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The cofinality of a linear order is the least cardinality of a cofinal subset. It is preserved by order isomorphisms. The order sum of ${\omega }_1$ copies of the rational order has cofinality ${\aleph }_1$, whereas appending one final rational block changes its cofinality to ${\aleph }_0$ without changing its cardinality or the dense-linear-order axioms.

Every lattice-ordered group embeds in a lattice-ordered permutation group on one totally ordered set. Choose a value in a lattice-ordered group for each nonidentity element, act on its ordered right cosets, and combine these chains in an order sum. The original element moves its own value's identity coset, making the action faithful. The coset lattice formulas make the embedding preserve both join and meet.

For each $g\ne 1$, choose a value in a lattice-ordered group $P_g$, using Zorn's lemma. Such a value is a prime convex lattice subgroup. Otherwise the last construction in Question 1(i) would give two larger convex lattice subgroups whose intersection is $P_g$; maximality among subgroups omitting $g$ would put $g$ in both, hence in their intersection. Thus the right coset set ${\Omega }_g=P_g\backslash G$ is a totally ordered set.

Right multiplication defines an action ${\rho }_g(a):P_{g}h\mapsto P_{g}ha$. It is well-defined, preserves the total order, and has inverse ${\rho }_g(a^{-1})$. The coset formulas from Question 1(i) show pointwise preservation of both join and meet, becauseand likewise for the meet.

Take the disjoint union of these chains, arranged in any fixed order sum of the indices $g\ne 1$. Let every $a\in G$ act on each component by ${\rho }_g(a)$. This gives a lattice-ordered group homomorphism into the order automorphisms of one chain. It is injective: a nonidentity $a$ moves the identity coset in ${\Omega }_a$, since $P_{a}a\ne P_a$. Its image is closed under pointwise join and meet, giving exactly the required lattice-ordered permutation group. This is the Holland representation theorem.

The inductive definition of ordinal addition is $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ for a nonzero limit ordinal $\lambda$. The synthetic definition is the order type of the order sum obtained by putting a copy of $\beta$ after a disjoint copy of $\alpha$.

These agree by transfinite induction on the second argument. Appending no points gives $\alpha$; appending one new greatest point gives a successor. At a limit stage, the order sum is exhausted by the increasing initial segments obtained by appending each $\beta <\lambda$, so its order type is their supremum. Thus the synthetic construction satisfies all three recursion clauses, and uniqueness of transfinite recursion proves equivalence.

The maximum is $2^{{\aleph }_1}$. We construct that many aleph-one-like linear orders, whose proper initial segments are countable, and prove that they are nonisomorphic. As usual the rational-type requirement concerns nonempty proper initial segments; the empty segment occurs in every order.

Let $D$ be the club set of nonzero limit ordinals below ${\omega }_1$. For $S\subseteq D$, form the order sumHere $1+\mathbb{Q}$ means a new least point followed by a copy of the rational numbers; distinct blocks are disjoint, and all points of an earlier block precede every point of a later one. Every block is countable, dense and has no greatest point. Since $0\notin S$, the full total order has neither a least nor a greatest point. It has size ${\aleph }_1$.

The sum is dense: within a block use density, and for points in different blocks choose a larger point in the earlier block. If $I$ is a proper initial segment and $y\notin I$ lies in block $\alpha$, then $I$ is contained in the union of the blocks through $\alpha$. There are only countably many such blocks, each countable. Thus $I$ is countable, rather than an uncountable interval in an aleph-one-dense order.

Every nonempty $I$ is dense and has no least point. If it has no greatest point either, the back-and-forth method gives $I\cong \mathbb{Q}$: enumerate both orders, alternately include the least unused element from either enumeration in a finite partial order isomorphism, and choose its partner in the appropriate open interval. Density and absence of endpoints always supply a new partner. The union is a bijective order isomorphism. If $I$ has a greatest point, remove it, apply the same argument, and reattach it. Hence every nonempty proper initial segment is isomorphic to $\mathbb{Q}$ or $\mathbb{Q}+1$.

The stationary information is recorded at the limit cuts. WriteThis is a continuous increasing kappa-filtration by countable initial segments. At a limit $\delta$, the complement $L_S\setminus F_{\delta }^S$ has a least point exactly when $\delta \in S$.

