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.

  • $\begingroup$ 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. $\endgroup$ – Chilote Feb 22 '18 at 0:43
  • 1
    $\begingroup$ Countably saturated really means $\aleph_1$-saturated, i.e., all types in countably many parameters are realized. Accordingly, $\mathbb R$ is not countably saturated. $\endgroup$ – Andrés E. Caicedo Feb 22 '18 at 0:54
  • $\begingroup$ Andrés, so how is the definition of countable saturation given by @Eric Wofsey related to the $\aleph_1$-saturation? $\endgroup$ – Chilote Feb 22 '18 at 1:15
  • $\begingroup$ I think that I got the answer once I learned about $\eta$-sets. en.wikipedia.org/wiki/%CE%97_set $\endgroup$ – Chilote Feb 22 '18 at 1:46
  • 1
    $\begingroup$ @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). $\endgroup$ – Andrés E. Caicedo Feb 22 '18 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.