\omega^{\kappa} + \xi$, which is a contradiction. We conclude that $\kappa\le \kappa'$, and therefore by interchanging the roles of $\kappa$ and $\kappa'$ we obtain that $\kappa = \kappa'$. In conclusion, $\omega^\kappa + \xi = \omega^\kappa + \xi'$ and therefore $\xi = \xi'$ as well.

We choose now a sequence $(\theta(\xi,n))_n$ of ordinal numbers increasing to $\omega^\xi$, so that $$\label{E:2.11}
\mathcal{S}_{\omega^{\omega^\kappa}\theta(\xi,n)}\subset \mathcal{S}_{\omega^{\omega^\kappa}\theta(\xi,n+1)}$$ and define $$\label{E:2.12} \lambda(\alpha,n) = \omega^{\omega^\kappa}\theta(\xi,n).$$ We describe how can be obtained. Start with an arbitrary sequence $(\theta'(\xi,n))_n$ increasing to $\omega^\xi$. We shall recursively choose natural numbers $(k_n)_{n\in{\mathbb{N}}}$, so that setting $\theta(\xi,n) = \theta'(\xi,n) + k_n$, is satisfied. Assuming that $k_1,\ldots,k_n$ have been chosen choose $k_{n+1}$,, as in the argument yielding , so that $$\mathcal{S}_{\omega^{\omega^\kappa}\theta(\xi,n)}\subset \mathcal{S}_{\omega^{\omega^\kappa}\theta'(\xi,n+1) + k_{n+1}}.$$ We will show that this $k_{n+1}$ is the desired natural number, i.e. that $$\mathcal{S}_{\omega^{\omega^\kappa}\theta(\xi,n)}\subset \mathcal{S}_{\omega^{\omega^\kappa}(\theta'(\xi,n+1) + k_{n+1})}.$$ First note that using finite induction and Proposition , it is easy to verify that for $\gamma<\alpha$, with $\gamma=\omega^\xi$, for some $\xi<\omega_1$, and $n{\!\in\!}{\mathbb{N}}$ $$\label{E:2.13}
{\mathcal S}_{\gamma\cdot n} = \underbrace{{\mathcal S}_{\gamma}[{\mathcal S}_{\gamma}\cdots{\mathcal S}_{\gamma}[{\mathcal S}_{\gamma}]]]}_{n\text{-times}}.$$ and thus $$\begin{aligned}
\mathcal{S}_{\omega^{\omega^\kappa}(\theta'(\xi,n+1) + k_{n+1})} &=
 ={\mathcal S}_{\omega^{\omega^\kappa} \Theta'(\xi,n+1)+\omega^{\omega^\kappa} k_{n+1}}=
 \underbrace{{\mathcal S}_{\omega^{\omega^\kappa}}[\cdots [{\mathcal S}_{\omega^{\omega^\kappa}}}_{k_{n+1}\text{-times}}[{\mathcal S}_{\omega^{\omega^\kappa}\theta'(\xi,n+1)}]]] \\
&\supset\underbrace{{\mathcal S}_{1}[\cdots [{\mathcal S}_{1}}_{k_{n+1}\text{-times}}[{\mathcal S}_{\omega^{\omega^\kappa}\theta'(\xi,n+1)}]]]
= \mathcal{S}_{\omega^{\omega^\kappa}\theta'(\xi,n+1) + k_{n+1}} \supset \mathcal{S}_{\omega^{\omega^\kappa}\theta(\xi,n)}.\end{aligned}$$ We point 