite cardinals:

1) if $k\leq x$, then $k^x=x+$

2) if $cfk\leq x$, then $k^x=k+$

3) if $x<cfk$, then $k^x=k$

So, I found that, $a)=\aleph_4$, $b)$ $\aleph(\aleph_{\aleph_1})$

But, in parts $c)$ and $d)$ I couldn't compute the cofinity of these numbers,

Can someone help me out? Thanks in advance for your time!

Answer 1

HINT

If $\alpha$ is a limit ordinal, then the cofinality of $\aleph_\alpha$ is the same as the cofinality of $\alpha.$ This will allow you to compute the cofinalities.

Your answer for (a) is correct, but I'm not sure I understand your notation for (b). Since $\aleph_{\omega}<\aleph_{\omega_1},$ we have $(\aleph_\omega)^{\aleph_{\omega_1}} = 2^{\aleph_{\omega_1}} = \aleph_{\omega_1+1}.$ Perhaps you are using $\aleph(\cdot)$ as a function to mean "cardinal successor" (I have seen this notation for Hartog's number, which I suppose is the same thing as the cardinal successor here.)

(Also on a minor note, as suggested by the notation I used there, indexing an aleph with something in cardinal notation (as in $\aleph_{\aleph_0}$ or $\aleph_{\aleph_1}$) doesn't really feel right to me. The index is best thought of as an ordinal, so I think notation should reflect that.)