Number of Non-Isomorphic linear orders on $\kappa$ elements

Question

After having proven that there is no $\kappa$-categorical axiomatizations of strict linear order for any infinite cardinality $\kappa$, a professor of mine mentioned in passing that it was actually the case that there are in fact $2^{\kappa}$ non-isomorphic SLOs on any infinite $\kappa$, in contrast to $\kappa +$ wellorders. He also mentioned that the assertion that number of wellorders is equal to the number of strict linear orders is equivalent to the continuum hypothesis. Sadly, class ended and I was unable to get clarification (and now am too interested to wait for the next office hours).

I am confused on a few points:

1) why are there $2^\kappa$ such non-isomorphic orders?

Any texts/reading that could be suggested to address any of these questions, or any explanation thereon, would be much appreciated. Thanks!

Update: 2, 3 should've been obvious. They're below in case anyone ever asks the same question. Still interested in 1.

2) what does $\kappa +$ mean?

3) how is the assertion that $2^\kappa = \kappa +$ equivalent to the continuum hypothesis?

Answer 1

Here's my shot at answering (1).

There cannot be more than $2^\kappa$ linear orders on a set of cardinality $\kappa$, since the number of binary relations on $\kappa$ is $2^{\kappa^2}$ and since $\kappa$ is infinite, $\kappa^2 = \kappa$ so there are a total of $2^\kappa$ binary relations on $\kappa$.

Since $\kappa$ is infinite, $\kappa = \kappa \times \aleph_0$, so if $S$ is of size $\kappa$, then the number of non-isomorphic linear orders on $S$ is the same as that on $S \times \mathbb{N}$. Now, for every map $f : S \to \{0, 1\} $, consider the order on $S \times \mathbb{N}$, where $\preceq$ is a fixed well ordering on $S$.

$(s_0, n_0) < (s_1, n_1)$ if $s_0 \prec s_1$

$(s, n_0) < (s, n_1)$ if $f(s) = 0$ and $n_0 < n_1$ or if $f(s) = 1$ and $n_0 > n_1$

This scheme should produce non isomorphic linear orders for any distinct $f_1, f_2$, since if $s$ is the least element under $\preceq$ of $S$ such that $f_1(s) \neq f_2(s)$, then any order isomorphism must map $\{s'\} \times \mathbb{N}$ to itself for each $s' \preceq s$, but this is impossible for $s$.

Thus, there is a distinct order for each function from $S$ to $\{0,1\}$, of which there are $2^\kappa$. This means that there are at least $2^\kappa$ orderings, and since there are no more than $2^\kappa$, this is the exact cardinality.

Answer 2

Given $X\subseteq \kappa$, let $slo(X)$ be the simple linear order that results from replacing every $\alpha \in X$ with a copy of $\omega^* + \omega$ (the order type of $\Bbb Z$). This is actually the ordered sum along $\kappa$ of linear orderings $R_{\alpha}$, where $R_{\alpha}$ is equal to $\omega^* + \omega$ for $\alpha \in X$, and equal to $1$ otherwise.

This order has $\kappa$-many elements, and we can ensure in a canonical way that all of its elements are members of $\kappa$. Clearly (yes?), $X\mapsto slo(X)$ is an injection on $\mathcal{P}(\kappa)$, and furthermore, if $X,Y\subseteq \kappa$ and $X\neq Y$ then $slo(X)$ is not isomorphic to $slo(Y)$.