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

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:

• (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 $$(I_i,{<})=\bigcup_{j, in the sense that $$I_i=\bigcup_{j and $${<}_{I_i}=\bigcup_{j;
• (Isomorphism) $$(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

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.

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 $$i. Define $$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.

• Nice! Now I wonder whether the result is actually independent.... Commented Dec 22, 2022 at 18:30