S}_\alpha\setminus \{\emptyset\}$ is not maximal, then $(A\cup\{k\})_{k>\max(A)}\subset {\mathcal S}_\alpha$.

Secondly, by transfinite induction we can easily prove that ${\mathcal S}_\alpha$ is “almost” increasing in $\alpha$, in the following sense:

\[P:2.2\] For all ordinals $\alpha<\beta<\omega$, there is an $n\in{\mathbb{N}}$ so that $${\mathcal S}_\alpha\cap[n,\infty)^{<\omega}\subset {\mathcal S}_\beta.$$

The following formula for ${\text{{\rm CB}}}({\mathcal S}_\alpha)$ is well know and can easily be shown by transfinite induction for all $\alpha<\omega_1$.

\[P:2.3\] For $\alpha<\omega_1$ we have ${\text{{\rm CB}}}({\mathcal S}_\alpha)=\omega^\alpha+1$.

We now make further assumptions on the approximating sequence $(\lambda(\alpha,n))\subset [1,\alpha)$, we had chosen to define the Schreier family $\ S_\alpha$, for limit ordinals $\alpha<\omega_1$. And we will choose $\big(\lambda(\alpha,n)\big)$ recursively. Assume that $\alpha$ is a countable limit ordinal and that we have defined $\big(\lambda(\gamma,n)\big)$, for all limit ordinals $\gamma<\alpha$, and thus, ${\mathcal S}_\gamma$ for all $\gamma<\alpha$.

Recall that $\alpha$, can be represented uniquely in its [*Cantor Normal Form* ]{} $$\label{E:2.7}
\alpha=\omega^{\xi_k} m_k+\omega^{\xi_{k-1}} m_{k-1}+\ldots+\omega^{\xi_1} m_1,$$ where $\xi_k>\xi_{k-1}>\ldots \xi_1$, $m_k,m_{k-1},\ldots, m_1\in {\mathbb{N}}$, and, since $\alpha$ is a limit ordinal, $\xi_1\ge 1$.

We distinguish between three cases:

Case 1. $k\ge 2$ or $m_1\ge 2$. In that case we put for $n\in{\mathbb{N}}$ $$\label{E:2.8}
 \lambda(\alpha,n)=\omega^{\xi_k} m_k+\omega^{\xi_{k-1}} m_{k-1}+\ldots+\omega^{\xi_1} (m_1-1) +\lambda(\omega^{\xi_1},n)$$ Before considering the next cases we need to make the following observation:

