# example of a universal structure

Let $L$ be the first order language with the binary relation  $\leq$. Consider the following sentences  :

• $\forall x,y((x\leq y\lor y\leq x)\land((x\leq y\land y\leq x)\leftrightarrow x% =y))$

• $\forall x,y,z(x\leq y\land y\leq z\rightarrow x\leq z)$

Any $L$-structure  satisfying these is called a linear order. We define the relation   $<$ so that $x iff $x\leq y\land x\neq y$. Now consider these sentences:

1. 1.

$\forall x,y((x

2. 2.

$\forall x\exists y,z(y

A linear order that satisfies 1. is called dense (http://planetmath.org/DenseTotalOrder). We say that a linear order that satisfies 2. is without endpoints. Let $T$ be the theory of dense linear orders without endpoints. This is a complete theory.

We can see that $(\mathbb{Q},\leq)$ is a model of $T$. It is actually a rather special model.

###### Theorem 1

Let $(S,\leq)$ be any finite linear order. Then $S$ embeds in $(\mathbb{Q},\leq)$.

Suppose that the statement holds for all linear orders with cardinality less than or equal to $n$. Let $|S|=n+1$, then pick some $a\in S$, let $S^{\prime}$ be the structure induced by $S$ on $S\setminus{a}$. Then there is some embedding  $e$ of $S^{\prime}$ into $\mathbb{Q}$.

• Now suppose $a$ is less than every member of $S^{\prime}$, then as $\mathbb{Q}$ is without endpoints, there is some element $b$ less than every element in the image of $e$. Thus we can extend $e$ to map $a$ to $b$ which is an embedding of $S$ into $\mathbb{Q}$.

• We work similarly if $a$ is greater than every element in $S^{\prime}$.

• If neither of the above hold then we can pick some maximum $c_{1}\in S^{\prime}$ so that $c_{1}. Similarly we can pick some minimum $c_{2}\in S^{\prime}$ so that $c_{2}. Now there is some $b\in\mathbb{Q}$ with $e(c_{1}). Then extending $e$ by mapping $a$ to $b$ is the required embedding. $\square$

###### Theorem 2

$(\mathbb{Q},\leq)$

Proof: The following type of proof is known as a back and forth argument. Let $S_{1}$ and $S_{2}$ be two finite substructures of $(\mathbb{Q},\leq)$. Let $e:S_{1}\to S_{2}$ be an isomorphism      . It is easier to think of two disjoint copies $B$ and $C$ of $\mathbb{Q}$ with $S_{1}$ a substructure of $B$ and $S_{2}$ a substructure of $C$.

Let $b_{1},b_{2},\ldots$ be an enumeration of $B\setminus S_{1}$. Let $c_{1},c_{2},\ldots,c_{n}$ be an enumeration of $C\setminus S_{2}$. We iterate the following two step process:

The $i$th forth step If $b_{i}$ is already in the domain of $e$ then do nothing. If $b_{i}$ is not in the domain of $e$. Then as in proposition  1, either $b_{i}$ is less than every element in the domain of $e$ or greater than or it has an immediate successor    and predecessor in the range of $e$. Either way there is an element $c$ in $C\setminus\operatorname{range}(e)$ relative to the range of $e$. Thus we can extend the isomorphism to include $b_{i}$.

The $i$th back step If $c_{i}$ is already in the range of $e$ then do nothing. If $c_{i}$ is not in the domain of $e$. Then exactly as above we can find some $b\in B\setminus\operatorname{dom}(e)$ and extend $e$ so that $e(b)=c_{i}$.

After $\omega$ stages, we have an isomorphism whose range includes every $b_{i}$ and whose domain includes every $c_{i}$. Thus we have an isomorphism from $B$ to $C$ extending $e$. $\square$

A similar back and forth argument shows that any countable dense linear order without endpoints is isomorphic to $(\mathbb{Q},\leq)$ so $T$ is $\aleph_{0}$-categorical.

 Title example of a universal structure Canonical name ExampleOfAUniversalStructure Date of creation 2013-03-22 13:23:16 Last modified on 2013-03-22 13:23:16 Owner uzeromay (4983) Last modified by uzeromay (4983) Numerical id 15 Author uzeromay (4983) Entry type Example Classification msc 03C50 Classification msc 03C52 Related topic Homogeneous4 Related topic KappaCategorical Related topic DifferentialEquation Related topic ExampleOfDefinableType Related topic RandomGraph Defines back and forth