On countably saturated linear orders and certain class of countably saturated graphs

Abstract

The idea of this paper is to explore the existence of canonical countably saturated models for different classes of structures. It is well-known that, under CH, there exists a unique countably saturated linear order of cardinality $\mathfrak {c}$. We provide some examples of pairwise non-isomorphic countably saturated linear orders of cardinality $\mathfrak {c}$, under different set-theoretic assumptions. We give a new proof of the old theorem of Harzheim, that the class of countably saturated linear orders has a uniquely determined one-element basis. From our proof it follows that this minimal linear order is a Fraïssé limit of certain Fraïssé class. In particular, it is homogeneous with respect to countable subsets. Next we prove the existence and uniqueness of the uncountable version of the random graph. This graph is isomorphic to $(H(\omega _1),\in \cup \ni )$, where $H(\omega _1)$ is the set of hereditarily countable sets, and two sets are connected if one of them is an element of the other. In the last section, an example of a prime countably saturated Boolean algebra is presented.

1 Introduction

It is a general theorem of model theory, that for any first-order theory which has infinite models, all countably saturated models of cardinality $\mathfrak {c}$ are isomorphic, as long as we assume CH. The definition of countable saturation and the proof of this theorem can be found in any textbook on model the ory, for example [7]. However, we will not bother with realization of types in general, and focus on some particular cases. These cases are:

• (Hausdorff) Assume CH. Then all countably saturated linear orders (see next section for the definition) of cardinality $\mathfrak {c}$ are isomorphic.

• Assume CH. There exists a unique up to isomorphism graph G of cardinality $\mathfrak {c}$, with the property that for any countable, disjoint subsets $A, B \subseteq G$, there exists a vertex in $G\setminus (A \cup B)$ connected to every vertex in A and none of the vertices in B.

• (Parovičenko) Assume CH. Then all Boolean algebras of cardinality $\mathfrak {c}$ with the strong countable separation property are isomorphic. (for the definition see [8] or Section 3 of this article).

All of the above claims are proved using so-called back-and-forth argument, and all of them fail without CH. The fact that Parovičenko Theorem is equivalent to CH was proved by van Douwen and van Mill in [15]. The fact that each of the other two claims is equivalent to CH is much easier to see. However, it turns out that some applications of back-and-forth argument can be carried out only assuming that $\mathfrak {c}$ is regular, or even in ZFC alone. For example, Avilés and Brech generalized Parovičenko Theorem, by introducing a property stronger than the countable separation property, which guarantees the uniqueness of certain Boolean algebra with this property, when $\mathfrak {c}$ is regular [1]. A different generalization was obtained by Dow and Hart [2]. Majority of structures obtained this way are examples of uncountable Fraïssé limits, which general theory is described in [9]. The aim of this paper is to study possible generalizations of the above results in the absence of CH.

Theorem 1

(Harzheim, [5]). There exists a unique up to isomorphism linear order of cardinality $\mathfrak {c}$, which is countably saturated, and embeds into any countably saturated linear order.

Theorem 2

There exists a unique up to isomorphism graph G of cardinality $\mathfrak {c}$, such that

• For any disjoint sets $A,B \subseteq G$, $|A|\le \omega$, $|B|<\mathfrak {c}$, there exist a vertex $g\in G\setminus (A\cup B)$, such that g is connected to every element of A and none of the elements in B,

• G embeds into any graph with the above property.

Theorem 3

There exists a unique up to isomorphism directed graph $\overleftarrow{G}$ of cardinality $\mathfrak {c}$, such that

• For any countable set $A \subseteq \overleftarrow{G}$ there exist $\mathfrak {c}$-many $g\in \overleftarrow{G}$, such that $A=\{a\in \overleftarrow{G}|\, (g,a)\in E(\overleftarrow{G})\}$,

• For any $g \in G$, the set $A=\{a\in \overleftarrow{G}|\, (g,a)\in E(\overleftarrow{G})\}$ is countable,

• $\overleftarrow{G}$ embeds into any graph with the above properties.

The author would like to thank Wiesław Kubiś for inspiring conversations and asking many good questions which laid foundation for this work, and the anonymous referee for pointing out numerous mistakes in the first version of the paper. Also some credit goes to Arturo Martinez Celis, who came up with the idea of representation of the graph from Theorem 2.

2 Countably saturated linear orders

In this section we give the definition and introduce some examples of countably saturated linear orders. Next, we introduce the notion of linear dimension, which we use to characterize those countably saturated linear orders, which are embeddable into any countably saturated linear order. Finally, we prove that the linear order with this property is unique up to isomorphism.

We use the distinction non-decreasing/increasing rather than increasing/strictly increasing. For linear orders $(K,\le _K)$, $(L,\le _L)$, we define $K+L$ as $\{0\}\times K \cup \{1\}\times L$, with the lexicographic order. For $A,B \subseteq L$, by $A<B$, we denote that $\forall _{x\in A}\forall _{y\in B} x<y$. Later, whenever a product of linear orders is considered, we equip it with lexicographic order without mentioning it. By symbol $K \hookrightarrow L$ we denote that there is an increasing function from K to L. By $L^*$ we denote L with the reversed order, i.e. $(L,\le )^*=(L,\ge )$.

A linear order is compact, if it is compact in the order topology, i.e. every set has both supremum and infimum, in particular both endpoints exist. We often call such orders compact lines.

A linear order is a linearly ordered continuum, if it is compact and connected in the order topology, i.e. it is compact and dense as a linear order.

A linear order L has character at most $\kappa$, if for each $l \in L$, there exists subsets $A,B \subset L$ of cardinality at most $\kappa$, such that $A<B$,

and $\{x\in L|\; A<\{x\}<B \}=\{l\}$. As to be expected, L has character $\kappa$, if $\kappa$ is the least cardinal, such that L has character at most $\kappa$. We will make use of the following observation a few times. The proof is left for the Reader.

Proposition 1

If a linear order L has uncountable character, then $\omega _1 \hookrightarrow L$ or $\omega _1^* \hookrightarrow L$.

2.1 Basic examples

Definition 1

We say that a linear order $(L,\le )$ is countably saturated, if for any countable linear orders a, b, and increasing functions $i:a\rightarrow b$, $f:a\rightarrow L$, there exists $\tilde{f}:b\rightarrow L$, such that $\tilde{f}\circ i=f$.

Moreover, L is prime countably saturated, if any countably saturated linear order contains an isomorphic copy of L.

The following simple Lemma provides more operational characterization, and justifies the name “saturated”—it is just saturation in the usual, model-theoretic sense.

Lemma 1

A linear order is countably saturated if and only if

• it is dense, without the least and the greatest element,

• no countable increasing sequence has supremum,

• no countable decreasing sequence has infimum,

• there are no $(\omega ,\omega )$-gaps: for any two sequences $\{x_n\}_{n<\omega }$, $\{y_n\}_{n<\omega }$ such that $\forall {n<\omega }\;{x_n<x_{n+1}<y_{n+1}<y_n}$, exists z s.t. $\forall {n<\omega }\;{x_n<z<y_n}.$

Proof

Suppose, that L satisfies all of the above, a and b are countable, and $f:a \rightarrow L$ is an embedding, which we want to extend to $\tilde{f}:b\rightarrow L$. Notice, that it is sufficient to show this in case, when b is a one-point extension of a, since then we can do so for any countable linear order by induction. But conditions from the Lemma assure exactly, that any one-point extension of a can be realized in L.

In the other direction, verification is straightforward. $\square$

Linear orders with this property were investigated already by Hausdorff [6], under the name $\eta _1$-orders, and later by Harzheim [5]. A brief survey with substantial bibliography can be found in Chapter 9 of the book by Rosenstein [13].

Since each one-point extension of a countable subset of countably saturated linear order is realised, each linear order of cardinality at most $\omega _1$ can be embedded into any countably saturated linear order. Moreover, the following stregthening of this observation is attributed to Hausdorff in [14], and [13] p.164.

Theorem 4

A countably saturated linear order contains a copy of any linear order, which doesn’t have a copy of $\omega _1$ or $\omega _1^*$.

In particular, each countably saturated linear order contains an isomorphic copy of the real line order type, therefore has cardinality at least $\mathfrak {c}$.

Let us now define a linear order, which will turn out to be prime countably saturated.

$$\begin{aligned} {\mathbb {L}}^{\omega _1}=\{x\in [-1,1]^{\omega _1}|\,|\{\alpha <\omega _1:\, x(\alpha )\ne 0\}|\le \omega \}. \end{aligned}$$

If D is a compact linear order and $d_0\in D$ is neither the least, nor the greatest element of D, then we define

$$\begin{aligned} {\mathbb {L}}_{(D,d_0)}^{\omega _1}=\{x\in D^{\omega _1}|\,|\{\alpha <\omega _1:\, x(\alpha )\ne d_0\}|\le \omega \}. \end{aligned}$$

The case $D=\{-1,0,1\}$ and $d_0=0$ is the classical construction by Hausdorff, and is described in [5]. Theorem 5 is therefore a slightly more general version of Theorem 3.22 therein, and Theorem 7—a version of Theorem 3.13.

