Is it possible to have cofinal and coinitial sets indexed by the same ordinal?

Question

Let $X$ be a totally ordered set without a maximum element and without a minimum element. As it is indicated here, there is a cofinal subset of $X$ indexed by an ordinal $\beta_1$, say $\{x_\alpha:\alpha<\beta_1\}$. Similarly, we can prove (again assuming the Axiom of Choice) the existence of a coinitial set $\{y_\alpha:\alpha<\beta_2\}$, for some ordinal $\beta_2$.

Are there sets $\{y_\alpha\in X:\alpha<\beta\}$ and $\{x_\alpha\in X:\alpha<\beta\}$ (for certain ordinal $\beta$) satisfying the following conditions?:

1. $\{y_\alpha:\alpha<\beta\}$ is coinitial in $X$ and $\{x_\alpha:\alpha<\beta\}$ is cofinal in $X$.
2. $y_\alpha<x_\alpha$ for all $\alpha<\beta$.
3. $\alpha_1<\alpha_2<\beta\Rightarrow [y_{\alpha_2}<y_{\alpha_1} \mbox{ and } x_{\alpha_1}<x_{\alpha_2}]$

Answer 1

No, in general. If $\kappa,\lambda$ are ordinals with different (infinite) cofinalities - say, $\kappa=\omega$ and $\lambda=\omega_1$ - then the linear order $\kappa^*+\lambda$ is a counterexample: any coinitial sequence has cofinality $cf(\kappa)$, while any cofinal sequence has cofinality $cf(\lambda)$.

(Here "$A^*$" denotes the reverse of the linear order $A$.)

Answer 2

Let X be the sum of the negative integers and $\omega_1$ (with their usual orderings and all negative integers less than all members of $\omega_1$). Every cofinal set has order-type $\omega_1$ and all coinitial sets have type $\omega$.