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 $\text{ED}(M)=\{\varphi (c_{m_1},\dots ,c_{m_n}):M\models \varphi [m_1,\dots ,m_n]\}$ 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 $N\models \text{ED}(M)$ admits an elementary embedding $j_{M,N}:M\to N$ (mapping $m\mapsto c_m^N$), such that if $j_{N,N^{\prime }}:N\to N^{\prime }$ is an elementary embedding between models of $\text{ED}(M)$, then $j_{N,N^{\prime }}\circ j_{M,N}=j_{M,N^{\prime }}$ (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 $\mathfrak{C}\models T$, $\mathbf{I}=\{a_i:i\in I\}$, 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 $M\cong j_{M,M_0}(M)⪯M_0$, the properties of $\text{ED}(M)$ from before imply that $M⪯M_i$ 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
$$I_i=\beta \cup \{j+\frac{1}{2}:j<i\},$$
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 ${\mathrm{\aleph }}_0$.

For $i<{\omega }_1$, we define the linear order
$$(I_i,<)=((1+i)\times \mathbb{Q},{<}_{\text{lex}}).$$
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 ${\mathrm{\aleph }}_0$, 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 ${\mathrm{\aleph }}_1$-dense set of reals. We'll show that $(X,<)$ is robust.

$\mathsf{\text{PFA}}$ implies $|\mathbb{R}|={\mathrm{\aleph }}_2$ ([3] Theorem 31.23), so we can pick ${\mathrm{\aleph }}_2$ 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 ${\mathrm{\aleph }}_1$-dense since $X\subseteq I_i$ is ${\mathrm{\aleph }}_1$-dense and $|I_i|={\mathrm{\aleph }}_1$, 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 ${\mathrm{\aleph }}_1$.

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 $a\upharpoonright (i+1){\le }_{\text{lex}}(f\upharpoonright i{)}^{⌢}f(i)$. 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 $\gamma =\underset{b\in B}{sup}{\gamma }_b$ 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 $\gamma =\underset{a\in A}{sup}{\gamma }_a<\kappa$ 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 $L\bigsqcup K=\{x_i:i<\kappa \}$.

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 $x_i\in L\setminus \text{dom}(f_i)$, we let $A=f_i[(-\mathrm{\infty },x_i)]$ and $B=f_i[(x_i,\mathrm{\infty })]$. 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 $\kappa$.