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!

  • 4
    $\begingroup$ I would comment that "${\aleph}_{\aleph_0}$ is very poor notation. The alephs are indexed by ordinals, not cardinals, and the alephs represent cardinals. So it should really be $\aleph_{\omega}$, not $\aleph_{\aleph_0}$. Note for example that while $\aleph_{\omega+1}$ is different from $\aleph_{\omega}$ (which is easy to spot since $\omega+1\neq\omega$ as ordinals), you would run into interpretation issues with $\aleph_{\aleph_0+1}$, since with cardinals, $\aleph_0+1=\aleph_0$. $\endgroup$ Commented May 27, 2019 at 3:56

1 Answer 1



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.)

  • $\begingroup$ So, for part $d)$ we will have $cf((\aleph_{\aleph_1}))=\aleph_1\leq \aleph_2$ and the answer is $\aleph(\aleph_{\aleph_1})$? $\endgroup$
    – Galymbek
    Commented May 27, 2019 at 3:07
  • $\begingroup$ @Galymbek Yes, that's right. $\endgroup$ Commented May 27, 2019 at 3:13
  • $\begingroup$ Yes, I used the notation above to mean the cardinal successor. But I have one more little question. What is the cardinal successor of $\aleph_{\aleph_1}$? $\endgroup$
    – Galymbek
    Commented May 27, 2019 at 3:20
  • $\begingroup$ @Galymbek $\aleph_\alpha$ is a sequence enumerating the cardinals. So the next cardinal is the next one in the enumeration. In other words, $\aleph_\alpha^+ = \aleph_{\alpha+1}.$ Or in this case $\aleph_{\omega_1}^+=\aleph_{\omega_1+1}$ $\endgroup$ Commented May 27, 2019 at 3:21
  • 2
    $\begingroup$ @Galymbek Yep, although this is a good example of why, as I mentioned in the answer, using cardinal notation in the index rubs me the wrong way. For the $\aleph_1+1$, we want ordinal sum, but the notation suggests cardinal sum. I would write it $\aleph_{\omega_1+1}.$ $\endgroup$ Commented May 27, 2019 at 3:24

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