the sequence $\big(\lambda(\alpha,n):n\in{\mathbb{N}}\big)$: $$\label{E:2.10}
{\mathcal S}_{\lambda(\alpha,n)}\subset {\mathcal S}_{\lambda(\alpha, n+1)}, \text{ for all $n\in{\mathbb{N}}$}.$$ We can assure as follows: first choose any sequence $\lambda'(\alpha,n)$, which increases to $\alpha$. Then we notice that Proposition \[P:2.2\] yields that for a fast enough increasing sequence $(l_n)\subset {\mathbb{N}}$, it follows that ${\mathcal S}_{\lambda'(\alpha,n)+l_n}\subset {\mathcal S}_{\lambda'(\alpha,n+1)+l_{n+1}}$. Indeed, we first note that the only set $A\in{\mathcal S}_\gamma$, $\gamma<\alpha$ which contains $1$, must be the singleton $A=\{1\}$. This follows easily by induction. Secondly we note that by it follows that $[\{2,3,\ldots,n\}]\subset {\mathcal S}_{\gamma+n}$, for each $\gamma<\alpha$, and $n\in{\mathbb{N}}$, which yields our claim.

The remaining case is the following

Case 3. $\alpha=\omega^{\omega^\kappa+\xi}$, where $1\le \xi\le \omega^\kappa$.

We first observe that in this case $\kappa$ and $\xi$ are uniquely defined.

\[L:2.5\] Let $\alpha$ be an ordinal number so that there are ordinal numbers $\kappa$, $\xi$ with $\xi \le \omega^\kappa$ and $\alpha = \omega^{\omega^\kappa + \xi}$. Then for every $\kappa'$, $\xi'$ with $\xi' \le \omega^{\kappa'}$ so that $\alpha = \omega^{\omega^{\kappa'}+\xi'}$, we have $\kappa = \kappa'$ and $\xi = \xi'$.

Let $\alpha = \omega^{\omega^\kappa + \xi} = \omega^{\omega^{\kappa'}+\xi'}$ be as above. By [@Su Theorem 41, §7.2] $\omega^\kappa+\xi = \omega^{\kappa'}+\xi'$. If $\kappa'<\kappa$, then $\omega^{\kappa'}+\xi' \le \omega^{\kappa'}2 < \omega^{\kappa'}\omega = \omega^{\kappa'+1} \le \omega^{\kappa} \le 