# number of linear orders

It is well known that for every infinite cardinal $\kappa$ the number of non-isomorphic total orders of cardinality $\kappa$ is $2^\kappa$. Who first proved this, and in what context? Was it proved for $\kappa=\aleph_0$ first, and then for uncountable $\kappa$, or for all $\kappa$ right away?

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Maybe you should provide a sketch of the proof or a reference to it. – Quinn Culver Feb 20 '12 at 14:30
Thank you for the suggestion. I am not sure what the original proof is, but I would prove it as follows: Take two sufficiently different countable linear orders $A$ and $B$. For every subset $S \subseteq \kappa$, replace each $i\in S$ by a copy of $A$, and each $j\in \kappa\setminus S$ by a copy of $B$; this will yield a linear order $L_S$, and if $A$ and $B$ were chosen suitably, all the different $L_S$ will be non-isomorphic. – g.castro Feb 20 '12 at 20:41
(continued) For example, one can choose $A$ to be $\omega+1$, and $B$ the converse ordering. From $L_S$ one can recover $S$ and $\kappa\setminus S$ by only looking at the non-isolated points in $L_S$, and checking whether they have an upper or lower neighbor. Or, as David Marker suggests in an exercise of his model theory book, let $A$ be $\mathbb Q + 1 + 1 + \mathbb Q$ (a copy of the rationals, followed by 2 discrete points, followed by another copy of the rationals), and let $B=\mathbb Q + 1 + 1 + 1 + \mathbb Q$. – g.castro Feb 20 '12 at 20:48

In short, the result for $\kappa=\aleph_0$ is due to Cantor (at least $2^{\aleph_0}$) and to Bernstein and Hausdorff, in 1901, independently (at most $2^{\aleph_0}$).