Which are the countable ordinals $\lambda$ such that, for every sequence of ordinals $\alpha_i\ (i\in\mathbb{N})$ such that
- $\ $it is strictly increasing for all sufficiently large $i$
- $\ $$\alpha_i<\lambda\ $ for all $i\in\mathbb{N}$
- $\ $$\sup\{ \alpha_i:i\in\mathbb{N}\}=\lambda$,
there exists an ordinal $\beta<\lambda$ such that there exists a sequence $\gamma_i$, with $\gamma_i+\beta=\alpha_i$ for all $i\in\mathbb{N}$, verifying 1.,2. and 3. above.
Furthermore, if such ordinals exist, how are they characterized? Thanks.
Let us say that an ordinal $\xi$ is very good if for any $(\alpha_i)_{i=1}^\infty \subset \xi=[0,\xi)$ such that there exists $i_0\in \mathbb{N}$ such that $(\alpha_i)_{i=i_0}^\infty$ is strictly increasing to $\xi$, there exist $j_0\geqslant i_0$, $\beta\in\xi$, and $(\gamma_i)_{i=j_0}^\infty\subset \xi$ such that $(\gamma_i)_{i=j_0}^\infty$ strictly increases to $\xi$ and $\alpha_i=\gamma_i+\beta$ for all $i\geqslant j_0$.
Recall that a subset $I\subset \xi$ is called cofinal in $\xi$ if $\sup I=\sup\xi$. The cofinality of $\xi$ is the minimum cardinality of a cofinal subset. An ordinal has cofinality $0$ iff it is $0$, it has cofinality $1$ iff it is a successor. In order for there to exist a sequence $(\alpha_i)_{i=1}^\infty$ which is strictly increasing to $\xi$, the cofinality of $\xi$ must be exactly $\aleph_0$. So if $\xi$ is any ordinal with cofinality not equal to $\aleph_0$, it will vacuously be very good, because there will be no $\alpha_i$ as in the hypothesis of the definition of very good to violate the required property. So, for the remainder of the problem, I will amend the definition of very good to require that the ordinal $\xi$ also have cofinality $\aleph_0$ (the non-vacuous case).
Proof: Assume $\xi=\nu+\omega$ and $\alpha_i\uparrow \xi$. Without loss of generality, we can assume $\alpha_i>\nu$ for all $i$, so $\alpha_i=\nu+m_i$ for some $m_i\uparrow \omega$. Let $\beta=1$ and $\gamma_i=\nu+(m_i-1)$. So $\alpha_i=\gamma_i+1=\gamma_i+\beta$ and $\gamma_i\uparrow \xi$. So $\nu+\omega$ is very good.
Assume $\xi$ has cofinality $\aleph_0$ and $$\xi\notin\{\nu+\omega:\nu\in \textbf{Ord}\}.$$ Fix $\mu_i\uparrow \xi$, which exists because of the cofinality of $\xi$. Let $\eta$ be the smallest exponent which appears in the Cantor normal form of $\xi$ and note that for some $\zeta$, $$\xi=\zeta+\omega^\eta.$$ Since we have assumed that $\xi\notin \{\nu+\omega:\nu\in \textbf{Ord}\}$, it must be the case that $\eta>1$. Since $\mu_i\uparrow \xi$, we can assume without loss of generality that $\mu_i>\zeta$ for all $i$, so we can write $\mu_i=\zeta+\rho_i$ for $\rho_i\uparrow \omega^\eta$. Since $\eta>1$, it follows that $\omega+1<\omega^\eta$. Since $\rho_i<\omega^\eta$, it follows that $\rho_i+\omega+1<\omega^\eta$ for all $i$. Since $\sup_i \rho_i=\omega^\eta$, it follows that for any $i$, there exists $j>i$ such that $\rho_j>\rho_i+\omega+1$. Recursively select $j_1<j_2<\ldots$ such that $j_1=1$ and $\rho_{j_i}+\omega+1<\rho_{j_{i+1}}$. By relabeling, we can assume without loss of generality that $j_i=i$, so that $\rho_i+\omega+1<\rho_{i+1}$ for all $i$. Define $$\alpha_{2i-1}=\zeta+\rho_i+\omega$$ and $$\alpha_{2i}=\zeta+\rho_i+\omega+1$$ for all $i$. By construction, $\alpha_i$ is strictly increasing. Since $\rho_i+\omega+1<\rho_{i+1}<\omega^\eta$, $$\alpha_{2i-1}<\alpha_{2i}=\zeta+\rho_i+\omega+1<\zeta+\omega^\eta=\xi.$$ So each $\alpha_i<\xi$. Since $\alpha_{2i-1},\alpha_{2i}>\zeta+\rho_i$ and $\zeta+\rho_i\uparrow \xi$, $\alpha_i\uparrow\xi$.
Suppose that there exists $\beta\neq 0$ such that for all sufficiently large $i$, there exists $\gamma_i$ such that $\alpha_i=\gamma_i+\beta$. Then if $2i-1$ is sufficiently large, $$\zeta+\rho_i+\omega=\alpha_{2i-1}=\gamma_{2i-1}+\beta,$$ which implies that $\beta$ must be a limit ordinal, since $\omega$ is. But then $$\zeta+\rho_i+\omega+1=\alpha_{2i}=\gamma_{2i}+\beta,$$ which implies that $\beta$ must be a successor, since $1$ is. This is a contradiction, an shows that $\xi$ cannot be very good.