数理解析研究所講究録 1450 巻 2005 年 75-82

75

On Predimensions of Finite Structures
神戸大学・工桔梗 宏孝 (Hirotaka Kikyo)

Faculty of Engineering, Kobe University
kikyo@kobe-u. ac. jp

September 7, 2005
Abstract
structures with a predimension function, inoffinitesomegiveWe
families
which we can construct various structures with desired predimensions.

1

Introduction

N. Peatfield and B. Zilber investigated a generic structure of certain class of finite
modification of
structures considered by Hrushovski and they showed that withanalytic Zariski structure on it. One
the structure, they can put a topology called
following.
is
theof their main tools
is elemetarily equivalent to athey considered, a structure
For the classsatisfies Axioms 1 attd 2:
generic structure of if and only if
belongs to K.
Axiom 1. Any finite substructure of$A\leq
B$ , if:
is an -embedding
such that
Axiom 2. For any ,.
:-embedding
toanthen it can be extended
Axiom 2 represents a strong form of the amalgamation property. For any classof finite structures, these axioms may have some meaning, This will be investigatedin a joint work with K. Ikeda and A. Tsuboi [2]. This paper will be a part of thiswork.
$\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}$$\mathrm{a}\lambda 1$$\mathrm{A}f$$\mathrm{K}$$\lambda I$$\mathrm{K}$

$A$

$A$

$\Lambda I$$f$$B\in \mathrm{K}$$f’$$\mathcal{L}$$Aarrow \mathrm{A}I$$\mathcal{L}$$Barrow\Lambda I$$\mathrm{K}$2

Preliminaries

Throughout this paper, is a finite relational language. If $M$ is an
is a relation with arguments tl en$\mathcal{L}$$\mathcal{L}$$R$$R(\Lambda I)$$n$,
Definition 2.1 Suppose
positive real numbers at most 1.
For a finite structure , let$\mathcal{L}=\{R_{1}$$\mathcal{L}$$R_{2}$,...,$=\{x\in\Lambda I^{n} :$R_{l}\}$, and$\alpha=$\Lambda I \models R(x)\}$structure alld.

$(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{l})$$A$$\delta(A)=|A|-|R_{1}(A)|-|R_{2}(A)|-\cdots-|R_{l}(A)|$

.

where

$\alpha_{i}\mathrm{a}1^{\backslash }\mathrm{e}$7$\mathrm{G}$is called a predimension function.
Let be an -structure andthenwheneverfor anyLet{ :$\delta$$B$$A\subseteq B$$\mathcal{L}$$\delta(A)$$A\underline{\subseteq}X\subseteq B$$\mathrm{K}_{\alpha}=$$A$. We write $A\leq B$ if for any subset$\leq\delta(X)$$\mathrm{R}(\mathrm{x})\geq 0$

$X$

of

$B$

,

.

$X\subseteq A$}.

be -structures, and let
andDefinition 2.2 (Asymmetric amalgam) Let$B\cup C$ and for each relation$A=B\cap C$ as a set.
withuniverse
is a structure
in , $R(B\oplus_{A}C)$ $=R(B)\cup R(C)-(R(C)\cap A^{n})$ where is the number of argumentsof .
is a substructure of
Note that $\delta(B\oplus_{A}C)=\delta(B)+\delta(C)-\delta(C|_{A})$ wherewith the universe .
for $B\oplus\{a\}C$ ,Furthermore, if $B\cap C=\{a\}$ (a singleton), then we write$B\oplusC$.
for, then we write
and if$C$$B$$\mathcal{L}$$B\oplus_{A}C$

$R$

$n$

$\mathcal{L}$

$R$

$C$

$C|_{A}$$A$$B\oplus_{a}C$$B\cap C=\emptyset$$B\oplus {}_{\emptyset}C$

elements in , and suppose that for
Definition 2.3 Let be an -structure,such that $A\models R(x)$ and, there is no tuple of elements in
any relation
contains alud .
$A/(a=b)$ is an -structure obtained from by identifying and . More precisely,
it is defined as follows: The universe, also denoted $A/(a=b)$ , is A $-\{b\}$ . Let: $Aarrow A/(a=b)$ be a projection map defined by $f(x)=x$ for $x\neq b$ , and $f(b)=a$ .
, $c_{n}\in A/(a=b)$ if an, $A/(a=b)\models R(c_{1}, \ldots, c_{n})$ for ,For any relation,, .
such that $f(c_{i}’)=c_{i}$ forfor some , . . . ,only if$A$$R\in \mathcal{L}$

