If HC (continuum hypothesis in French) holds, then some of those sequences are cofinal whereas some are not.

Indeed, HC implies that the corresponding ultrapower$\ ^*\mathbb{R}$ of $\mathbb{R}$ is a saturated ordered field with cardinal $\omega_1$. This is the same for the field $\mathbf{No}(\omega_1)$ of surreal numbers with countable birthday. So they are isomorphic.

In $\mathbf{No}(\omega_1)$, there is a strictly increasing and cofinal embedding $x \mapsto \omega^x: \mathbf{No}(\omega_1) \rightarrow \mathbf{No}(\omega_1)^{>0}$ which satisfies in particular $\forall x,y(0\leq x<y\Longrightarrow \omega^x+1<\omega^y)$. So taking integer parts in $^*\mathbb{N}$, we obtain a cofinal order embedding $ x \mapsto \left\lfloor \omega^x \right\rfloor: \mathbf{No}(\omega_1)^{\geq 0} \longrightarrow\ ^*\mathbb{N}$.
Since there are copies of $\omega_1$ in $\mathbf{No}(\omega_1)^{\geq 0}$ which are cofinal, and others which are bounded, this yields cofinal and bounded $\omega_1$-sequences in$\ ^*\mathbb{N}$.

In ZFC, I think (but I am not sure) that it is consistent that the cofinality of$\ ^*\mathbb{N}$ be $\omega_2$, meaning that each $\omega_1$-sequence would be bounded.

cofinalityascardinality. $\endgroup$8more comments