Theorem 5

$\mathbb {L}^{\omega _1}$ and $\mathbb {L}_{(D,d_0)}^{\omega _1}$ are countably saturated.

This follows in fact almost immediately from a theorem of Novák.

Theorem 6

(Novák, [11]) Let L be a compact line. Then for any ordinal $\alpha$, $L^\alpha$ is compact. If L was moreover a continuum, then so is $L^\alpha$.

Proof

We prove the theorem by induction on $\alpha$. Case $\alpha =1$ is trivial, and for the succesor step it is sufficient to show that the product of two compact linear orders is compact, and the product of two continua is a continuum.

Let $(K,\le _K)$ and $(L,\le _L)$ be compact lines. Surely $(\min {K},\min {L})$ is the minimal element of $K\times L$, and $(\max {K},\max {L})$ is the greatest element. Let $S\subseteq K \times L$ be a non-empty set. We aim to find the least upper bound for S. If projection on the first coordinate ${\text {proj}}_K{S}$ has no maximal element, then $(\sup {{\text {proj}}_K{S}},\min {L})$ is the least upper bound for S. If ${\text {proj}}_K{S}$ has a maximal element, say $s_K$, then let $s_L$ be the least upper bound of $\{x \in L|\; (s_K,x) \in S\}$. Then $(s_K,s_L)$ is the least upper bound for S in $K\times L$. In a similar way we prove that S has the greatest lower bound.

Assume now that $\alpha$ is a limit ordinal, and for every $\beta <\alpha$, each lexicographic power $L^\beta$ is compact. Then the space ${\prod _{\beta <\alpha }\; L^{\beta }}$ with the product topology (not the lexicographic power) is a compact Hausdorff space, and so the set

$$\begin{aligned} P=\{{\overline{x}}\in \prod _{\beta<\alpha }L^\beta |\; \forall {\beta _1<\beta _2<\alpha } {\overline{x}}(\beta _2)\restriction \beta _1 = {\overline{x}}(\beta _1) \} \end{aligned}$$

is also compact, as a closed subset of a compact space. It is then sufficient to show that the lexicographic power $L^\alpha$ is a continuous image of P. The obvious function witnessing that is given by $\pi (\overline{x})(\beta )=\overline{x}(\beta +1)(\beta )$. To see that $\pi$ is continuous, it is enough to check that $\pi ^{-1}[\{x\in L^\alpha |\; x>a \}]$ is open for every $a \in L^\alpha$. But since inequality $x>a$ must be true already between some initial segments of x and a, it is easy to check that

$$\begin{aligned} \pi ^{-1}[\{x\in L^\alpha |\; x>a \}]=\{{\overline{x}} \in \prod _{\beta<\alpha }L^\beta |\; \exists {\gamma <\alpha }\; \overline{x}(\gamma )>a\restriction \gamma \} \cap P. \end{aligned}$$

The latter set is clearly open in P. The claim about continua is straightforward, given that a compact line is a continuum if and only if it is dense in itself. $\square$

Proof of Theorem 5

Notice, that $\mathbb {L}^{\omega _1}={\bigcup _{\alpha <\omega _1}{[-1,1]^\alpha }}$. We check conditions from Lemma 1. Density is clear. For verification of the last condition, take two sequences $\cdots<x_n<x_{n+1}<\cdots<y_{n+1}<y_n<\cdots$. There is some level $\alpha$, such that all $x_n$ and $y_n$ belong to $[-1,1]^\alpha$. Since it is compact, we can take supremum of $\{x_n\}_{n<\omega }$ in this set, and it will clearly separate each $x_n$ from each $y_n$.

The only problematic case is when we want to separate an increasing sequence $\{x_n\}_{n<\omega }$ from $\tilde{x}=\sup _{n<\omega }{x_n}$, where supremum is taken in some $[-1,1]^\alpha$, big enough to contain $\tilde{x}$ and each $x_n$. But then it is clear, that $\tilde{x}^\frown {-1} \in [-1,1]^{\alpha +1}$ is a good separating element. Proof for $\mathbb {L}_{(D,d_0)}^{\omega _1}$ goes exactly the same way. $\square$

Theorem 7

$\mathbb {L}^{\omega _1}$ is prime countably saturated. Likewise, if D is a separable compact line, and $d_0 \in D$ is neither the least, nor the greatest element, then $\mathbb {L}^{\omega _1}_{(D,d_0)}$ is prime countably saturated.

$\mathbb {L}^{\omega _1}=\{x\in [-1,1]^{\omega _1}|\,|\{\alpha <\omega _1:\, x(\alpha )\ne 0\}|\le \omega \}$ is (isomorphic to) an increasing sum $\bigcup _{\alpha <\omega _1}{[-1,1]^{\alpha }}$. If X is any countably saturated linear order, we build an embedding $\mathbb {L}^{\omega _1}\rightarrow X$ by induction on $\alpha$, using the Lemma below.

Lemma 2

Let $(X,\le _X)$ be any countably saturated linear order, $(L,\le _L)$ a linear order of countable character, and $(E,\le _E)$ be separable linear order with distinguished element $e_0$. Then each embedding $i:\, L\simeq L\times \{e_0\}\rightarrow X$ extends to $\tilde{i}:L\times E\rightarrow X$.

Proof

Det $E_0\subseteq E$ be a countable dense subset. Fix $l\in L$ and sequences $\alpha _n$ increasing, $\beta _n$ decreasing, witnessing countable character of l. Then, there exists an extension:

where $j(l)=e_0$, $j(\alpha _n)=\alpha _n$. and $j(\beta _n)=\beta _n$ for every n.

Next, we extend

$$\begin{aligned} i_l:(\{\alpha _n\}_{n<\omega },\le _L)+E_0+(\{\beta _n\}_{n<\omega },\le _L) \rightarrow X, \end{aligned}$$

to

$$\begin{aligned} \tilde{i}_l: (\{\alpha _n\}_{n<\omega },\le _L)+E+(\{\beta _n\}_{n<\omega },\le _L). \end{aligned}$$

This, in turn, can be done, because $E_0$ is countable and dense in E. Finally, we set $\tilde{i}(l,e)=i_l(e)$. $\square$

Now we only need to know, that the sets $[-1,1]^\alpha$ have countable character.

Proposition 2

If $\alpha$ is a countable ordinal, then $[-1,1]^\alpha$ doesn’t contain any copy of $\omega _1$ or $\omega _1^*$. In particular, it has countable character.

Proof

It is sufficient to prove that $2^\gamma$ doesn’t contain copy of $\omega _1$, for any countable ordinal $\gamma$. Assume otherwise, and let $\gamma <\omega _1$ be minimal, such that $\omega _1 \hookrightarrow 2^\gamma$.

Suppose that $\gamma$ is limit. It is easy to check that the set

$$\begin{aligned} 2^{<\gamma }=\{x\in 2^\gamma |\, \exists _{\beta<\gamma }\forall {\beta<\delta<\gamma }\; x(\delta )=0 \}=\bigcup _{\delta <\gamma }{2^\delta } \end{aligned}$$

is dense in $2^\gamma$, and so if $2^\gamma$ has an uncountable well-ordered sequence, so does $2^{<\gamma }$. But then for some $\beta <\gamma$, $2^\beta$ has uncountably many elements of that sequence, which itself constitute a copy of $\omega _1$. This contradicts the minimality of $\gamma$.

Suppose now that $\gamma$ is successor. If $2^\gamma$ has an uncountable well-ordered sequence, then either $\omega _1$ of its elements has 0 on the last coordinate, or $\omega _1$ of its elements has 1 on the last coordinate. Either way, their restriction to the first $\gamma -1$ coordinates constitute an uncountable well-ordered sequence in $2^{\gamma -1}$. This again contradicts the minimality of $\gamma$. $\square$

We address the question of uniqueness of a prime countably saturated order. This question is completely settled under CH, and it was proved by Hausdorff. It is in fact a particular case of a much more general phenomenon. Namely, given any first order theory, all $\kappa$-saturated models of cardinality $\kappa$ are pairwise isomorphic. For the proof of this well-known fact, we refer to for example [7], or to [9] for a more general approach, using the language of category theory.

Theorem 8

Assume CH. Then every two countably saturated linear orders of cardinality $\mathfrak {c}$ are isomorphic.

