# Building a special cofinal set in a totally ordered set

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?:

1. $\hat{\beta}\leq\beta$.
2. $\alpha_1<\alpha_2<\hat{\beta}\Rightarrow y_{\alpha_1}<y_{\alpha_2}$
3. $\{y_\alpha\in X:\alpha<\hat{\beta}\}$ is cofinal in $X$.
4. 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.
• I see the following problem with your answer: $\mathbb{R}$ is countably saturated and $\mathbb{N}$ is a countable cofinal set satisfying my conditions. In other words, $\hat{\beta}$ does not have to be uncountable. – Chilote Feb 22 '18 at 0:43
• Countably saturated really means $\aleph_1$-saturated, i.e., all types in countably many parameters are realized. Accordingly, $\mathbb R$ is not countably saturated. – Andrés E. Caicedo Feb 22 '18 at 0:54
• Andrés, so how is the definition of countable saturation given by @Eric Wofsey related to the $\aleph_1$-saturation? – Chilote Feb 22 '18 at 1:15
• I think that I got the answer once I learned about $\eta$-sets. en.wikipedia.org/wiki/%CE%97_set – Chilote Feb 22 '18 at 1:46
• @Chilote For ordered sets, Eric's description is $\aleph_1$-saturation. Another way of produce an example is to start with, say, $\mathbb R$ and form its ultrapower via a nonprincipal ultrafilter (this is one of the standard presentations of the hyperreals). The advantage of this construction is that it gives you an $\aleph_1$ saturated model ${}^*\mathbb R$ even if you consider $\mathbb R$ as a structure in a larger language (in particular, as an ordered field). – Andrés E. Caicedo Feb 22 '18 at 4:31