1
$\begingroup$

Here's Attempt #2 at asking this question.

Suppose $X$ is a set (not necessarily finite), and let $T$ denote the collection of all totally ordered sets $(A,\leq)$ such that $A \subseteq X.$ Now let $\cong$ denote the unique relation on $T$ such that $(A,\leq_A) \cong (B,\leq_B)$ holds iff they're isomorphic as totally ordered sets.

Question. What is the cardinality of $T /\!\cong$ in terms of the cardinality of $X$?

If $X$ is finite, this is easy: the answer is $|T/\!\cong\!| = |X|+1.$ So its really the infinite case that I'm interested in. For the infinite case, all I know is that $|X|^+ \!\leq |T/\!\cong\!|,$ where $\kappa^+$ denotes the cardinal successor of $\kappa$. This follows more or less from the definition of cardinal successor.

$\endgroup$
  • 2
    $\begingroup$ See here. $\endgroup$ – Andrés E. Caicedo Apr 18 '14 at 4:28
  • $\begingroup$ @AndresCaicedo, thanks. I get it now. $\endgroup$ – goblin Apr 18 '14 at 4:36
  • $\begingroup$ @AndresCaicedo, hey any idea what the answer is for scattered linear orderings? $\endgroup$ – goblin Apr 18 '14 at 17:47
  • $\begingroup$ I haven't thought much about it... The counting in that case should probably be based on Hausdorff's result, see here. $\endgroup$ – Andrés E. Caicedo Apr 18 '14 at 18:43
  • $\begingroup$ @AndresCaicedo, okay I'll have a think about it. Probably just end up asking another question at some point. $\endgroup$ – goblin Apr 18 '14 at 19:03
2
$\begingroup$

OK, we get $2^\kappa$ orders, as explained here.

The answer does not change if we restrict our attention to scattered orders: Given any element of $(\kappa^+)^{<\kappa^+}$, say $(\alpha_\iota\mid \iota<\tau)$, simply considered the ordered (transfinite) sum $$\alpha_0+(\omega^*+\omega)+\alpha_1+(\omega^*+\omega)+\dots$$ Different sequences give rise to non-isomorphic scattered orders, so we get at least $|(\kappa^+)^{<\kappa^+}|=2^\kappa$ orders this way.

(Related.)

$\endgroup$

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.