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An aleph-one-like linear order has size ${\aleph }_1$ and every point has only countably many predecessors. Consequently every proper initial segment is countable, since it is bounded by any point outside it. This differs from requiring every nonempty interval to be uncountable.

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

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.