Define $\tau: \mathbf{Ord} \to \mathbf{Ord}$ such that $\tau(\alpha)$ is the order type of the minimal set $S$ of ordinals such that $\alpha \in S$ and $S$ is closed under ordinal exponentiation.


We have the following:
$\tau(0)=2, \tau(1)=1$ and for all $\alpha \in [2, \omega), \tau(\alpha)=\omega$.


*Question*: What are the values $\tau(\alpha)$ for $\alpha\ge\omega$?



Order type of the minimal set closed under ordinal exponentiation
Take $\alpha$ to be any ordinal greater than or equal to $\omega$. The set of ordinals $S(\alpha)$ obtained generated from $\alpha$ using ordinal exponentiation are the ordinals of the form $\alpha^{\alpha^{E(\alpha)}}$ where $E(\alpha)$ is an exponential polynomial over the base $\alpha$. By "exponential polynomial over the base $\alpha$" I mean the set of ordinals defined by:


$E\_0(\alpha)$ = the finite ordinals


$E\_{n+1}(\alpha) = \lbrace 0 \rbrace \cup \lbrace\alpha^{\beta\_1} + \alpha^{\beta\_2} + \ldots + \alpha^{\beta\_n} | \beta\_i \in E\_n(\alpha) \& \beta\_1 \geq \beta\_2 \geq \ldots \geq \beta\_n \rbrace$


$E(\alpha) = \bigcup E\_n(\alpha)$


First, observe that $\alpha = \alpha^{\alpha^0}$, and for arbitrary $\beta, \gamma$, $(\alpha^{\alpha^\beta})^{\alpha^{\alpha^\gamma}} = \alpha^{\alpha^{\beta + \alpha^\gamma}}$. So every element of $S(\alpha)$ is of the form $\alpha^{\alpha^\beta}$. It remains to show that the set of ordinals $T(\alpha)$ generated from $0$ using the function $\beta, \gamma \rightarrow \beta + \alpha^\gamma$ is exactly the exponential polynomials. It is easy to see that the exponential polynomials are closed under $\beta, \gamma \rightarrow \beta + \alpha^\gamma$; going the other direction, $0 + \alpha^0 + \ldots \alpha^0 = n$, so $E\_0(\alpha) \subset T(\alpha)$. Assume $E\_n(\alpha) \subset T(\alpha)$; for $\beta\_1, \beta\_2, \ldots, \beta\_n \in E\_n(\alpha) \subset T(\alpha)$, we have $0 + \alpha^{\beta\_1} + \alpha^{\beta\_2} + \ldots + \alpha^{\beta\_n} \in T(\alpha)$. So $E\_{n+1}(\alpha) \subset T(\alpha)$, and by induction, $E(\alpha) \subset T(\alpha)$. So $T(\alpha) = E(\alpha)$, and $S(\alpha)$ is as we described.


By an extension of Cantor's Normal Form Theorem, for any ordinal $\alpha$ and any ordinal $\beta$, $\beta$ is uniquely expressible in the form


$\alpha^{\beta\_1} \gamma\_1 + \alpha^{\beta\_2} \gamma\_2 + \ldots + \alpha^{\beta\_n} \gamma\_n$, where $\beta\_1 > \beta\_2 > \ldots > \beta\_n$ and $\gamma\_i < \alpha$ for all $1 \leq i \leq n$,


and more importantly, different values of $\beta\_i$ and $\gamma\_i$ lead to different values of $\beta$.


It follows that the set $E(\alpha)$, when iteratively described using the above form, has a unique expression for each ordinal, and different expressions lead to different ordinals. Indeed, $E\_0(\alpha)$ has a different expression $\alpha^0 + \ldots + \alpha^0$ for each finite ordinal $n$. Next assume that different expressions lead to different ordinals in $E\_n(\alpha)$. Given an pair of expressions $\beta = \alpha^{e(\beta\_1)} + \ldots + \alpha^{e(\beta\_n)}$ and $\gamma = \alpha^{e(\gamma\_1)} + \ldots + \alpha^{e(\gamma\_n)}$ for $\beta\_i \in E\_n(\alpha), \gamma\_i \in E\_n(\alpha)$, $\beta\_i$ and $\gamma\_i$ weakly decreasing, and $e(\beta\_i)$ and $e(\gamma\_i)$ representing expressions for $\beta\_i$ and $\gamma\_i$; if the two expressions are different, then for some $i$ we must have $e(\beta\_i) \neq e(\gamma\_i)$, and by the induction hypothesis $\beta\_i$ is different from $\gamma\_i$. But then by the extended Cantor Normal Form Theorem $\beta$ and $\gamma$ are different. So different expressions of $E\_{n+1}(\alpha)$ represent different ordinals, so by induction different expressions of $E(\alpha)$ lead to different ordinals.


So for any ordinal $\alpha \geq \omega$, we can define an order-preserving bijection $E(\alpha) \rightarrow \varepsilon\_0$ by simply replacing $\alpha$ with $\omega$ for every appearance in the iterative normal form expression. So $\tau (\alpha) = \varepsilon\_0$.