Let us note, that this is not true without CH. Namely, if $2^\omega \ge \omega _2$, then both

$$\begin{aligned} \{x\in [-1,1]^{\omega _1}|\,|\{\alpha <\omega _1:\, x(\alpha )\ne 0\}|\le \omega \}, \end{aligned}$$

and

$$\begin{aligned} \{x\in [-1,1]^{\omega _2}|\,|\{\alpha <\omega _2:\, x(\alpha )\ne 0\}|\le \omega \}, \end{aligned}$$

are countably saturated linear orders of cardinality continuum, but are not isomorphic, since the latter contains an isomorphic copy of the ordinal $\omega _2$. This was noted already in [4]. In many cases we can provide somewhat better example. For the forcing terminology we refer the reader to [10].

Theorem 9

It is relatively consistent with ZFC that there exist two non-isomorphic, countably saturated linear orders of cardinality $\mathfrak {c}$, none of which contains copy of neither $\omega _2$ nor $\omega _2^*$.

Proof

We start with some countable transitive model $\mathbb {V} \models ZFC+CH$. The idea is that we will find two linear orders satisfying conditions of the Theorem, but only one will contain a copy of some linear order of cardinality $\omega _2$ from $\mathbb {V}$. Let $\mathbb {P} \in \mathbb {V}$ be any c.c.c. forcing notion which forces $\mathfrak {c}>\omega _1$ (for example a finite support iteration of the Cohen forcing, of length $\omega _2$). Denote by $\mathbb {V}^\mathbb {P}$ the corresponding generic extension.

Lemma 3

In $\mathbb {V}^\mathbb {P}$, $\mathbb {L}^{\omega _1}$ doesn’t contain copy of any linear order of cardinality $\omega _2$, which is in $\mathbb {V}$.

Proof

Suppose that $(\omega _2,\preccurlyeq ) \in \mathbb {V}$ is a linear order, and that

$$\begin{aligned} \mathbb {P} \Vdash ``\dot{f}:(\omega _2,\preccurlyeq )\hookrightarrow \mathbb {L}^{\omega _1}\text {is increasing''}. \end{aligned}$$

For any $\alpha \in \omega _2$, in the generic extension ${f}(\alpha )$ is a sequence of reals of length $\omega _1$, which is constantly equal zero from some point. From the Maximum Principle, there exists a name $\dot{b_\alpha }$ for which $\mathbb {P} \Vdash \forall {\omega _1>\delta >\dot{b_\alpha }}\; \dot{f}(\alpha )(\delta )=0.$ Since $\mathbb {P}$ is c.c.c. there are at most countably many possible values of the ordinal $\dot{b_\alpha }$, and so we can define $c_\alpha =\sup {\{ \beta <\omega _1|\; \exists {p\in \mathbb {P}}\; p \Vdash \dot{b_\alpha }=\beta \}}$.

There is some $\beta < \omega _1$, for which the set $\{ \alpha < \omega _2|\; c_\alpha =\beta \}$ has cardinality $\omega _2$. Denote this set by S. Note, that in this case

$$\begin{aligned} \mathbb {P} \Vdash \dot{f}[S]\hookrightarrow [-1,1]^\beta . \end{aligned}$$

The set S was defined in $\mathbb {V}$, so $S \in \mathbb {V}$, and given that $|S|=\omega _2$, S must contain an uncountable well-ordered or reversed well-orered sequence (this is a standard application of the Erdös-Rado Theorem in $\mathbb {V}$). But $[-1,1]^\beta$ doesn’t contain uncountable (reversed) well-ordered sequences, by Proposition 2. This is a contradiction. $\square$

The Theorem is now proved as follows. We work in $\mathbb {V}^\mathbb {P}$. As one example take $\mathbb {L}^{\omega _1}$. For the other, let $R_0 \subseteq (2^{\omega _1})^\mathbb {V}$ be of cardinality $\omega _2$, and $R_0 \in \mathbb {V}$. Define inductively a sequence $(R_\alpha )_{\alpha \le \omega _1}$, such that for each $\alpha <\omega _1$, $R_{\alpha +1}$ patches $R_\alpha$ (see Definition 3 in the next section), and $R_\alpha ={\bigcup _{\beta <\alpha }{R_\beta }}$ for limit $\alpha$. Moreover, we assure that $|R_\alpha |\le \mathfrak {c}$ for each $\alpha$. It is clear, that $R_{\omega _1}$ is countably saturated linear order of cardinality $\mathfrak {c}$. $\square$

2.2 Linear dimension

We will use the notion of dimension, introduced by Novák in [11], for further classification of linear orders. Some time ago, it was also investigated by Giarlotta, under the name representability number [3]. Let 2 denote the 2-element linear order.

Definition 2

(Novák, [11]) Let L and X be linear orders, and $|L|\ge 2$. The dimension of X with respect to L is defined as

$$\begin{aligned} {\text {L-dim}}{X}=\min \{\alpha \in ON|\; X\hookrightarrow L^\alpha \}. \end{aligned}$$

Let us write down some easy observations.

Proposition 3

For any linear orders $L,L_1,L_2,X$, the following holds.

• If $X_1 \hookrightarrow X_2$, then ${\text {L-dim}}{X_1}\le {\text {L-dim}}{X_2}$.

• If $L_1\hookrightarrow L_2$, then $\text {L}_1-\text {dim}{X}\ge \text {L}_2-\text {dim}{X}$.

• If $L_1\hookrightarrow L_2$ and $L_2\hookrightarrow L_1$, then for every X, $\text {L}_1-\text {dim}{X}=\text {L}_2-\text {dim}{X}$.

In particular, the notions of $2^\omega -\text {dim}{X}, {\text {I-dim}}{X}$, and ${\mathbb {R}}-\text {dim}{X}$ coincide. We will denote them ${\text {I-dim}}{X}$.

Proposition 4

If A and B are subsets of some linear order, ${\text {I-dim}}{A}<\omega _1$, and ${\text {I-dim}}{B}<\omega _1$, then ${\text {I-dim}}{(A\cup B)}<\omega _1$.

Proof

Because $A\cup B=(A\setminus B) \, \cup (A\cap B) \, \cup (B \setminus A)$, and the sum is clearly disjoint, we can restrict ourselves to the case when A and B are disjoint. For $b_0,b_1 \in B$ we set $b_0 \sim b_1$ iff $[b_0,b_1]\cap A =\emptyset$ ($[b_0,b_1]$ actually means $[\min \{b_0,b_1\},\max \{b_0,b_1\}]$, but this is a harmless abuse of notation, and we won’t bother with it anymore). This is a convex equivalence relation on B. Let $i_A:A\rightarrow I^\alpha$, and $i_B:B\rightarrow I^\beta$ be embeddings. We aim to extend $i_A$ to $\tilde{i}:A\cup B\rightarrow I^{\alpha +1+\beta }$. It is sufficient to define $\tilde{i}$ on each equivalence class separately. Let $S\subseteq B$ be a selector of $\sim$. For $s\in S$ we choose $i'(s) \in I^{\alpha +1}$, so that

$$\begin{aligned} \{i_A(a)|\, a<s\}<\{i'(s)\}<\{i_A(a)|\, s<a\}. \end{aligned}$$

Keeping in mind that $I^\alpha$ is a linearly ordered continuum of countable character, it requires only standard verifiation, that this can be done, and it is left to the Reader. Let $i'(a)=i_A(a)$ for $a\in A$. Then, having $i':A\cup S \rightarrow I^{\alpha +1}$, we extend it to $\tilde{i}:A\cup B \rightarrow I^{\alpha +1+\beta }$, via the formula

$$\begin{aligned} \tilde{i}(b)=i'(s_b)^\frown i_B(b), \end{aligned}$$

for $s_b\sim b$, and $s_b\in S$. $\square$

Theorem 10

(Novotný [12], Novák [11]) Let L be a linearly ordered continuum. Then for any ordinal $\alpha$, ${\text {L-dim}}{L^\alpha }=\alpha$.

Once we are familiar with the theorem of Novák, this result is an immediate consequence of a Lemma due to Novotný, which has quite pleasant proof. Since it was originally published in Czech, we quote it here.

Lemma 4

(Novotný, [12]) Let $(X,\le )$ be a linearly ordered continuum, and $\mathcal {A}$ be a disjoint family of closed, not-single-point intervals in X. The relation $\le$ induces a linear order $\tilde{\le }$ on $\mathcal {A}$, via the formula $I\tilde{\le }J$ iff $\max {I}<\min {J}$. Then $(X,\le )$ does not embed into $(\mathcal {A},\tilde{\le })$.

Proof

Suppose otherwise, that $i:(X,\le )\rightarrow (\mathcal {A},\tilde{\le })$ is an embedding. Let $A=\{x\in X|\; x<\min {i(x)} \}$. A is nonempty, because X has the least element, so let $a=\sup {A}$. We have two cases.

