If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?
Apologies if this isn't a sensible question, I really don't know too much about these infinite cardinals aside from the basics. I did, however, think that the idea of the "aleph-null"th aleph number was interesting enough to base my username on and my own attempts did not prove fruitful, so I was wondering if anyone here could shed some light. Thanks!
For clarity: I'm asking about $\aleph_{\aleph_0}$ . Thanks!
P.S. I was somewhat unsure about the tags for this, sorry if I accidentally placed it in the wrong category.
Best Answer
I'd like to elaborate on some of the fine points that Arthur raised.
The $\aleph$ numbers (also the $\beth$ numbers) are used to denote cardinals. However one of the key features of cardinals is that we can say "the next cardinal", and we can say which cardinal came first and which came second. These are ordinal properties.
Note that the least cardinal greater than $\aleph_{\aleph_0}$ also has countably many [infinite] cardinals smaller than itself. But since $\aleph_0+1=\aleph_0$, what sense would that make?
So we are using the ordinals. It's a fine point, because the finite notions coincide, the finite cardinals are the finite ordinals, and it's not until we reach the infinite ordinals that we run into the difference between $\omega$ and $\aleph_0$.
Therefore, instead of $\aleph_{\aleph_0}$ we have $\aleph_\omega$, then we have $\aleph_{\omega+1}$ and so on and so forth. After we have gone through uncountably many of these we finally have $\aleph_{\omega_1}$, where $\omega_1$ is the least uncountable ordinal -- which corresponds to $\aleph_1$.
And so on and so forth. For every ordinal $\alpha$ we have $\aleph_\alpha$ is the unique cardinal that the infinite cardinals below it have the same order type as $\alpha$.