TITLE: Intuition for $\omega^\omega$
QUESTION [24 upvotes]: I'm trying to understand the ordinal number $\omega^\omega$ and I'm having a hard time. I think I understand what $\omega^2$ is. It's what I would get if I took countably many copies of $\omega$ and order the set of those copies like $\omega.$ $\omega^3$ is a bit more difficult because I need to take countably many copies of $\omega^2$ now and do the same with them. I understand that the next ones are made in the same way, recursively. But understanding that the definition has meaning doesn't help me very much with visualizing $\omega^{10}$. And I become completely helpless when I try to tell myself what $\omega^\omega$ is in plain words. Is some explanation possible that is not just the definition (which I know)?

REPLY [8 votes]: Here's how I visualize it. Consider this subset of $[0,1]$:
$$E_1 \;\; = \;\; \left\{\frac{1}{2}, \;\; \frac{2}{3}, \;\; \frac{3}{4}, \;\; \frac{4}{5}, \;\; \frac{5}{6}, \;\; \frac{6}{7}, \;\; \ldots \right \}$$
This is $\omega$.
More precisely, when ordered by the usual ordering for real numbers, $E_1$ becomes a representation for the order type $\omega.$
Now consider the union of $E_1$ and its $1$-unit translate (a subset of $[1,2]$), which will be a subset of $[0,2]$.
This is $\omega 2$.
By taking the unions of various appropriate integer translates of $E_1$, we get $\omega 3, \; \omega 4, \; \omega 5, \; \ldots$
By putting an integer translate of $E_1$ into the interval $[n, \; n+1],$ for each non-negative integer $n,$ and taking the union of these countable many translates of $E_1,$ we get $\omega \omega = {\omega}^2.$
If we take this last set, the set representing ${\omega}^2$, and scale/translate it into the interval $[0,1]$ (do this in a way the preserves order, which will be the case if you use a positive scale factor), we'll have a representation of ${\omega}^2$ that is a subset of $[0,1].$ Call this last set $E_2.$
The union of $E_2$ with its $1$-translate will be ${\omega}^2 2$.
By taking the unions of various appropriate integer translates of $E_2$, we get ${\omega}^2 3, \; {\omega}^2 4, \; {\omega}^2 5, \; \ldots$
You can also get things like ${\omega}^2 3 + \omega 2 + 4$ by taking the union of a set in $[0,3]$ that represents ${\omega}^2 3$ with a set in $[4,5]$ that represents $\omega$ with a set in $[5,6]$ that represents $\omega$ with the set $\left\{6 + \frac{1}{2}, \; 6 + \frac{2}{3}, \; 6 + \frac{3}{4}, \; 6 + \frac{4}{5} \right\}.$
By putting an integer translate of $E_2$ into the interval $[n, \; n+1],$ for each non-negative integer $n,$ and taking the union of these countable many translates of $E_2,$ we get ${\omega}^2 \omega = {\omega}^3.$
If we take this last set, the set representing ${\omega}^3$, and scale/translate it into the interval $[0,1]$ (do this in a way the preserves order, which will be the case if you use a positive scale factor), we'll have a representation of ${\omega}^3$ that is a subset of $[0,1].$ Call this last set $E_3.$
Keep going in this manner, obtaining $E_4 \subseteq [0,1]$ that is a representation of ${\omega}^4,$ $E_5 \subseteq [0,1]$ that is a representation of ${\omega}^5,$ 
$E_6 \subseteq [0,1]$ that is a representation of ${\omega}^6, \; \ldots$ 
Then ${\omega}^{\omega}$ can be represented by
$$ E_1 \; \cup \; \left(E_2 + 1 \right) \; \cup \; \left(E_3 + 2 \right)\; \cup \; \left(E_4 + 3 \right)\; \cup \; \left(E_5 + 4 \right) \; \cup \; \ldots,$$
where $E_{n+1} + n$ represents the $n$-translate of $E_{n+1}$ for $n = 1, \; 2, \; 3, \; \ldots$