• (a) Assume $a\in A$. Then $a<\min {i(a)}$, but for any $a'\in i(a)$ this is the case as well. Since $a'\in i(a)$ is greater than a, this contradicts the definition of a.

• (b) Assume $a\notin A$. Then $a\ge \min {i(a)}$. If $a<\max {i(a)}$, then let $a' \in (a,\max {i(a)})$. Then $a'<\max {i(a)}<\min {i(a')}$, so $a'\in A$, and this is a contradiction. The only way out is $a\ge \max {i(a)}$. But notice, that in this case for any $a' \in (\min {i(a)},a)$, we have $a' \notin A$. Therefore $\sup {A}\le \min {i(a)}<a$, which is again a contradiction.

$\square$

Corollary 1

If $\alpha$ is an ordinal with the property, that $\omega \cdot \alpha =\alpha$, then ${\text {I-dim}}{2^\alpha }=\alpha$.

Proof

Clearly $2^\omega \hookrightarrow I$, so

$$\begin{aligned} 2^\alpha \simeq 2^{\omega \cdot \alpha } \simeq (2^\omega )^\alpha \hookrightarrow I^\alpha \end{aligned}$$

In the other direction,

$$\begin{aligned} I^\alpha \hookrightarrow (2^\omega )^\alpha \simeq 2^{\omega \cdot \alpha } \simeq 2^\alpha . \end{aligned}$$

$\square$

Theorem 11

Assume $\mathfrak {c}=2^{\omega _1}$. Let $X=I^{\omega _1}$. Then $\mathbb {L}^{\omega _1}_{(X,0)}$ is a countably saturated linear order of cardinality $\mathfrak {c}$, character $\omega _1$, and ${\text {I-dim}}=\omega _1^2$. In particular $\mathbb {L}^{\omega _1}_{(X,0)}$ is not isomorphic to $\mathbb {L}^{\omega _1}$.

Proof

For every ordinal $\alpha <\omega _1$, $\mathbb {L}^{\omega _1}_{(X,0)}$ contains a copy of $X^\alpha$. $X^{\alpha }$ is a linearly ordered continuum, so ${\text {I-dim}}{X^\alpha }= {\text {I-dim}}{I^{\omega _1\cdot \alpha }}=\omega _1\cdot \alpha$. Since $\alpha$ was arbitrary, this shows that ${\text {I-dim}}{\mathbb {L}^{\omega _1}_{(X,0)}} \ge \omega _1^2$.

In the other direction, $\mathbb {L}^{\omega _1}_{(X,0)}\hookrightarrow X^{\omega _1} \hookrightarrow I^{\omega _1\cdot \omega _1}$, so ${\text {I-dim}}{\mathbb {L}^{\omega _1}_{(X,0)}}=\omega _1^2$. $\square$

2.3 Uniqueness of the prime countably saturated linear order

The assumption of CH is not needed if we take into account only linear orders which are in some sense minimal.

Theorem 12

(Harzheim, [5]) All prime countably saturated linear orders are isomorphic.

The order type of the unique prime countably saturated order is in fact the order type $h_1$ in the language of [5]. Theorem 12 follows from Theorems 3.13 and 3.17 there. We give a proof in a slightly different language.

Definition 3

For two linear orders $A\subseteq B$, we will say, that B patches A, if for any two countable sets $a_0, a_1 \subseteq A$, if $a_0<a_1$, there exists $b\in B$, such that $a_0<\{b\}<a_1$.

One can easily verify, that this comes down to patching gaps of four different types: $(1,1),\;(\omega ,1),\;(1,\omega ),$ and $(\omega ,\omega )$, which correspond to situations, where respectively

• $a_0$ has the greatest element, and $a_1$ has the least element,

• $a_0$ doesn’t have the greatest element, but $a_1$ has the least element,

• $a_0$ has the greatest element, but $a_1$ doesn’t have the least element,

• neither $a_0$ has the greatest element, nor $a_1$ has the least element.

Lemma 5

If ${\text {I-dim}}{A} < \omega _1$, then there exists $B\supseteq A$ patching A, such that ${\text {I-dim}}{B}<\omega _1$.

Proof

Let $\alpha ={\text {I-dim}}{A}$. Then $A \hookrightarrow I^{\alpha }$, so we can assume that $A \subseteq I^{\alpha }$. This order clearly patches gaps of the form (1, 1) and $(\omega ,\omega )$. To take care of $(1, \omega )$- and $(\omega ,1)$-gaps, we replace every point of $I^{\alpha }$ with a unit interval.

$$\begin{aligned} A\subseteq I^{\alpha }\simeq I^{\alpha }\times \{0\}\subseteq I^{\alpha +1}. \end{aligned}$$

$\square$

Definition 4

If L is a countably saturated linear order, we define a filtration of L, as a sequence of subsets of L, $\{L_\alpha \}_{\alpha <\omega _1}$, with the following properties:

• $\{L_\alpha \}_{\alpha <\omega _1}$ is increasing with respect to inclusion,

• for each $\alpha <\omega _1$, $L_{\alpha +1}$ patches $L_\alpha$,

• ${\bigcup _{\alpha <\omega _1}{L_\alpha }=L}$,

• for each $\alpha <\omega _1$, ${\text {I-dim}}{L_\alpha }<\omega _1$.

The next Lemma is the key tool in the proof of Theorem 12. It should be noted, that its first part as actually a direct consequence of Theorem 3.6 p. 81 [5].

Lemma 6

(On Bounded Injectivity) Assume, that A and B are linear orders with ${\text {I-dim}}{A}, {\text {I-dim}}{B}<\omega _1$, and L is countably saturated. Let $f:A\hookrightarrow L$, and $i:A\hookrightarrow B$ be increasing functions. Then

• (a) There exists an increasing mapping $\tilde{f}:A \hookrightarrow L$, such that $\widetilde{f}\circ i=f$.

  

• (b) If L has a filtration $\{L_\alpha \}_{\alpha <\omega _1}$, and there is an index $\alpha < \omega _1$, for which $f[A] \subseteq L_\alpha$, then we can choose $\tilde{f}$ bounded, i.e. $\tilde{f}[B]\subseteq L_\beta$, for some $\beta <\omega _1$.

Proof

Without loss of generality we may assume that $A\subseteq B$, and i is the identity mapping. For $b_0,b_1 \in B \setminus A$, we set $b_0 \sim b_1$ iff $[b_0,b_1]\cap A=\emptyset$. This clearly defines a convex equivalence relation on $B\setminus A$, and it is sufficient to define $\tilde{f}$ on each equivalence class separately. a) For any $b\in B \setminus A$,

$$\begin{aligned} \{a\in A|\,a<b\}<[b]_{\sim }<\{a\in A|\,b<a\}. \end{aligned}$$

Since any linear order of countable ${\text {I-dim}}$ has character $\omega$, both these sets have countable cofinality and coinitiality respectively. Therefore, this is also the case for

$$\begin{aligned} \{f(a)|\,a<b, \, a \in A\}<\{f(a)|\,b<a,\, a\in A\}, \end{aligned}$$

so L will contain a point between them. Since it will contain a point, it will contain an interval, and this interval will be countably saturated itself. Since it will be countably saturated, it will contain a copy of $\mathbb {L}^{\omega _1}$, and in turn a copy of any order of countable I-dimension, in particular $[b]_{\sim }$. We set $\tilde{f}$ to isomorphically map $[b]_{\sim }$ onto that copy.

The “bounded” variant will require more care.

Sublemma

Let $\{L_\alpha \}_{\alpha <\omega _1}$ be a filtration. Then for every $\alpha <\omega _1$, $\omega \le \gamma <\omega _1$, and for all $a_0 < a_1 \in L_\alpha$, $(a_0,a_1)\cap L_{\alpha +\gamma +1}$ contains a copy of $2^\gamma$.

Proof

We proceed by induction on $\gamma$.

