${\omega }^2$, ${\omega }^{\omega }$, and ${\epsilon }_0=\sup \{\omega ,{\omega }^{\omega },{\omega }^{{\omega }^{\omega }},\dots \}$ satisfying ${\omega }^{{\epsilon }_0}={\epsilon }_0$

The first few powers of $\omega$ are

${\omega }^1=\omega ,{\omega }^2=\omega \cdot \omega ,{\omega }^{\omega }=\sup \{{\omega }^n:0<n<\omega \},$

and each is strictly larger than the one before (${\alpha }^{\beta +\gamma }={\alpha }^{\beta }\cdot {\alpha }^{\gamma }$ and $({\alpha }^{\beta }{)}^{\gamma }={\alpha }^{\beta \cdot \gamma }$; and for $\alpha >1$ exponentiation is strictly increasing with $\beta \le {\alpha }^{\beta }$, clause (b)). Iterating the exponential produces the $\omega$-tower

$t_0=\omega ,t_{n+1}={\omega }^{t_n},$

so $t_1={\omega }^{\omega }$, $t_2={\omega }^{{\omega }^{\omega }}$, and so on. Its supremum

${\epsilon }_0:=\sup \{t_n:n\in \omega \}$

is a limit ordinal satisfying

${\omega }^{{\epsilon }_0}={\epsilon }_0,$

so it is a fixed point of $\beta \mapsto {\omega }^{\beta }$. This is exhibited here by hand: the tower is written down, its supremum is taken, and the fixed point equation is proved from continuity at limits. No fixed-point theorem is used, and none that this library proves applies here: every fixed-point theorem on disk is stated for a set carrying an order or a metric, whereas $\beta \mapsto {\omega }^{\beta }$ is a class operation on the ordinals, which are not a set.

Why Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal and not the recursion theorem over $\mathbb{N}$. The published The recursion theorem builds $g:\mathbb{N}\to A$ from a function $f:A\to A$ on a set $A$. Here the step is $\beta \mapsto {\omega }^{\beta }$, a class operation with no set-sized codomain available at this point, so the recursion theorem does not apply as stated. Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal is exactly the class-valued version, and Replacement then makes the range $\{t_n:n\in \omega \}$ a set.

What is proved and what is not. That ${\epsilon }_0$ is a fixed point of $\beta \mapsto {\omega }^{\beta }$ is proved above. That it is the least such fixed point is true and is not proved here; it would follow from the observation that any fixed point is closed under the tower, and it needs nothing new, but nothing on these pages uses it. No general theory of normal functions or of the Veblen hierarchy is developed, and none is needed for the statement above.

${\epsilon }_0$ and the Cantor normal form. By Cantor normal form: every nonzero ordinal is ${\omega }^{{\beta }_0}\cdot c_0+\cdots +{\omega }^{{\beta }_{k-1}}\cdot c_{k-1}$ with ${\beta }_0>\cdots >{\beta }_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way every nonzero ordinal has a unique base-$\omega$ normal form. For ${\epsilon }_0$ that form is ${\omega }^{{\epsilon }_0}\cdot 1$, whose exponent is ${\epsilon }_0$ itself, so the normal form does not reduce ${\epsilon }_0$ to strictly smaller data. Below ${\epsilon }_0$ it always does, and that is the sense in which ${\epsilon }_0$ is where base-$\omega$ notation runs out.

Cardinality is a separate question, and this page does not settle it. Nothing above says how large ${\omega }^{\omega }$ or ${\epsilon }_0$ is as a set. Showing them at most countable would need the countable ordinals to be closed under ordinal exponentiation, which is not proved anywhere in this library; the natural route runs through Assuming countable choice: every at most countable subset of ${\omega }_1$ is bounded below ${\omega }_1$, so no at most countable subset of ${\omega }_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable and a transfinite induction that no item here carries out. What the tower demonstrates is growth in order type, which is the invariant this page is about.