0
$\begingroup$

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. $\alpha_1<\alpha_2<\beta\Rightarrow [y_{\alpha_2}<y_{\alpha_1} \mbox{ and } x_{\alpha_1}<x_{\alpha_2}]$
$\endgroup$
1
  • 3
    $\begingroup$ I think this should be impossible if the cofinality and coinitiality disagree, say $X$ is $\omega^\ast + \omega_1$ (an infinite descending sequence followed by the first uncountable ordinal $\omega_1$). This would satisfy the hypotheses of having no maximum or minimum element. Because the cofinality is $\omega_1$, you would need $\beta=\omega_1$, but there is no $\omega_1$-descending sequence. $\endgroup$ – Hayden Feb 5 '18 at 16:36
2
$\begingroup$

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 $A$.)

$\endgroup$
3
  • $\begingroup$ Why isn't $\omega$ a counterexample? $\endgroup$ – Asaf Karagila Feb 5 '18 at 16:46
  • $\begingroup$ @AsafKaragila "Let $X$ be a totally ordered set without a maximum element and without a minimum element." $\endgroup$ – Noah Schweber Feb 5 '18 at 16:56
  • $\begingroup$ Oh. Yeah. That's why. :D $\endgroup$ – Asaf Karagila Feb 5 '18 at 16:59
2
$\begingroup$

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 $\omega$.

$\endgroup$

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.