Suppose $f:L_S\to L_T$ is an order isomorphism. Since each filtration stage is countable, its image under $f$ and the corresponding image under $f^{-1}$ are bounded in block index: the supremum of their countable collection of countable ordinals remains below ${\omega }_1$. Choose $g(\alpha )<{\omega }_1$ so thatThe setis a club set. To see unboundedness, start above any given ordinal and recursively choose ${\gamma }_{n+1}>{\gamma }_n$ larger than every $g(\alpha )$ for $\alpha \le {\gamma }_n$. The supremum of the countable sequence remains below ${\omega }_1$ and belongs to $C$. Closure follows because below a limit of members of $C$, each fixed $\alpha$ lies below one of those members.

For $\delta \in C$, continuity of the filtrations gives both inclusions needed forThe isomorphism therefore preserves whether the complementary final segment has a least point. ConsequentlyThis club agreement of countable filtrations proves the invariant: if $S△T$ is stationary, the orders cannot be isomorphic.

We also construct enough pairwise different stationary encodings. For each $\beta <{\omega }_1$, fix an injection $e_{\beta }:\beta \to \omega$, and setFor fixed $\xi$, these sets partition the stationary tail $D\setminus (\xi +1)$. At least one $A_{\xi ,n}$ is stationary: otherwise choose a club avoiding each one and intersect the countably many clubs to obtain a club avoiding the entire tail. The countable-intersection fact follows directly: above any starting point choose an increasing sequence, visiting each club infinitely often; its countable supremum belongs to every club by closure. For each $\xi$ let $n(\xi )$ be the least such $n$. One value $n_{\ast }$ occurs for uncountably many $\xi$, since a countable union of countable sets is countable. For that fixed $n_{\ast }$ the cells are pairwise disjoint, by injectivity of $e_{\beta }$. Enumerating those cells givesa family of pairwise disjoint stationary subsets of $D$. This is the explicit Ulam matrix on omega-one proof of disjoint stationary subsets of omega-one.

For $A\subseteq {\omega }_1$ put $S(A)={\bigcup }_{\eta \in A}{S}_{\eta }$. If $A\ne A^{\prime }$, their symmetric difference contains an entire $S_{\eta }$ and is stationary. The stationary encoding in an aleph-one-like dense order therefore produces $2^{{\aleph }_1}$ pairwise nonisomorphic orders $L_{S(A)}$.

Finally, on a fixed carrier of size ${\aleph }_1$, every total order is a subset of its Cartesian square. That square has size ${\aleph }_1$, so there are at most $2^{{\aleph }_1}$ possible relations. Thus

Use the one-relation language $\{<\}$ and the total orderswhere the second is the order sum: every point in its natural-number block lies below every point in its integer block.

To show II wins every finite Ehrenfeucht-Fraïssé game, use the interval threshold strategy for discrete linear orders. With $k$ rounds left, corresponding open intervals and end rays between already matched points have either the same size below $h_k=2^k-1$, or both have at least $h_k$ elements. Their sizes may be infinite. The chosen-point correspondence is order preserving. Initially the unique interval is infinite in both orders, so the invariant holds.

If I chooses a repeated point, II repeats its counterpart. If I chooses in a short interval, II chooses the point with the same finite rank in the matching interval. In a long interval, let $h=h_{k-1}$. If one side of the chosen point has fewer than $h$ elements, reproduce that size in the other interval; the opposite sides then both have at least $h$ elements. Otherwise choose a response having at least $h$ elements on each side. Such a response exists because the interval has at least $2h+1=h_k$ elements. The endpoint and successor/predecessor properties of these two discrete orders allow the finite small-side counts to be reproduced. This establishes the invariant with $k-1$ rounds left and gives a finite-game winning strategy. Therefore II wins the timed game.

In the infinite Ehrenfeucht-Fraïssé game, I first chooses a point $b$ in the integer block of $N$. It has infinitely many predecessors. II must respond with some $m\in \mathbb{N}$, which has exactly $m$ predecessors. I then chooses $m+1$ distinct predecessors of $b$. To preserve order and equality, II would need $m+1$ distinct points below $m$, which is impossible. ThusI chooses the required finite length only after seeing II's first response, which explains why this does not contradict the timed-game result.

For a subset $S$ of the nonzero countable limit ordinals, let $B_{\alpha }=1+\mathbb{Q}$ when $\alpha \in S$ and $B_{\alpha }=\mathbb{Q}$ otherwise. The order sum is an aleph-one-like linear order whose nonempty proper initial segments are $\mathbb{Q}$ or $\mathbb{Q}+1$. At the limit cut before block $\delta$, the complementary final segment has a least point exactly when $\delta \in S$. Club agreement of countable filtrations makes this condition invariant modulo a nonstationary difference. Disjoint stationary subsets of omega-one consequently supply $2^{{\aleph }_1}$ nonisomorphic orders.