Intuitively, $(\mathbb{R};<)$ defines the set of real numbers without specifying where either $0$ or $1$ is, i.e. we can neither say the origin nor the scale. Therefore, for a given set there is no way to say, for example, whether $0$ belongs to it or not. Formally, we can use the corollary of the homomorphism theorem on page 98 to argue against some possibilities. Namely, suppose that $R$ is a definable unary relation on $\mathbb{R}$ (a subset of $\mathbb{R}$ ). If $a,b\in\mathbb{R}$ then consider the automorphism $h(x)=x+b-a$ such that $h(a)=b$ . The corollary implies that if $R$ is definable, then both $a$ and $h(a)=b$ either belong to $R$ or not. Since we have chosen $a$ and $b$ arbitrary, this implies that either all points of $\mathbb{R}$ are in $R$ , or $R$ is empty. Hence, the subsets of $\mathbb{R}$ that are definable in $(\mathbb{R};<)$ are $R=\emptyset$ and $R=\mathbb{R}$ (both subsets are indeed definable via, for example, $\forall v_{2}v_{2}<v_{2}$ and $\forall v_{2}v_{2}=v_{2}$ , correspondingly, and we have shown that there cannot be any other definable subsets). Note, that alternatively we could have said that there is one subset of $\mathbb{R}$ , namely, $A_{1}=\mathbb{R}$ , such that a) $A_{1}$ is definable, and b) if any point of $A_{1}$ is in $R$ , then all points of $A_{1}$ are in $R$ , and to construct all possible definable subsets of $\mathbb{R}$ we either take this subset or we don’t, which gives as exactly $2^{1}=2$ possibilities.

Intuitively, with $\mathbb{R}\times\mathbb{R}$ the situation is slightly different. We still cannot tell where $0$ or any other point is, however, given a point $(x,y)$ we at least can say whether $x<y$ , or $x=y$ , or $x>y$ . Therefore, we at least can distinguish subsets having points $(x,y)$ such that, say, $x<y$ , and those that do not have such points. However, if we know that $<x,y>\in R$ and $x<y$ , we still cannot say whether this point is, for example, $<0,1>$ or any other point $<a,b>$ satisfying $a<b$ . Formally, we again use the corollary to argue against most of subsets. Consider two points $<a,b>$ and $<c,d>$ such that $a\neq b$ . Then $h(x)=\frac{d-c}{b-a}(x-a)+c$ is an automorphism as long as $d-c$ and $b-a$ have the same sign. Note that $h(a)=c$ and $h(b)=d$ . Therefore, according to the corollary, for any definable $R\subseteq\mathbb{R}^{2}$ , and $<a,b>$ and $<c,d>$ such that either $a<b$ and $c<d$ or $a>b$ and $c>d$ , either $<a,b>,<c,d>\in R$ or they both are not in $R$ . Moreover, for two points $<a,a>$ and $<b,b>$ we can consider the automorphism $h(x)=x+b-a$ such that $h(a)=b$ , and as before argue that either both such points are in $R$ or they both are not in $R$ . Now we have three subsets $A_{1}=\{<x,y>|x<y\}$ , $A_{2}=\{<x,y>|x=y\}$ and $A_{3}=\{<x,y>|x>y\}$ such that a) all these subsets are definable, and b) if any point of $A_{i}$ is in $R$ , so are all other points of $A_{i}$ . This gives us exactly $2^{3}$ definable subsets of $\mathbb{R}^{2}$ , namely, the empty set, $A_{i}$ , $i\in\{1,2,3\}$ , their pairwise unions, and the whole plane $\mathbb{R}^{2}$ .

We can proceed this way further, even though the remark warns us against doing so. We can argue in a similar way, that if $R$ is a definable $n$ -ary relation on $\mathbb{R}$ , then for two points $<x_{1},\ldots,x_{n}>$ and $<y_{1},\ldots,y_{n}>$ such that for every $1\le k<m\le n$ , $x_{m}-x_{k}$ and $y_{m}-y_{k}$ are either both zero or have the same sign, either both points are in $R$ or they both are not in $R$ . Therefore, we can distinguish several subsets $A_{i}$ of $\mathbb{R}^{n}$ such that a) $A_{i}$ is defined through specific relations among coordinates (such as $x_{k}=x_{m}$ or $x_{k}<x_{m}$ or $x_{k}>x_{m}$ ), b) if any point of $A_{i}$ is in a definable relation $R$ , so are all other points. Then, any definable $n$ -ary relation on $\mathbb{R}$ is either empty or a union of some of these sets. And if there are $S(n)$ such sets, then we have $2^{S(n)}$ definable $n$ -ary relations. The only question now is how we determine $S(n)$ . Let $S(n,k)$ be the number of possible combinations of relations among coordinates such that the coordinates form a set of $k$ distinct values. For example, $S(2,1)=1$ as the only possibility here is when $x_{1}=x_{2}$ , $S(2,2)=2$ as we have two cases $x_{1}<x_{2}$ and $x_{1}>x_{2}$ , and $S(3,2)=6$ as we can have $x_{1}=x_{2}<x_{3}$ , $x_{1}=x_{2}>x_{3}$ and 4 more possibilities, 2 when $x_{1}=x_{3}$ , and 2 when $x_{2}=x_{3}$ . Note that, $S(n,k)=k(S(n-1,k-1)+S(n-1,k))$ where $S(n,0)=S(n,n+1)=0$ for all $n$ (to produce $k$ different values with $n$ coordinates we can either take any combination of $n-1$ coordinates with $k-1$ distinct values and add the $n$ th coordinate above all, below all, or between any two coordinates, or take any combination of $n-1$ coordinates with $k$ distinct values and add the $n$ th coordinate by requiring it to be equal to one of those values). Now, the total number of possible combinations of relations among $n$ coordinates is $S(n)=\sum_{k=1}^{n}S(n,k)$ . This recursive formula produces the following results.

$n$	$k=1$	$2$	$3$	$4$	$5$	$S(n)$
$1$	$1$					$1$
$2$	$1$	$2$				$3$
$3$	$1$	$6$	$6$			$13$
$4$	$1$	$14$	$36$	$24$		$75$
$5$	$1$	$30$	$150$	$240$	$120$	$541$

Therefore, we have $2^{1}=2$ definable unary relations, $2^{3}=8$ definable binary relations, $2^{13}$ definable ternary relations, $2^{75}$ definable quaternary relations, $2^{541}$ definable quinary relations etc.