Robustly Isomorphic Models

Let $M=(\mathbb{Q},{<})$. Then the increasing continuous chain $\{M_i=(I_i,{<}):i<\alpha\}$ of linear orders is what we wanted.

Let $\text{ED}(M)=\{\phi(c_{m_1},\ldots,c_{m_n}):M\vDash\phi[m_1,\ldots,m_n]\}$ be the elementary/complete diagram of $M$, so that $\text{ED}(M)$ is a countable complete first-order theory extending the theory $\text{Th}(M)$ of $M$, and every model $N\vDash\text{ED}(M)$ admits an elementary embedding $j_{M,N}:M\to N$ (mapping $m\mapsto c_m^N$), such that if $j_{N,N'}:N\to N'$ is an elementary embedding between models of $\text{ED}(M)$, then $j_{N,N'}\circ j_{M,N}=j_{M,N'}$ (see [1] Lemma 2.3.3).

Let $T\supseteq\text{ED}(M)$ be a countable complete first-order extension (in an expanded language) with built-in Skolem functions, and let $\mathfrak{C}\vDash T$, $\mathbf{I}=\{a_i:i\in I\}$, and $\text{EM}({\cdot})$ be as in the Ehrenfeucht–Mostowski theorem, where $I=\bigcup_{i<\alpha}I_i$.

For every $i<\alpha$, let $\mathbf{I}_i=\{a_i:i\in I_i\}$, and let $M_i=\text{EM}(\mathbf{I}_i)$. We can check that $\{M_i:i<\alpha\}$ is the elementary chain that we wanted:

1. By the Ehrenfeucht–Mostowski theorem part (2), $\{M_i:i<\alpha\}$ forms an $\mathcal{L}(T)$-elementary chain. So their reducts $\{M_i\upharpoonright\mathcal{L}(M):i<\alpha\}$ form an $\mathcal{L}(M)$-elementary chain.
2. By the Ehrenfeucht–Mostowski theorem part (1), every model $M_i$ is a model of $\text{ED}(M)$ of size $\kappa$. So by identifying the model $M$ with its isomorphic image $M\cong j_{M,M_0}(M)\preceq M_0$, the properties of $\text{ED}(M)$ from before imply that $M\preceq M_i$ for all $i<\alpha$ under this identification.
3. From the assumption that $\mathbf{I}$ is a sequence of order indiscernibles, one can prove that $\text{EM}(\mathbf{J})\cap\mathbf{I}=\mathbf{J}$ for all $\mathbf{J}\subseteq\mathbf{I}$. This implies that since $\{\mathbf{I}_i:i<\alpha\}$ is increasing, the chain $\{M_i:i<\alpha\}$ is also increasing.
4. From the definition of Skolem hulls, one can prove that if $\{\mathbf{J}_i:i<\beta\}$ is a chain of subsets $\mathbf{J}_i\subseteq\mathbf{I}$ (such that $i\le j$ implies $\mathbf{J}_i\subseteq\mathbf{J}_j$), then $\bigcup_{i<\beta}\text{EM}(\mathbf{J}_i)=\text{EM}(\bigcup_{i<\beta}\mathbf{J}_i)$. This implies that since $\{\mathbf{I}_i:i<\alpha\}$ is continuous, the chain $\{M_i:i<\alpha\}$ is also continuous.
5. By the Ehrenfeucht–Mostowski theorem part (4), since the linear orders $\{(I_i,{<}):i<\alpha\}$ are all isomorphic, the models $\{M_i:i<\alpha\}$ are also all isomorphic.

Assume $\alpha<\kappa^+$. Let $\beta$ be a limit ordinal such that $\max\{\kappa,\alpha\}\le\beta<\kappa^+$. For $i<\alpha$, we can define the linear order
$$I_i=\beta\cup\{j+\tfrac{1}{2}:j< i\},$$
such that $(I_i,{<})$ is a well-order, and $\beta$ being a limit ordinal implies that $(I_i,{<})\cong(\beta,{<})$ for all $i<\alpha$. It is easy to check that $\{I_i:i<\alpha\}$ is an increasing continuous chain of linear orders of size $\kappa$ which are all isomorphic to $\beta$.

We will show that $(\mathbb{Q},{<})$ is robust of size $\aleph_0$.

For $i<\omega_1$, we define the linear order
$$(I_i,{<})=\left((1+i)\times\mathbb{Q},{<}_{\text{lex}}\right).$$
This $I_i$ is constructed by replacing every point in the linear order $1+i$ with a copy of $\mathbb{Q}$. Then every $I_i$ is a dense linear order without endpoints of size $\aleph_0$, so by Cantor's back-and-forth method (see [1] Theorem 2.4.1), we get an isomorphism $(I_i,{<})\cong(\mathbb{Q},{<})$.

It's easy to see that $\{I_i:i<\omega_1\}$ forms an increasing continuous chain, witnessing robustness of $(\mathbb{Q},{<})$.

Let $X$ be an $\aleph_1$-dense set of reals. We'll show that $(X,{<})$ is robust.

$\textsf{PFA}$ implies $|\mathbb{R}|=\aleph_2$ ([3] Theorem 31.23), so we can pick $\aleph_2$ many distinct reals $\{x_i\in\mathbb{R}:i<\omega_2\}$ that are not in $X$. For $i<\omega_2$, define
$$I_i=X\cup\{x_j:j< i\}.$$
Then every $I_i$ is $\aleph_1$-dense since $X\subseteq I_i$ is $\aleph_1$-dense and $|I_i|=\aleph_1$, and so Baumgartner's theorem implies $(I_i,{<})\cong (X,{<})$.