$\gamma =\omega$. $L_{\alpha +\omega } \cap (a_0,a_1)$ contains copy of rationals, so $L_{\alpha +\omega +1} \cap (a_0,a_1)$ contains a copy of reals, and in particular $2^\omega$.

successor step. Assume that $i:2^\gamma \hookrightarrow (a_0,a_1)\cap L_{\alpha +\gamma +1}$.For every $x\in 2^\gamma$, there exists $v_x \in L_{\alpha +\gamma +2}$, such that

$$\begin{aligned} i(x)<v_x<i[\{y\in 2^\gamma |\,y>x\}], \end{aligned}$$

because the set on the right has countable coinitiality. We set $\tilde{i}(x^\frown 0)=i(x)$, and $\tilde{i}(x^\frown 1)=v_x$.

limit step. Fix an increasing sequence $\{\gamma _n\}_{n<\omega }$, cofinal on $\gamma$. Let

$$\begin{aligned} 2^{<\gamma }=\{x\in 2^\gamma |\,\exists {\beta<\gamma }\forall {\beta<\delta<\gamma }\; x(\delta )=0\}=\bigcup _{n<\omega }{2^{\gamma _n}}. \end{aligned}$$

By induction we define embeddings $j_n:2^{\gamma _n}\hookrightarrow (a_0,a_1)\cap L_{\alpha +\gamma _n+1}$, such that $j_n\subseteq j_{n+1}$. $j_0:2^{\gamma _0}\rightarrow (a_0,a_1)\cap L_{\alpha +\gamma _0+1}$ exists by induction hypothesis. Assume, that we have $j_n$ defined, and look at $x\in 2^{\gamma _n}$. Since $L_{\alpha +\gamma _n+2}$ patches $L_{\alpha +\gamma _n+1}$, there exist $v_x$, such that

$$\begin{aligned} j_n(x)<v_x<j_n[\{y\in 2^{\gamma _n}|\,y>x\}], \end{aligned}$$

and $(j_n(x),v_x)\subseteq (a_0,a_1)$ (this is automatic, unless $\{y\in 2^{\gamma _n}|\,y>x\}=\emptyset$). By the induction hypothesis we can find $l:2^{\gamma _{n+1}\setminus \gamma _n}\hookrightarrow (j_n(x),v_x)\cap L_{\alpha +\gamma _{n+1}+1}$. We define, for $w\ne 0$, $j_{n+1}(x^\frown w)=l(w)$. Notice, that $x^\frown w \in 2^{\gamma _n}\times 2^{\gamma _{n+1}\setminus \gamma _n}=2^{\gamma _{n+1}}$.

Finally, we see that ${\bigcup _{n<\omega }{ j_n:2^{<\gamma }\hookrightarrow L_{\alpha +\gamma }}}$, and so $2^\gamma \hookrightarrow L_{\alpha +\gamma +1}$. $\square$

b) Assume that $\{L_\alpha \}_{\alpha <\omega _1}$ is a filtration of L, and $f[A]\subseteq L_\beta$. Then there exists $u,v \in L_{\beta +2}$, such that

$$\begin{aligned} \{f(a)|\,a<b\}<u<v<\{f(a)|\,b<a\}. \end{aligned}$$

By the Sublemma, $(u,v)\cap L_{\beta +\omega \cdot \gamma +1}$ contains a copy of $2^{\omega \cdot \gamma }$, so also $I^\gamma$, for any given $\gamma$. If $\gamma =\max \{{\text {I-dim}}{B},\omega \}$, then in particular $(u,v)\cap L_{\beta +2+\omega \cdot \gamma +1}$ contains a copy of $[b]_{\sim }$. We define $\tilde{f}$ the same way as before, but making sure, that $\tilde{f}[B]\subseteq L_{\beta +\omega \cdot \gamma +1}$. $\square$

Proposition 5

A countably saturated linear order is prime if and only if it has a filtration.

Proof

Assume, that L is prime countably saturated. Then without loss of generality, we can assume, that $L \subseteq \mathbb {L}^{\omega _1}$. Denote

$$\begin{aligned} I_\alpha =\{x\in \mathbb {L}^{\omega _1}|\; \forall {\beta \ge \alpha } \;x(\beta )=0 \}. \end{aligned}$$

Notice, that the sequence $\{I_\alpha \}_{\alpha <\omega _1}$ constitutes a filtration of $\mathbb {L}^{\omega _1}$. We define a filtration $\{L_\alpha \}_{\alpha <\omega _1}$ of L by induction.

• $L_0 = I_0 \cap L$,

• Assume we have $L_\gamma$ constructed. Using Lemma 5, we can find $\tilde{L_\gamma } \subseteq \mathbb {L}^{\omega _1}$, with ${\text {I-dim}}{\tilde{L_\gamma }}<\omega _1$, which patches $L_\gamma$. Define $L_{\gamma +1}=\tilde{L_\gamma } \cup (L\cap I_{\gamma })$. By Proposition 4, ${\text {I-dim}}{L_{\gamma +1}}<\omega _1$.

• When $\gamma <\omega _1$ is limit, we set $L_\gamma ={\bigcup _{\alpha <\gamma }{L_\alpha }}.$

Suppose now that L is countably saturated, and has a filtration $\{L_\alpha \}_{\alpha <\omega _1}$. Let X be any countably saturated linear order. Using Lemma On Bounded Injectivity, we can easily build an $\subseteq$-increasing sequence of increasing mappings $i_\alpha :L_\alpha \hookrightarrow X$. Its sum will be an embedding of L into X. $\square$

Proof of Theorem 12

Consider two prime countably saturated linear orders. They both admit filtrations, so, using the Lemma On Bounded Injectivity, we can use back-and-forth argument, to inductively build an isomorphim between them. $\square$

We can characterize the prime countably saturated linear order using the dimension.

Proposition 6

Let $(L,\le )$ be a countably saturated linear order. The following are equivalent:

• L is prime countably saturated;

• L is an increasing sum $\bigcup _{\alpha <\omega _1}{L_\alpha }$, where ${\text {2-dim}}{L_\alpha }<\omega _1$, for each $\alpha <\omega _1$;

• L is an increasing sum $\bigcup _{\alpha <\omega _1}{L_\alpha }$, where ${\text {I-dim}}{L_\alpha }< \omega _1$, for each $\alpha <\omega _1$.

Proof

• $1 \Rightarrow 2$. If L is prime countably saturated, then

• $L \simeq \mathbb {L}^{\omega _1}= \bigcup _{\alpha<\omega _1}{[-1,1]^\alpha }\hookrightarrow \bigcup _{\alpha <\omega _1}{2^{\omega \cdot \alpha }}$.

• $2 \Rightarrow 3$. Clear.

• $3 \Rightarrow 1$. If X is any countably saturated linear order, we construct an embedding

• $L \hookrightarrow X$ by induction, using the Lemma on Bounded Injectivity. $\square$

This should by compared with the Characterization Theorem of Harzheim [5].

Proposition 7

(Harzheim, [5]) Let $(L,\le )$ be a countably saturated linear order. The following are equivalent:

• L is prime countably saturated;

• L is an increasing sum $\bigcup _{\alpha <\omega _1}{L_\alpha }$, where $L_\alpha$ doesn’t contain a copy of $\omega _1$ or $\omega _1^*$, for each $\alpha <\omega _1$.

There is one more propery of $\mathbb {L}^{\omega _1}$, which follows from the Lemma on Bounded Injectivity.

Proposition 8

Any isomorphism between subsets of $\mathbb {L}^{\omega _1}$ of countable ${\text {I-dim}}$ extends to an automorphism of $\mathbb {L}^{\omega _1}$. In particular, any automorphism between countable suborders extends to an automorphism of $\mathbb {L}^{\omega _1}$.

Proof

Let $\phi :X \hookrightarrow Y$ be in isomorphism, where $X,Y \subseteq \mathbb {L}^{\omega _1}$, and ${\text {I-dim}}(X)<\omega _1$, ${\text {I-dim}}(Y)<\omega _1$. We can define two filtrations $(X_\alpha )_{\alpha <\omega _1}$, $(Y_\alpha )_{\alpha <\omega _1}$, such that $X=X_0$, $Y=Y_0$. Then we successively extend the automorphism, using the Lemma on Bounded Injectivity. $\square$

As a matter of fact, Lemma on Bounded Injectivity basically proves that $\mathbb {L}^{\omega _1}$ is a Fraïssé limit of length $\omega _1$, as defined in [9], of the class of linear orders of countable ${\text {I-dim}}$.

3 $(\omega _1,\mathfrak {c})$-saturated graphs

We investigate existence and uniqueness of graphs with certain homogeneity properties. All graphs denoted by G or H are undirected, those ones denoted by $\overleftarrow{G}$ or $\overleftarrow{H}$ are directed, and if $\overleftarrow{G}$ is a directed graph, then G is the corresponding undirected graph, i.e. $\{a,b\}\in E(G)$ if and only if $(a,b)\in E(\overleftarrow{G})$ or $(b,a)\in E(\overleftarrow{G})$. By N(g), where g is a vertex of some graph, we denote the set of all vertices connected to g (in whichever direction, in case the graph is directed). By $N_z(g)$ we denote the set of all vertices connected with g by an arrow starting in g.

Definition 5

Let $\lambda$ and $\kappa$ be any cardinal numbers. A graph G is $(\lambda ,\kappa )$-saturated if for any $A,B\subset G$, $|A|<\lambda$, $|B|<\kappa$, $A\cap B=\emptyset$, there exists a vertex $v\in G$ such that $A\subseteq N(v)$, $B\cap N(v)=\emptyset$.

