# How many ways can we totally order a set $X$ (but not necessarily the whole set $X$, perhaps just a subset) up to isomorphism?

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.

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

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.