It's easy to see that $\{I_i:i<\omega_2\}$ is increasing continuous, which witnesses that $(X,{<})$ is robust of size $\aleph_1$.

We can define a supremum $f=\sup(A)\in 2^\kappa$ recursively, such that for all $i<\kappa$, $f(i)\in 2$ is the smallest such that for all $a\in A$, there is $a\upharpoonright(i+1)\le_\text{lex}(f\upharpoonright i)^\frown f(i)$. It is not hard to show that for all $a\in A$, there is $a\le_\text{lex} f$, and for all $g<_\text{lex}f$ there exists some $a\in A$ such that $g<_\text{lex}a$.

We split into cases.

1. If $f=\sup(A)\in A$, then for all $b\in B$, since $f< b$, we can pick some $\gamma_b<\kappa$ such that $f\upharpoonright\gamma_b< b\upharpoonright\gamma_b$. Then let $\gamma=\sup_{b\in B}\gamma_b$ so that $\gamma<\kappa$ by regularity, and we see that $g=(f\upharpoonright\gamma)(1)^\kappa$ satisfies $g\upharpoonright\gamma< b\upharpoonright\gamma$ for all $b\in B$.
   Since $f\le g$, $f\in A\subseteq\mathcal{I}$, and $g\notin\mathcal{I}$, we have $f< g$, so that $f\upharpoonright\delta< g\upharpoonright\delta$ for some $\gamma\le\delta<\kappa$. Letting $c=(g\upharpoonright\delta)(01)^\kappa$, we have $f< c$, so that for all $a\in A$, $a\upharpoonright\delta\le f\upharpoonright\delta< c\upharpoonright \delta$, and for all $b\in B$, $c\upharpoonright\delta= g\upharpoonright\delta< b\upharpoonright\delta$. By $c\in\mathcal{J}$, we've seen that $c$ is what we wanted.
2. Similarly if $\inf(B)\in B$, then a desired $c\in \mathcal{J}$ exists.
3. Assume $\sup(A)\notin A$ and $\inf(B)\notin B$.
   First we can show that $f=\sup(A)$ is eventually constantly $0$. For every $a\in A$, fix $\gamma_a<\kappa$ such that $a\upharpoonright\gamma_a< f\upharpoonright\gamma_a$, and let $\gamma=\sup_{a\in A}\gamma_a<\kappa$ by regularity. Then $g=(f\upharpoonright\gamma)(0)^\kappa$ is such that $g\le f$, but $a< g$ for all $a\in A$. Since $f$ is a supremum, we must have $f=g$, and so $f$ is eventually constantly $0$ as claimed.
   Similarly $h=\inf(B)$ is eventually constantly $1$.
   From $A< B$ we know that $f\le h$. But since $f\ne h$, we have $f< h$. So we can fix some $\delta<\kappa$ such that $f\upharpoonright\delta< h\upharpoonright\delta$, $f=(f\upharpoonright\delta)(0)^\kappa$, and $h=(h\upharpoonright\delta)(1)^\kappa$. Then $c=(f\upharpoonright\delta)(01)^\kappa$ satisfies $f< c< h$ and $c\in\mathcal{J}$, so $c$ is what we wanted.

Let $L\sqcup K=\{x_i:i<\kappa\}$.

We can build a chain of partial isomorphisms $\{f_i:i<\kappa\}$ such that $\text{dom}(f_i)\subseteq L$ and $\text{img}(f_i)\subseteq K$ both have size $<\kappa$, $f_i$ is an isomorphism between $(\text{dom}(f_i),{<}_L)$ and $(\text{img}(f_i),{<}_K)$, and $x_i\in\text{dom}(f_{i+1})\cup\text{img}(f_{i+1})$.

1. When $i=0$ let $f_0=\varnothing$.
2. When $i<\kappa$ is limit, let $f_i=\bigcup_{j< i}f_j$.
3. When $i+1$ is a successor stage and $x_i\in L\smallsetminus\text{dom}(f_i)$, we let $A=f_i[(-\infty,x_i)]$ and $B=f_i[(x_i,\infty)]$. By saturation of $K$, there is $c\in K$ such that $A< c< B$, and we can define $f_{i+1}=f_i\cup\{(x_i,c)\}$.
4. The other cases are similar, and it's never hard to see that $f_i$ has the desired properties for all $i<\kappa$.

Finally, we get an isomorphism $f=\bigcup_{i<\kappa}f_i$ from $(L,{<})$ to $(K,{<})$.

We'll show the linear order $(\mathcal{J},{<})$ is robust of size $\kappa$.

Since $|\mathcal{I}|=2^\kappa$, we can pick $\kappa^+$ many distinct elements $\{x_i\in\mathcal{I}:i<\kappa^+\}$ that are not in $\mathcal{J}$. For $i<\kappa^+$, define
$$I_i=\mathcal{J}\cup\{x_j:j< i\}.$$
By the saturation lemma part 1, every $I_i$ is saturated of size $\kappa$, because $\mathcal{J}\subseteq I_i$. By the saturation lemma part 2, $(I_i,{<})\cong(\mathcal{J},{<})$.

It's easy to see that $\{I_i:i<\kappa^+\}$ is increasing continuous, which shows that $(\mathcal{J},{<})$ is robust of size $\kappa$.