If $f,g:\Bbb R \to \Bbb R$ define $f<g \iff \exists M:\forall x>M$, $f(x)<g(x)$ [Does this mean that $g$ dominates $f$].
Now define $\hat f$ to be the smallest set of functions containing $\Bbb N[x]$ and closed under the operation $f\mapsto x^{f}$ and $(f,g)\mapsto f+g$
[e,g $x^{x^{3x+2}+5x^{x}}+2x+4\in \hat f$, but $(x+1)^{x} \notin \hat f$] I want to prove that $<$ defined above on $\hat f$ is a well-ordering of order type $\epsilon_0$.
My attempt:
Recall that a set $x$ is said to be well oredered by $<$ if $x$ is linearly ordered by $<$ and has the property if $A\subseteq X$ non- empty then $A$ contains a least element i,e $\exists x \in A : x\le a ,\forall a \in A$.
I need to show that $(\hat f,<)\cong (\epsilon_{0},<)$ to ensure that it is a well ordering of type $\epsilon_0$ that is i need to find a Bijection?
If $\hat{f}$ is also stable under sum, then you could try to prove that the map recursively defined on $\varepsilon_0$ by $\varphi(0) = 0$ and $\varphi(\sum \limits_{k=0}^p \omega^{\alpha_k}.n_k) := \sum \limits_{k=0}^p x^{\varphi(\alpha_k)}.n_k$ is an isomorphism.
The best way to do this is to prove it by induction on the least integer $n_{\lambda}$ such that $\lambda \in x_n:= \omega^{\omega^{... \omega}}$ where $\omega$ appears $(n+1)$-times: prove that $\varphi$ is well-defined on $x_n$, strictly increasing, and that $\hat{f} = \bigcup \limits_{n \in \mathbb{N}} \varphi(x_n)$.