If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?
Solution 1:
Yes, mostly. It's the least cardinal for which there are infinitely many infinite cardinals below. And it is usually denoted $\aleph_\omega$. (Where $\omega$ denotes the least infinite ordinal; of course, $\omega = \aleph_0$, but we use $\omega$ to indicate that we are interested in ordinal properties of this object. Another description for $\aleph_\omega$ is that it is the $\omega$'th infinite cardinal, or the unique cardinal such that the order-type of the family of all infinite cardinals below it is $\omega$.)
Solution 2:
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 $\alpha$.
Solution 3:
I realize this question has been answered but I would like to provide a way of visualizing $\aleph_\omega$ and $\aleph_{\omega_\omega}$:
Start with $\aleph_0$ and consider the sequence $\aleph_0, \aleph_1,... \aleph_n,...$ for $n \in \omega.$
This is a strictly monotone increasing sequence that has a limit, namely $\aleph_\omega$, which would make $\aleph_\omega$ the smallest uncountable limit cardinal. This number does have applications. Example: if GHC holds and $\aleph_\lambda = \beth_\lambda$ for all $\lambda$, then $|V_{\omega +\omega}|=\beth_\omega=\aleph_{\omega}$. In other words $\aleph_\omega$ would be the cardinality of the von Neumann universe $V_{\omega +\omega}$ which is a universe sufficient for most ordinary mathematics, and is a model of Zermelo set theory.
As to $\aleph_{\omega_\omega}$, consider the sequence $\aleph_\omega, \aleph_{\omega+1},...,\aleph_{\omega^2},...,\aleph_{\omega^\omega},...,\aleph_{\epsilon_0},...$
Notice that this also is a strictly monotone increasing sequence which is countable, the supremum of which is a $\aleph_{\omega_1}$. Continuing on like this yields $\aleph_{\omega_1}, \aleph_{\omega_2},...,\aleph_{\omega_n},...$ the limit of which is $\aleph_{\omega_\omega}$