Continuous chain of $\kappa^+$ isomorphic linear orders on $\kappa\ge\aleph_1$

Question

For $\kappa\ge\aleph_0$ an infinite cardinal, and for $\kappa<\alpha\le\kappa^+$ an ordinal, we ask the existence of a chain $\left\langle(I_i,{<}):i<\alpha\right\rangle$ of linear orders such that, for all $i<j<\alpha$:

• (Cardinality) $\left|I_i\right|=\kappa$;
• (Increasing) $(I_i,{<})\subsetneqq(I_j,{<})$, in the sense that $I_i\subsetneqq I_j$ and ${<}_{I_i}={<}_{I_j}\upharpoonright{I_i}$
• (Continuity) If $i>0$ is limit, then $(I_i,{<})=\bigcup_{j<i}(I_j,{<})$, in the sense that $I_i=\bigcup_{j<i}I_j$ and ${<}_{I_i}=\bigcup_{j<i}{<}_{I_j}$;
• (Isomorphism) $(I_i,{<})\cong(I_0,{<})$.

Examples

1. For $(\kappa,\alpha)=(\aleph_0,\omega_1)$, we can take $I_i$ to be the transformation of $(1+i,{<})$ under replacing each point by a copy of $\mathbb{Q}$.
2. For $\kappa<\alpha<\kappa^+$ any limit ordinal, we can take, for all $i<\alpha$, $$(I_i,{<})=\left(\alpha\cup\left\{j+\tfrac{1}{2}:j<i\right\},{<}\right)\cong(\alpha,{<})$$

Question

Can such a chain $\left\langle(I_i,{<}):i<\kappa^+\right\rangle$ exist for when $\kappa\ge\aleph_1$ and $\alpha=\kappa^+$?

The question was originally motivated by considering EM models. I assumed this has some connection to certain special trees of size $\kappa$, but it turned out trickier than I had thought. Much thanks in advance.

Answer 1

It is consistent that there is a $(\aleph_1, \omega_2)$ chain.

Assume Proper Forcing Axiom (so $|ℝ|=\aleph_2$), enumerate intervals with rationals end points $(J_i\mid i\in ω)$.

For each $i<\aleph_0$ let $A_{i}$ be a set of $\aleph_1$ real numbers such that $A_i\subseteq J_i$ and $A_i\cap A_j=\emptyset$ for each $i<j<\omega$. Define $A=\bigcup A_i$.

Let $B=\{b_i\}⊆ℝ$ a set of cardinality $\aleph_2$ such that $B\cap A=\emptyset$.

Lastly we define $(I_i\mid i\in \aleph_2)$ with $I_{α}=A\cup\{b_{β}\mid β<α\}$ (All with the usual order of the real numbers), this is trivially increasing continuous sequence.

Clearly, each $\aleph_1=|A|\le|I_α|=|A|+|\alpha|\le2\aleph_1=\aleph_1$.

Now take an interval $J⊆ℝ$, and let $J_k\subseteq J$ be an interval with rational end points, then by construction $A_k\subseteq J_k∩I_{α}\subseteq J∩I_{α}$ hence $|J∩I_{α}|=\aleph_1$, in particular each $I_α$ is a $\aleph_1$-dense set.

Baumgartner has proven in "All $ℵ_1$-dense sets of reals can be isomorphic" that under Proper Forcing Axiom, indeed all such sets are order isomorphic.