It is long and well-known, that all countable $(\omega ,\omega )$-saturated graphs are isomorphic. Under suitable cardinal arithmetic this result can be generalized to higher cardinalities. In particular if CH holds, then all $(\omega _1,\omega _1)$-saturated graphs of cardinality $\mathfrak {c}$ are isomorphic.

Definition 6

We introduce some classes of graphs, useful for our purposes.

• A graph G is of the first type, if there exists a bijective enumeration $G=\{g_\alpha |\, \alpha <\mathfrak {c}\}$, such that:

  • 1. $|N(g_\alpha )\cap \{g_\beta |\beta <\alpha \}|\le \omega ,$ for all $\alpha <\mathfrak {c};$

  • 2. for each countable $A\subseteq G$, there are continuum many indices $\alpha$, such that $N(g_\alpha )\cap \{g_\beta |\beta <\alpha \}=A$.

• A graph $\overleftarrow{G}$ is strictly saturated, if

  • 1. $|N_z(v)|\le \omega$, for each $v\in \overleftarrow{G}$;

  • 2. for each countable $A\subseteq \overleftarrow{G}$, there are $|\overleftarrow{G}|$-many $g \in \overleftarrow{G}$, such that $N_z(g)=A$.

As in the case of linear orders, a graph from some class of graphs $\mathcal {C}$ is prime, if it embeds into every graph from $\mathcal {C}$. We aim to prove the existence and uniqueness of the prime $(\omega _1,\mathfrak {c})$-saturated graph.

Recall, that the colouring number, denoted ${\text {Col}}(G)$ of a graph G is the least cardinal $\kappa$, such that there exists a well-ordering $(G,\preccurlyeq )$, with the property

$$\begin{aligned} |\{ h\preccurlyeq g|\, \{g,h\}\in E(G) \}|<\kappa , \end{aligned}$$

for all $g \in G$. In follows from the definition, that any graph of the first type has colouring number $\omega _1$.

Proposition 9

If a graph G is of the first type, then the edges of G can be directed in such a way, that the resulting directed graph is strictly saturated, and doesn’t have an infinite path.

Proof

Fix a well-ordering, which witnesses that G is of the first type, and then direct every edge in a decreasing manner with respect to that ordering. $\square$

This is an important observation, and this is why.

Lemma 7

If $\overleftarrow{G}$ and $\overleftarrow{H}$ are strictly saturated graphs of cardinality $\mathfrak {c}$ without infinite paths, then they are isomorphic.

The proof will slightly resemble the proof of Mostowski Collapse Lemma.

Proof

Let $\{g_\alpha |\alpha <\mathfrak {c}\}$ and $\{h_\alpha |\alpha <\mathfrak {c}\}$ be any bijective enumerations of $\overleftarrow{G}$ and $\overleftarrow{H}$ respectively. By induction we construct an isomorphism $\phi :\overleftarrow{G}\rightarrow \overleftarrow{H}$. Assume, that we are in step $\eta <\mathfrak {c}$.

• If $\eta$ is even, then choose the least $\alpha$ such that $\phi (g_\alpha )$ has not yet been defined, but $\phi \restriction N_z(g_{\alpha })$ is defined (such element always exists, because there’s no infinite path). By strict saturation we can choose $h\in \overleftarrow{H}$, for which $\phi ^{-1}(h)$ is not yet defined, and $N_z(h)=\phi [N_z(g_{\alpha })]$. Set $\phi (g_{\alpha })=h$.

• If $\eta$ is odd, then choose the least $\alpha$, such that $\phi ^{-1}(h_\alpha )$ is has not been defined, but $\phi ^{-1}\restriction N_z(h_{\alpha })$ is defined. Again, by strict saturation we can choose $g\in \overleftarrow{G}$, which hasn’t yet been used, and $N_z(g)=\phi ^{-1}[N_z(h_{\alpha })]$. Set $\phi (g)=h_\alpha$.

A standard verification shows that $\phi$ is indeed a bijection between $\overleftarrow{G}$ anf $\overleftarrow{H}$. Indeed, suppose that some vertex $x_0 \in \overleftarrow{G}$ somehow evaded being assigned the image $\phi (x_0)$. Then, there must be some $x_1\in \overleftarrow{G}$, such that $x_0\rightarrow x_1$, and which shared this fate. If not, then $\phi \restriction N_z(x_0)$ would have been defined, and since ${\text {cf}}{\mathfrak {c}}> \omega =|N_z(x_0)|$, it would have been defined at a certain step. From that moment on, $x_0$ would be ready to be chosen next. Since there are less than $\mathfrak {c}$ elements with priority below $x_0$, and $\mathfrak {c}$ future steps, its turn would eventually arrive. But then we apply the same argument to $x_1$, getting $x_1 \rightarrow x_2$, and so on, producing an infinite path. Because of symmetry of the construction, the same argument shows, that $\phi$ is onto.

We will check, that it is a homomorphism. Fix an arrow $g_\alpha \rightarrow g_{\beta }$ in $\overleftarrow{G}$. The construction assures, that $\phi (g_\beta )$ is defined before $\phi (g_\alpha )$.

• 1. Suppose, that we are defining $\phi (g_\alpha )$ in an even step. Then $\phi (g_\alpha )\rightarrow \phi (g_\beta )$ is in $\overleftarrow{H}$, by the definition of $\phi (g_\alpha )$.

• 2. Suppose, that we are defining $\phi (g_\alpha )$ in an odd step. Then, in this step we are setting $\phi ^{-1}(h_\gamma )=g_\alpha$ for some $\gamma <\mathfrak {c}$. Then $g_\beta \in N_z(g_\alpha )=\phi ^{-1}[N_z(h_\gamma )]$. Therefore, $\phi (g_\beta )\in N_z(h_\gamma )$, and $h_\gamma =\phi (g_\alpha )$. So $\phi (g_\alpha )\rightarrow \phi (g_\beta )$ is in $\overleftarrow{H}$.

Because the construction was symmetric, exactly the same argument shows that if $g_\alpha$ and $g_\beta$ were not connected, $\phi (g_\alpha )$ and $\phi (g_\beta )$ are not connected either. This finishes the proof. $\square$

We can now give the internal characterization of prime strictly saturated graphs.

Theorem 13

A strictly saturated graph of cardinality $\mathfrak {c}$ is prime if and only if it has no infinite path.

Proof

• $\Rightarrow .$ The property of having no infinite path is hereditary, so if some strictly saturated graph has this property (and we will soon see, that it has), then so does every prime one.

• $\Leftarrow .$ We do the same construction as in the proof above, but only in one direction, so that we obtain an embedding instead of an isomorphism. The only thing which is not immediate, is why if $g_\alpha$ and $g_\beta$ are not connected, $\phi (g_\alpha )$ and $\phi (g_\beta )$ are not connected either. But we are choosing $\phi (g_\alpha )$ in a way that $N_z(\phi (g_\alpha ))=\phi [N_z(g_\alpha )]$. $g_\beta \notin N_z(g_\alpha )$, so $\phi (g_\beta ) \notin N_z(\phi (g_\alpha )).$ $\square$

Corollary 2

All prime strictly saturated graphs are isomorphic.

From Proposition 9 follows

Corollary 3

All graphs of the first type are isomorphic.

Theorem 14

There exists a prime strictly saturated graph without an infinite path. Moreover, a graph G is of the first type if and only if it can be assigned a direction of edges, such that the resulting directed graph is prime strictly saturated.

Proof

All graphs of the first type are isomorphic, and also all prime strictly saturated graphs are isomorphic. We therefore need only to find some strictly saturated graph $\overleftarrow{G}$ with no infinite path, such that G is of the first type. We apply a bookkeeping argument, similar to the one used in the proof of existence of Avilés–Brech Boolean algebra in [1]. Let $\{\Phi _\alpha |\alpha <\mathfrak {c}\}$ be a partition of $\mathfrak {c}$ into sets of cardinality $\mathfrak {c}$, and such that $\min {\Phi _\alpha }\ge \alpha$.

• $G_0=\{0\}$, and for $\gamma \in \Phi _0$, we set $S_\gamma =\{0\}$.

• If $\alpha$ is a limit ordinal, let $G_\alpha =\bigcup \{G_\beta |\beta < \alpha \}$, and $\{S_\gamma |\gamma \in \Phi _\alpha \}$ be an enumeration of all countable subsets of $G_\alpha$, in which every set appears $\mathfrak {c}$ many times.

• Assume that $G_\alpha$ and $S_\gamma$, for $\gamma \in \bigcup {\{\Phi _\beta |\beta \le \alpha \}}$, have been defined. We set $G_{\alpha +1}=G_{\alpha }\cup \{\alpha \}$, and $\{\alpha ,g\}\in G_{\alpha +1}$ iff $g\in S_\alpha$, where $g\in G_\alpha$. Finally, we use $\Phi _{\alpha +1}$ to list all countable subsets of $G_{\alpha +1}$, and each of them $\mathfrak {c}$-many times.

