Arithmetic on infinite cardinal numbers

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 $$x, then $$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!

• 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, $\aleph_0+1=\aleph_0$. Commented May 27, 2019 at 3:56

1 Answer

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.)

• So, for part $d)$ we will have $cf((\aleph_{\aleph_1}))=\aleph_1\leq \aleph_2$ and the answer is $\aleph(\aleph_{\aleph_1})$? Commented May 27, 2019 at 3:07
• @Galymbek Yes, that's right. Commented May 27, 2019 at 3:13
• Yes, I used the notation above to mean the cardinal successor. But I have one more little question. What is the cardinal successor of $\aleph_{\aleph_1}$? Commented May 27, 2019 at 3:20
• @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 $\aleph_{\omega_1}^+=\aleph_{\omega_1+1}$ Commented May 27, 2019 at 3:21
• @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 $\aleph_{\omega_1+1}.$ Commented May 27, 2019 at 3:24