TITLE: Intuition about ordinal fixed points $\alpha = \aleph_\alpha$
QUESTION [11 upvotes]: I wanted to ask for your intuition about ordinal fixed points $\alpha = \aleph_\alpha$, where $\aleph_\alpha$ stands for the $\alpha$-th Aleph number in the Aleph sequence of cardinalities.
For background why I am asking this. I was surprised when I first learned $|\mathbb{Q}| = |\mathbb{N}|$ and $|{\cal P}(A)|>|A|$ for all sets $A$, but gained an intuition over time. It took me a bit longer to understand the convergence proof for Goodstein sequences, because initially I did not understand why a strictly decreasing sequence of ordinals is zero after finitely many steps; I had the wrong intuition about well-ordering (I thought "going downstairs" would be symmetric to "going upstairs").
Now, I am still unable to find the right intuition for ordinal fixed points, especially for the Aleph sequence. I am aware of the fixed-point lemma for normal functions of ordinals from Veblen. But I have not really gained an intuition from knowing the proof. In a sense, I can understand the proof formally only, but not "morally".
In my intuition (which might be wrong), I am starting from
$0\mapsto\aleph_0\\
1\mapsto\aleph_1\\
2\mapsto\aleph_2$
and so on. The difference in size between the ordinal index $\alpha$ and the Aleph number $\aleph_\alpha$ gets enormously big in a very fast way. My intuition is, the index $\alpha$ can never catch up with $\aleph_\alpha$, even when $\alpha$ is a limit ordinal. In my mental picture, a limit ordinal $\alpha$ can be an extremely "high jump", but it can never really catch up with all the extremely high jumps of the Aleph sequence $(\aleph_\alpha)$ which happen in every single step.
Please could you help me find the right intuition, or maybe point me to the error in my current mental picture? I might be overlooking something obvious!

REPLY [11 votes]: Your intuition is finitary, and therefore wrong. Compare, for example, the two sequences:

$\alpha_n=n$, and
$\beta_n=2^n$.

It is easy to see that $\alpha_n<\beta_n$ for all $n$. We even know from elementary calculus that the rate of change between them is growing very fast as well, so there is no possible way for $\alpha_n$ to be equal to $\beta_n$. Game over, then, I suppose. We can all stop reading this and go get a beer.
But wait a minute, you might say, what about their limit? What is $\sup\alpha_n$ and how does it compare to $\sup\beta_n$? Well, both are $\omega$.
See, limit steps are a great opportunity for the slow and steady to catch up with the rapid. As long as the two sequences are continuous and increasing, they will catch up with one another at some limit points.
And so indeed, setting $\alpha_0=\omega$ and $\alpha_{n+1}=\omega_{\alpha_n}$, will have you with $\alpha=\sup\alpha_n=\omega_\alpha$. And therefore $\alpha=\aleph_\alpha$. The key point here is that the sequence is increasing incredibly fast. Enough to catch up at the limit point. And of course we may replace $\alpha_0$ with any ordinal as our starting point.