A natural ordering of G given by the well-ordering of the ordinal $\mathfrak {c}$ witnesses, that G is of the first type, and therefore can be directed into $\overleftarrow{G}$, strictly saturated graph without infinite path, by directing every edge downwards with respect to the well-ordering of $\mathfrak {c}$. $\square$

Note, that this argument actually shows, that for any cardinal $\kappa$ such that $\kappa ^\omega =\kappa$, there exists a strictly saturated graph of cardinality $\kappa$ without an infinite path, and all such graphs are isomorphic.

We obtain an internal characterization of prime $(\omega _1,\mathfrak {c})$-saturated graphs.

Theorem 15

An $(\omega _1,\mathfrak {c})$-saturated graph is prime $(\omega _1,\mathfrak {c})$-saturated if and only if it is of the first type.

Proof

$\Leftarrow .$ Let G be the graph of the first type. For simplifying notation, let us assume that $V(G)=\mathfrak {c}$, and its natural well-ordering witnesses the first type of G.

First, we check that if G is of the first type, it is actually $(\omega _1,\mathfrak {c})$-saturated. Let $A\subseteq G$ be countable and $B\subseteq G$ be of cardinality less than $\mathfrak {c}$, disjoint with A. For all $b\in B$, let $C_b=N(b)\cap b$, and $S=\bigcup \{C_b|\, b\in B\}$. $|S|<\mathfrak {c}$, so there exists $g \in G\setminus S$, such that $N(g)\cap g=A$. Clearly $N(g)\cap B=\emptyset$.

Verification that G embeds into any $(\omega _1,\mathfrak {c})$-saturated graph is proved by a standard transfinite induction argument.

$\Rightarrow .$ Consider a prime $(\omega _1,\mathfrak {c})$-saturated graph. Without loss of generality we can assume that it is an induced subgraph of G. Our goal for the moment is to prove, that we can direct its edges in such a way, that we obtain the prime strictly saturated graph. The property of being of the first type is not hereditary, but $(\omega _1,\mathfrak {c})$-saturated subgraph of G will be satisfying some property close enough to being of the first type.

Lemma 8

Let G be $(\omega _1,\mathfrak {c})$-saturated, such that $V(G)=\mathfrak {c}$, and for any $\alpha \in G$, $|N(\alpha )\cap \alpha | \le \omega$. Then G can be directed into the prime strictly saturated graph.

Proof

At the beginning, we direct all edges of G downwards with respect to the well-ordering of $\mathfrak {c}$. We will inductively redirect some edges, ensuring that every countable subset will appear as $N_z(\alpha )$, but no infinite path will be added.

So assume that $\overleftarrow{G}$ is our graph G with edges directed downwards. Denote by $A(\beta )$ the set of all vertices accessible via a finite, decreasing path from $\beta$ (note that adjective “decreasing” may sound superfluous, since all edges are directed downwards, but by the inductive procedure we will perform, some arrows will be reversed). It is clearly countable. Let $\{C_\alpha \}_{\alpha <\mathfrak {c}}$ be an enumeration of all countable subsets of $\mathfrak {c}$, such that each set appears $\mathfrak {c}$ many times. By induction we will choose the sequence $(x_\alpha )_{\alpha <\mathfrak {c}}$, satisfying what follows.

• 1. $x_\alpha \notin \{x_\beta |\,\beta <\alpha \}$;

• 2. $C_\alpha \subseteq N(x_\alpha )\cap x_\alpha$;

• 3. $N(x_\alpha )\cap { \bigcup _{\eta \le \alpha } \bigcup _{\beta \in C_\eta }{(A(\beta ) \setminus C_\alpha )} }=\emptyset$;

• 4. $(N(x_\alpha )\setminus C_\alpha ) \cap \bigcup \{N(x_\beta )\cap x_\beta |\,\beta <\alpha \}=\emptyset$;

• 5. $x_\alpha \notin {\bigcup _{\beta <\alpha }{N(x_\beta )\cap x_\beta }}$;

• 6. $N(x_\alpha )\cap (\{x_\beta |\,\beta <\alpha \}\setminus C_\alpha )=\emptyset$.

At each step the choice can be made because of $(\omega _1,\mathfrak {c})$-saturation. Once $x_\alpha$ has been chosen, we reverse all arrows $c\leftarrow x_\alpha$ for $c\in N(x_\alpha )\cap x_\alpha \setminus C_\alpha$. It follows from 2. that every countable subset of $\mathfrak {c}$ will appear as $N_z(\alpha )$ for $\mathfrak {c}$ many $\alpha$. The only things which are left to check, is that we won’t produce an infinite path, and that we will preserve the property $|N_z(\alpha )|\le \omega$ for $\alpha <\mathfrak {c}$. Since in the $\alpha$-th step we reverse only arrows from $N(x_\alpha )\cap x_\alpha \setminus C_\alpha$, for the second part, it is sufficient to prove

Claim 1

The sets $N(x_\alpha )\cap x_\alpha \setminus C_\alpha$ are pairwise disjoint for $\alpha <\mathfrak {c}$.

Proof

Follows directly from 4. $\square$

Claim 2

Each arrow was reversed no more than once.

Proof

If we are in step $\alpha$ and the arrow $c\leftarrow x_\alpha$ is about to be reversed for the second time, then clearly $c=x_\beta$ for some $\beta <\alpha$. But this contradicts 6. $\square$

Claim 3

There is no situation where $\alpha<\beta<\gamma <\mathfrak {c}$, and $\alpha \rightarrow \beta \rightarrow \gamma$.

Proof

Suppose otherwise. Then $\beta =x_\delta$ and $\gamma =x_\eta$, for some $\delta ,\eta <\mathfrak {c}$. If $\delta <\eta$, then $\beta \in (\{x_\rho |\,\rho <\eta \}\setminus C_\eta )\cap N(x_\eta )$, contradicting 6. If $\eta <\delta$, then $\beta =x_\delta \in N(x_\eta )\cap x_\eta$, which contradicts 5. $\square$

Claim 4

If $\alpha <\beta$, $\gamma <\beta$, $\gamma <\delta$, and we have arrows $\alpha \rightarrow \beta$, $\gamma \rightarrow \delta$, and a finite, decreasing path from $\beta$ to $\gamma$. Then there exists $\eta ,\theta < \mathfrak {c}$, such that $x_\eta =\beta$, $x_\theta =\delta$, and $\theta <\eta$.

Proof

Existence is clear, since otherwise the corresponding arrows wouldn’t be reversed. Suppose, that $\theta \ge \eta$. Then $\gamma \in N(x_\theta )\cap {\bigcup _{\epsilon \in C_\eta }{A(\epsilon )\setminus C_\theta }}$. But this contradicts 3. $\square$

Claim 5

The graph G has no infinite path.

Proof

Follows from the previous three Claims, and the fact that $\mathfrak {c}$ is well-ordered. $\square$

This concludes the proof of the last Claim. $\square$

Theorem 15 follows at once from Lemma 8. $\square$

Putting everything into one place, we obtain

Theorem 16

Let G be an $(\omega _1,\mathfrak {c})$-saturated graph. The following are equivalent.

• 1. G is prime $(\omega _1,\mathfrak {c})$-saturated.

• 2. ${\text {Col}}(G)=\omega _1$.

• 3. G is of the first type.

• 4. G has a direction of edges such that $\overleftarrow{G}$ is prime strictly saturated.

• 5. G has a direction of edges such that $\overleftarrow{G}$ is strictly saturated, without an infinite path.

Moreover, any of these properties determines the graph G uniquely up to isomorphism.

The natural question that arises, is whether similar conclusion holds for $(\omega _1,\omega _1)$-saturated graphs.

Theorem 17

The prime $(\omega _1,\omega _1)$-saturated graph exists only if $\mathfrak {c}=\omega _1$.

Proof

Let $\bar{\cdot }$ denote the complement of a graph. Let G be the prime $(\omega _1,\mathfrak {c})$-saturated graph. Of course both G and $\bar{G}$ are $(\omega _1,\omega _1)$-saturated, so it is enough to prove, that no graph of cardinality $\omega _2$ appears as an induced subgraph of both of them. Notice, that if S were such graph, then ${\text {Col}}(S)\le \omega _1$ and ${\text {Col}}(\bar{S})\le \omega _1$. So it is sufficient to show, that this cannot be the case.

