Let $X$ be a totally ordered set without a maximum element. As it is indicated here, there exists a cofinal subset of $X$ indexed by an ordinal $\beta$, say $\{x_\alpha:\alpha<\beta\}$, satisfying $\alpha_1<\alpha_2<\beta\Rightarrow x_{\alpha_1}<x_{\alpha_2}$.
Is there a set $\{y_\alpha\in X:\alpha<\hat{\beta}\}$ satisfying the following conditions?:
- $\hat{\beta}\leq\beta$.
- $\alpha_1<\alpha_2<\hat{\beta}\Rightarrow y_{\alpha_1}<y_{\alpha_2}$
- $\{y_\alpha\in X:\alpha<\hat{\beta}\}$ is cofinal in $X$.
- For each limit ordinal $\delta<\hat{\beta}$, $\sup\{y_\alpha:\alpha<\delta\}$ exists in $X$ and equals $y_\delta$.
I think that the answer is positive but I find difficult to find a rigorous proof. My idea is to use transfinite induction to build the set $\{y_\alpha:\alpha<\hat{\beta}\}$ by using $\{x_\alpha:\alpha<\beta\}$.
For $\delta<\beta$, if $\delta$ is a sucessor ordinal then define $y_\delta:=x_\delta$. Now suppose that $\delta$ is a limit ordinal. If $\sup\{x_\alpha:\alpha<\delta\}$ exists, then we define $y_\delta:=\sup\{x_\alpha:\alpha<\delta\}$. If $\sup\{x_\alpha:\alpha<\delta\}$ does not exist, then choose a sucessor ordinal $\delta_1<\delta$ and put $y_\alpha=x_\delta$ for all $\alpha$ such that $\delta_1+1\leq\alpha<\delta$.
With this idea, in the inductive step, I may need to redefine the values of $y_\alpha$ when $\delta_1+1\leq\alpha<\delta$ (when $\sup\{x_\alpha:\alpha<\delta\}$ does not exist), and I do not know if this generates a problem. Also, I am violating the condition 2 while I am pursuing the condition 3.
No, this is not possible in general. For instance, suppose $X$ is countably saturated, meaning that for any countable subsets $A,B\subset X$ with $a<b$ for all $a\in A$ and $b\in B$, there exists $c\in X$ with $a<c<b$ for all $a\in A$ and $b\in B$. (Such a totally ordered set can be constructed by a recursion of length $\omega_1$, where at each step you add a new element $c$ for each pair of countable subsets $A$ and $B$ as above.)
If $\{y_\alpha:\alpha<\hat{\beta}\}$ is a subset of $X$ meeting your requirements, then $\hat{\beta}$ must have uncountable cofinality (otherwise you could take $A$ to be a countable cofinal subset and $B=\emptyset$). In particular, $\hat{\beta}>\omega$. But then taking $A=\{y_n:n<\omega\}$ and $B=\{y_\omega\}$, we find that $y_\omega$ cannot be the supremum of the $y_n$ for $n<\omega$ without violating countable saturation of $X$. Thus no subset satisfying your requirements can exist.