Limits of transfinite numbers

Question

Does it makes sense to talk about:

$$ \lim_{i\to \aleph_0} \aleph_i $$

What type of infinity does it approach?

Maybe finding a limit of that doesn't make sense.

What about $\aleph_{\aleph_0}$? What type of infinity is that?

Answer 1

Yes it makes perfect sense. And $\aleph$ "commutes with lim" so that

$$ \begin{align} \lim_{n\to \aleph_0} \aleph_n &= \aleph_{lim_{n\to \aleph_0}} \\ &= \aleph_{\aleph_0} \end{align} $$

which is the least cardinal greater than all $\aleph_n$ for $n < \aleph_0$. Amongst the infinite cardinals, it's the least singular cardinal; all smaller infinite cardinals are regular. (See https://en.wikipedia.org/wiki/Regular_cardinal for definitions.) One usually writes it as $\aleph_{\omega}$ – as you probably know, $\omega = \aleph_0$, it's $\aleph_0$ regarded as an ordinal, and $\aleph_{\alpha}$ is defined for all ordinals $\alpha$ (and therefore for all cardinals).

Another characteristic of $\aleph_{\omega}$, which follows from it being singular: it's a limit cardinal, an infinite cardinal with no immediately smaller cardinal, i.e. it's not the successor of another cardinal. Every limit cardinal is of the form $\aleph_{\lambda}$ for $\lambda$ either $0$ or a limit ordinal. Thus $\aleph_0$ is a the first limit cardinal, and $\aleph_{\omega}$ is the second. For every ordinal $\alpha$ (in particular for every integer), $\aleph_{\alpha +1}$ is the successor of ${\aleph_{\alpha}}$, and all successors are regular.