Suppose otherwise, that S is a graph of cardinality $\omega _2$, with ${\text {Col}}(S)\le \omega _1$ and ${\text {Col}}(\bar{S})\le \omega _1$. Fix well-orderings $S=\{s_\alpha |\, \alpha<\omega _2\}=\{t_\alpha |\, \alpha <\omega _2\}$ witnessing the colouring number of S and $\bar{S}$ respectively, and denote by $\sigma :\omega _2 \rightarrow \omega _2$ the bijection given by $\forall _{\alpha <\omega _2} s_\alpha =t_{\sigma (\alpha )}$. Finally, fix an ordinal $\gamma >\omega _1$, for which also $\sigma (\gamma )>\sigma [\omega _1]$.

Let $A=\{\alpha < \omega _1|\, \{s_\alpha ,s_\gamma \}\in E(S)\}$, and $B=\{\alpha < \omega _1|\, \{s_\alpha ,s_\gamma \}\notin E(S)\}$. Clearly A is countable. Also $B=\{\alpha <\omega _1|\{t_{\sigma (\alpha )},t_{\sigma (\gamma )}\}\notin E(S)\}= \sigma ^{-1}[\{\alpha \in \sigma [\omega _1]|\{t_\alpha ,t_{\sigma (\gamma )}\}\notin E(S)\}]$, which is countable. But $A\cup B = \omega _1$, which is a contradiction. $\square$

3.1 The concrete representation

Consider the set of all hereditarily countable sets $H(\omega _1)$, i.e. the countable sets, of which every element is countable, and every element of element is countable, and so forth. This structure, and its higher analogs $H(\kappa )$, appear in the axiomatic set theory, because $(H(\kappa ),\in )$ is a natural model of ZFC without the powerset axiom, for any uncountable, regular $\kappa$. For more elaborated introduction, we refer the reader to [10]. It was noticed by Arturo Martinez-Celis, that the prime $(\omega _1,\mathfrak {c})$-saturated graph we have been studying is isomorphic to the graph where set of vertices is $H(\omega _1)$, and an edge is drawn between x and y if and only if $x\in y$ or $y\in x$.

Theorem 18

Let $(H(\omega _1),\in \cup \ni )$ be the graph defined above. It is prime $(\omega _1,\mathfrak {c})$-saturated.

Proof

First, let us check, that it is $(\omega _1,\mathfrak {c})$-saturated. Let $A,B \subseteq H(\omega _1)$, $|A|\le \omega$, $|B|<\mathfrak {c}$. We aim to find a countable set $A'\supseteq A$, disjoint with B, and not belonging to $\bigcup {B}$. But $|\bigcup {B}|<\mathfrak {c}$, and there are $\mathfrak {c}$-many different hereditarily countable sets having A as a subset, and disjoint with B.

We now prove, that it is prime. It is sufficient to prove that for any strictly saturated graph G, we have an embedding $\phi :(H(\omega _1),\in ) \hookrightarrow G$. We define $\phi$ by the induction, with respect to the $\in$ relation. Suppose that $x \in H(\omega _1)$, and for every $y\in x$, $\phi (y)$ is defined. Set $\phi (x)=g$, where $N_z(g)=\phi [x]$. It is routine to check, that this is an embedding. $\square$

Corollary 4

$|H(\omega _1)|=\mathfrak {c}.$

Note, that it is not even a priori clear, that $H(\omega _1)$ is a set, not a proper class.

4 Boolean algebras with the strong countable separation property

In this section letters A, B, C always denote Boolean algebras.

Definition 7

An infinite Boolean algebra A has the strong countable separation property if the following assertion holds:

Suppose $F,G \subseteq A$ are countable sets with the property that for all nonempty, finite subsets $f \subseteq F$, $g\subseteq G$, $\bigvee {f}<\bigwedge {g}$. Then there exist $a\in A$, such that for all $x\in F$, and $y\in G$, $x<a<y$.

This is equivalent to being injective with respect to countable subalgebras.

Proposition 10

(5.29 in [8]) Assume that A has the strong countable separation property, $f:B\rightarrow A$ is an embedding, and $B\subseteq C$ is countable. Then f extends to an embedding $\tilde{f}:C\rightarrow A$.

For exposition of general theory of Boolean algebras with the strong countable separation property, we refer the reader to [8]. In case of Boolean algebras, we start from the classical characterization of $\mathcal {P}(\omega ) /\text {Fin}$ by Parovičenko.

Theorem 19

(Parovičenko) Under CH, $\mathcal {P}(\omega ) /\text {Fin}$ is the unique Boolean algebra of cardinality $\mathfrak {c}$, with the strong countable separation property.

In fact, this statement is equivalent to CH, as was proved by van Douwen and van Mill [15]. In [1] Avilés and Brech consider Boolean algebras which realize extensions by so-called posex’ (push-out separable extensions).

Definition 8

(Avilés–Brech, [1]) An embedding $f:A\rightarrow B$ is a posex, if the following holds:

• for all $b\in B\setminus f[A]$, the ideal $\{a\in A|\, f(a) <b\}$ is countably generated,

• B is countably generated over f[A] (i.e. there exist a countable set $S \subseteq B$, such that B is generated by $f[A]\cup S$).

Definition 9

(Avilés–Brech, [1]) A Boolean algebra A is tightly $\sigma$-filtered if there exists a sequence of Boolean algebras $(A_\eta )_{\eta \le \zeta }$, such that

• $A_0=\{0,1\}$,

• $A_\eta \subseteq A_{\eta +1}$ is a posex, for $\eta <\zeta$,

• $A_\eta =\bigcup _{\alpha <\eta }{A_\alpha }$, for limit $\eta <\zeta$,

• $A_\zeta =A$.

We call the sequence $(A_\eta )_{\eta <\zeta }$ a filtration of A.

Theorem 20

(Avilés–Brech, [1]) Assume, that $\mathfrak {c}$ is regular. Then there exists a unique Boolean algebra A, such that:

• $|A|=\mathfrak {c}$,

• A is tightly $\sigma$-filtered,

• For any embedding $f:B\rightarrow A$, where $|B|<\mathfrak {c}$, and any posex $C\supseteq B$, there exists an embedding $\tilde{f}:C\rightarrow A$, extending f.

We will call an algebra with these properties an Avilés–Brech algebra. Since any extension between countable Boolean algebras is posex, the algebra defined above clearly has the strong countable separation property. Moreover, it is prime with respect to this property.

Theorem 21

The Avilés–Brech algebra defined above, embeds into any Boolean algebra having the strong countable separation property.

Proof

Let B be any Boolean algebra with the strong countable separation property. Let A be an Avilés–Brech algebra, with filtration $(A_\alpha )_{\alpha <\mathfrak {c}}$. We construct an increasing sequence of embeddings $f_\beta : A_\beta \rightarrow B$, by induction on $\alpha$, using the Corollary 5.8 from [8], quoted below.

Lemma 9

Let A be any Boolean algebra and A(x) its extension generated by $A\cup \{x\}$. Let $f:A\rightarrow B$ be an embedding, and let $y \in B$. Then f extends to an embedding $\tilde{f}:A(x)\rightarrow B$, with $\tilde{f}(x)=y$ if and only if for all $a,a' \in A$, $a \le x \le a'$ implies $f(a)\le y \le f(a')$.

Suppose that $f_\beta :A_\beta \rightarrow B$ is defined for all $\beta<\alpha <\mathfrak {c}$. If $\alpha$ is limit, we set $f_\alpha =\bigcup \{f_\beta :\, \beta <\alpha \}$, so suppose that $\alpha = \beta +1$. Let $S=\{s_0, s_1, \ldots \}$ be a set such that $A_\alpha = \langle A_\beta \cup S \rangle$. We successively extend $f_\beta$ to each of the generators using the Lemma. More precisely, let $f_\beta ^0=f_\beta$, and assume that $f_\beta ^i : \langle A_\beta \cup \{s_0,\ldots ,s_{i-1}\} \rangle \rightarrow B$ is defined. Let $\{a_n|\, n<\omega \}$ be a set of generators of the ideal $\{a \in \langle A_\beta \cup \{s_0,\ldots ,s_{i-1}\}\rangle |\, f_\beta ^i(a) < s_i \}$, and let $\{a'_n |\, n<\omega \}$ be a set of generators of the ideal $\{a \in \langle A_\beta \cup \{s_0,\ldots ,s_{i-1}\}\rangle |\, f_\beta ^i(a) < -s_i \}$. Because of the strong countable separation property, there exists $b \in B$ such that for all $n<\omega$

$$\begin{aligned} f_\beta ^i(a_n)<b<-f_\beta ^i(a'_n). \end{aligned}$$

We extend $f_\beta ^i$ to $f_\beta ^{i+1}$, so that $f_\beta ^{i+1}(s_i)=b$. Finally, let $f_\alpha = \bigcup _{i<\omega }{f_\beta ^i}$. $\square$

A natural question arises.

Problem 1

Let A be a Boolean algebra with the strong countable separation property, and such that whenever B has the strong countable separation property, then A embeds into B. Does it follow that A is the Avilés–Brech algebra?