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Let $D$ be a dense linear order with $|D|=\kappa$. Is there a way to define an equivalence relation such that $\forall a,b \in D$ there are $\{c_k : k\in \kappa\}$ s.t. $a<c_k<b$ and $[c_i]\not=[c_j] \text{ if } i\not=j$ ?

I want to use this to show this: let $T_0$ be the theory which says that $r$ is an equivalence relation and $<$ is a linear order, there exists a rich model of $T_0$ (i.e. a model $N$ s.t. if $\phi\colon M \rightarrow N$ is a partial isomorphism with $|\phi|<|N|$, $M$ also a model for $T_0$, $\forall m\in M$ I can extend the isomorphism to $m$).

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  • $\begingroup$ Are you sure this is true? Consider $D = \kappa \times \mathbb{Q}$ ordered lexicographically, with $\kappa > \aleph_0$. Then for each $i \in \kappa$ and each $a,b \in \mathbb{Q}$ there are only countably many elements of $D$ strictly between $(i,a)$ and $(i,b)$. $\endgroup$ – Clive Newstead Jun 24 at 16:54
  • $\begingroup$ What definition of "partial isomorphism" do you use? $\endgroup$ – Mark Kamsma Jun 24 at 17:42
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    $\begingroup$ Never mind the equivalence relation; from "$D$ is a dense linear order with $|D|=\kappa$" it does not follow that, for all $a,b\in D$, there are $\kappa$ elements between $a$ and $b$. $\endgroup$ – bof Jun 24 at 18:07
  • $\begingroup$ @MarkKamsma a partial map st. if $x,y \in dom(\phi)$ then $x r y \rightarrow \phi(x) r \phi(y)$ and $x<y \rightarrow \phi(x)<\phi(y)$ $\endgroup$ – MarcoM Jun 25 at 9:06
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As pointed out in the comments, you certainly can't do this for any dense linear order of cardinality $\kappa$. However, if $D$ is a dense linear order of cardinality $\kappa$ in which every interval has cardinality $\kappa$ (i.e. for all $a,b\in D$, $(a,b) = \{c\in D\mid a < c < b\}$ has cardinality $\kappa$), you can just take the equivalence relation to be the equality relation, and you're done.

And it's easy to build such a dense linear order by hand, for any infinite $\kappa$: Start with any order $M_0$ of size $\kappa$. Enumerate the pairs $(a,b)$ with $a<b$; there are $\kappa$-many of these. For each such pair, add $\kappa$-many new elements between $a$ and $b$. After having done this for every pair, you have an order $M_1$ of cardinality $\kappa$ with the property that every interval with endpoints in $M_0\subseteq M_1$ has cardinality $\kappa$. Now repeat, getting orders $M_2$, $M_3$, etc. The union $\bigcup_{n\in \omega} M_n$ has the desired property.

Great! Except that I see no connection at all between the question in your first paragraph and the goal in your second paragraph. The construction above will not help you get a "rich" model. (By the way, where did you find this terminology? I haven't seen it before.)

Let's forget about the equivalence relation for a moment and just think about the linear order. A rich model of the theory of linear order must be a dense linear order without endpoints (a model of the theory $\text{DLO}$), since density and the lack of endpoints can be characterized by the ability to extend any partial isomorphism from a finite linear order by one more point (between two elements of the domain, greater than all elements of the domain, or less than all elements of the domain). And indeed, the unique countable dense linear order without endpoints is a rich model of the theory of linear orders.

But when we move to uncountable linear orders of size $\kappa>\aleph_0$, your original question is about having large intervals (between any two points $a$ and $b$, there are $\kappa$-many points), while richness is about filling cuts: for any sets $A$ and $B$ with $|A\cup B|<\kappa$ and $a<b$ for all $a\in A$ and $b\in B$, there is some point $c$ with $a<c<b$ for all $a\in A$ and $b\in B$. Here $A$ and $B$ are the set of points in $M$ which are less than and greater than the new point $m$, respectively. It's much harder to fill all cuts than to have large intervals!

Actually, we can rephrase richness in more model-theoretic terminology: since $\text{DLO}$ has quantifier elimination and every linear order embeds in a model of $\text{DLO}$, a model $M\models \text{DLO}$ is rich if and only if it is saturated. The unique countable model $(\mathbb{Q},\leq)$ is saturated, but you can't prove in $\text{ZFC}$ that $\text{DLO}$ has any uncountable saturated models. You can build a saturated model if you assume $2^\kappa = \kappa^+$ for some cardinal (an instance of GCH) or the existence of inaccessible cardinals.

The situation is exactly the same for your theory $T_0$ of linear orders equipped with an equivalence relation. The finite models of $T_0$ form a Fraïssé class, and its Fraïssé limit $M$ is the unique countable rich model of $T_0$. More generally, a model of $T_0$ is rich if and only if it is a saturated model of the theory $T = \text{Th}(M)$, and we cannot prove in $\text{ZFC}$ that such models exist, other than the countable one.

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  • $\begingroup$ I forgot to say that also $|N|=\kappa$, so I want to extend partial isomorphism $\phi$ with $|\phi| < |N|$. $\endgroup$ – MarcoM Jun 25 at 9:03
  • $\begingroup$ If the statement in the question is wrong, you should fix the question! I'll edit it for you this time. $\endgroup$ – Alex Kruckman Jun 25 at 13:12
  • $\begingroup$ @MarcoM I've added more to my answer. $\endgroup$ – Alex Kruckman Jun 25 at 13:28

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