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I recently came upon the following theorem which was attributed to J. Conway:

For each $A\subset \omega_1$, let $\Phi(A)$ be a linear ordering of type $\sum_{\alpha<\omega_1} \tau_\alpha$, where $\tau_\alpha$ is $\eta$ (the order-type of the rationals) whenever $\alpha\notin A$, and $1+\eta$ whenever $\alpha\in A$. The theorem asserts that for every $\omega_1$-like dense linear ordering is isomorphic to some $\Phi(A)$, $A\subset\omega_1$.

My questions:

(1) Does the theorem generalize to higher cardinalities?

(2) The reference given was J. Conway's Ph.D. Thesis, but I did not manage to find anything on the web. Any help? Thank you in advance.

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    $\begingroup$ What would count as a generalization? There is a trivial "generalization" that every $\kappa^+$-like $\kappa$-dense linear ordering is of the form $\sum_{\alpha\lt\kappa^+} \tau_\alpha$, where each $\tau_\alpha$ is a $\kappa$-dense linear order type of size $\kappa$. But that's not very useful without a classification of $\kappa$-dense linear orders of size $\kappa$. $\endgroup$
    – François G. Dorais
    Commented Aug 7, 2012 at 21:20
  • $\begingroup$ Two related math.SE questions: math.stackexchange.com/questions/174065/… and math.stackexchange.com/questions/174043/…. $\endgroup$
    – Joel David Hamkins
    Commented Aug 8, 2012 at 1:46
  • $\begingroup$ He didn't say so, but probably part of what François meant was that the literal generalization of the theorem, considering $\kappa$-like orders, but still using $\eta$ and $1+\eta$ in the pattern of $A\subset \kappa$, is simply false, since every $\Phi(A)$ has an $\omega_1$-like initial segment, but not every $\kappa$-like order need have this. $\endgroup$
    – Joel David Hamkins
    Commented Aug 8, 2012 at 1:56
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    $\begingroup$ Also, for example, every $\Phi(A)$ defined using $\eta$ and $1+\eta$ has many intervals that look like $\eta$, but a $\kappa$-like dense order for larger $\kappa$ need not have any intervals of type $\eta$. $\endgroup$
    – Joel David Hamkins
    Commented Aug 8, 2012 at 2:41
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    $\begingroup$ Yes, there has been much work on that but there is no simple classification under isomorphism. The best we have are fascinating results of Baumgartner (All $\aleph_1$-dense sets of reals can be isomorphic, Fund. Math. 79 (1973), 101-106) and Moore (A five element basis for the uncountable linear orders, Ann. of Math. 163 (2006), 669-688). See also Todorcevic's survey Trees and linearly ordered sets in the Handbook of Set-Theoretic Topology. $\endgroup$
    – François G. Dorais
    Commented Aug 8, 2012 at 16:53

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Concerning question 2, the theorem that you mention in the case of $\omega_1$ is not actually difficult. I had briefly sketched a proof of it at the conclusion of my answer to the math.SE question Linearly ordered sets somewhat similar to $\mathbb{Q}$, which is concerned with these types of orders---what I called $\mathcal{Q}_A$ is the same as your $\Phi(A)$. Here is a complete argument:

Theorem. Every $\omega_1$-like dense linear ordering is isomorphic to $\Phi(A)$ for some $A\subset\omega_1$.

Proof. Suppose that $L$ is an $\omega_1$-like dense linear order. Let $\langle x_\alpha\mid \alpha\lt\omega_1\rangle$ be any increasing cofinal $\omega_1$-sequence in $L$, containing none of its limit points (i.e., scattered). Let $\tau_\alpha$ be the interval of points above $\cup_{\beta\lt\alpha}\tau_\beta$ and below the point $x_\alpha$. This is either $\eta$ or $1+\eta$, depending on whether $\cup_{\beta\lt\alpha}\tau_\beta$ has a supremum in $L$ or not. These intervals therefore realize $L$ as $\Sigma_{\alpha\lt\omega_1}\tau_\alpha$, as desired. Thus, $L$ is $\Phi(A)$, where $A$ is the set of $\alpha$ where that supremum exists. QED

In that previous answer, I proved that the $\Phi(A)$ orders are determined up to isomorphism essentially by the equivalence of $A$ modulo the club filter.

Theorem. $\Phi(A)$ is isomorphic to $\Phi(B)$ if and only if $A$ and $B$ agree on having $0$ and also agree modulo the club filter, meaning that there is a closed unbounded set $C\subset\omega_1$ such that $A\cap C=B\cap C$. In other words, this is if and only if $A$ and $B$ agree on $0$ and are equivalent to $B$ in $P(\omega_1)/\text{NS}$, as subsets modulo the nonstationary ideal.

It would seem to be an interesting question to inquire in your style whether one may extend this beyond $\omega_1$ to higher cardinals.

The anwer, unfortunately, is negative. Suppose $\kappa\gt\omega_1$ is any cardinal. Let $A$ be the empty set, and let $B$ be any nonstationary set containing some ordinals with uncountable cofinality; for example, consider the singleton set $B=\{\ \omega_1\ \}$. These two sets agree on a club, since both omit a club. But meanwhile, $\Phi(A)$ and $\Phi(B)$ are not isomorphic, because the former has all points having cofinality $\omega$, but the latter has points of uncountable cofinality.

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  • $\begingroup$ Joel, very nice answer. Let me be a little skeptic before I accept your answer: In the proof of the second theorem (I looked your answer at StackExchange.com) the direction ($A$ and $B$ agree on 0 and also agree modulo the club filter) implies $\Phi(A)$ and $\Phi(B)$ are isomorphic, depends on the fact that every countable dense linear ordering without endpoints is isomorphic to the rationals. In particular, for $\beta<\omega_1$, $\eta+\sum_{\alpha<\beta} \tau_\alpha$ is always isomorphic to $\eta$, no matter what the choices for the $\tau_\alpha$ are. ...(cont.) $\endgroup$
    – Ioannis Souldatos
    Commented Aug 8, 2012 at 15:09
  • $\begingroup$ ...(cont.) On the other hand, if $\beta$ is allowed to be $\ge\omega_1$, the argument fails. So, I wouldn't expect the second theorem to hold "as is" in higher cardinalities. If we also use more options for $\tau_\alpha$ than $\eta$ and $1+\eta$ as suggested above, then the situation is even more different. $\endgroup$
    – Ioannis Souldatos
    Commented Aug 8, 2012 at 15:26
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    $\begingroup$ I agree; the proof uses exactly that feature about $\omega_1$, that we understand all the countable dense linear orders, and perhaps this is difficulty with extending it beyond $\omega_1$, and the heart of François's comment, since we lack the analogous facts for uncountable dense orders. So as you say, the proof of the theorem definitely does not extend to higher cardinals, and the counterexamples show that the statement fails for the higher cardinals. $\endgroup$
    – Joel David Hamkins
    Commented Aug 8, 2012 at 15:32
  • $\begingroup$ I am not sure if this is a comment or a separate question, but can anyone help us about what is known for the $\kappa$-dense linear orderings of size $\kappa$? $\endgroup$
    – Ioannis Souldatos
    Commented Aug 8, 2012 at 16:50
  • $\begingroup$ For the case of $\omega_1$, my math.SE answer (follow link above) shows that there are $2^{\omega_1}$ many distinct isomorphism classes, even among the $\omega_1$-like dense linear orders. It seems that that part of the argument does generalize to show that there are at least $2^\kappa$ many distinct $\kappa$-like dense orders of size $\kappa$. $\endgroup$
    – Joel David Hamkins
    Commented Aug 8, 2012 at 17:59

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