$a$

$A$

$a_{2}b$$\mathcal{L}$

$A$

$x$

$x$

$b$

$A$

$\mathcal{L}$

$b$

$a$

$f$

$\mathrm{z}\mathrm{d}$$R\in \mathcal{L}$$c_{1}$$A\models R(d_{1^{*}}\ldots, c_{n}’)$3$\ldots$$\mathrm{i}=1$$c_{n}’\in A$$c_{1}’$$\ldots$$n$

Arithmetic of Structures

for some tuple of positive real numbers at most 1.
In this section, we work inof desired
to get structures in
We show that there are a plenty of structures inranks.
$\mathrm{K}_{\alpha}$$\alpha$$\mathrm{K}_{\alpha}$$\mathrm{K}_{\alpha}$

Definition 3.1 Let be a real number such that $0\leq s\leq 2$ . A triple $(E, a, b)$ is an-component if is ati -structure, , $b\in E$ , $a\neq b$ , $\delta(\{a\})=\delta(\{b\})=1$ , and thefollowing hold :For any noll-elllpty substructure $X$ of ,
$s$

$E$

$s$

$\mathcal{L}$

$a$

$E$

(1) $s\leq\delta(X)$ if , $b\in X$ ,$a$(2) $1\leq\delta(X)$ if$a\not\in X$

or$b\not\in X$, and(3) $\delta(E)=s$ .

We also say thatcomponent.

$E$

is an -component with joints
$s$

$a$

and , or just $E$ is an s-

Lemma 3.2 Let s. t. u be real numbers such that $0\leq s$ , t, u$b$$\leq 1$.

77(1) Suppose $(A, a, b)$ is an $(1+s)$ -component, $(B, b, c)a(1+t^{\backslash }/$ -component, $s+t\leq 1$ ,and $C=A\oplus_{b}$ B. Then $(C,a,c)$ is an $(1+s+t)$ -component.

(2) Suppose (A., , ) is an $(1+s)$ -component, $(B, b, c)a(1+t)$ -component, $(C, c, d)$$a(1+u)$ -component, $s+t\leq 1$ , $D=A\oplus_{b}B\oplus_{c}C$ , and $\overline{D}=D/(a=d)$ . Thenis an $(s+t+u)$ -component.
$a$

$b$

$(\overline{D}, c, a)$, rk-iare non-negative real numbers such thatand an -component exists for each . Then there
an $(r_{1}+r_{2}+\cdots+r_{k-1}+r_{k})$ -component.

(3) Suppose $k\geq 3$ ,$r_{1}$,$r_{2}$,$r_{k}$$\ldots$$r_{1}+r_{2}+\cdots+r_{k-1}\leq 1_{j}$$\mathrm{i}$$\iota s$$r_{i}$.
Proof. (1) Suppose $X\subseteq C=A\oplus_{b}B$ alld
We check the condition (1) of the definition of an $(1+s+t)$ -component first.
Assume that , $c\in X$ . Consider the case $b\in X$ . Then , $b\in X\cap A$ , , $c\in X\cap B$ ,mid $X=(X\cap A)\oplus_{b}(X\cap B)$ . Hence,$X\neq\emptyset$

$b$

$a$

$a$

$5(\mathrm{X})=\delta(X\cap A)+\delta(X\cap B)\geq(1+s)+(1+t)-1=1+(s+t)$ .

Now, consider the case
Hence,$b\not\in X$. Then $b\not\in X\cap A_{7}b\not\in X\cap B$ , and $X=(X\cap A)\oplus(X\cap B)$ .

$5(\mathrm{X})=\delta(X\cap A)+\delta(X\cap B)\geq 1+1\geq 1+(s+t)$ .

Therefore, , $c\in X$ implies $\delta(X)\geq 1+s$ $+t$ .
Now, we check the condition (2) of the definition of aza $(1+s+t)$ -component.
Suppose $X$ is a non-empty subset of . Then $\delta(X)\geq\delta(X\cap A)+\delta(X\cap B)-1\geq$$1+1-1=1$ .
(2) Since there is no relation on a tuple containing and , going from to ,, $\delta(X/(a=d))=$the number of relations does not change. Therefore, for any
$d\in
X$
.
, and $\delta(X/(a=d))=\delta(X)$ if
if ,
. Let
be such that $X=Y/(a=d)$ . We can assume
Suppose ,that , $d\in Y$ .
Consider the case $b\in Y$ . Since , , , $d\in Y$ , we have $\delta(Y\cap A)$ $\geq 1+s$ ,$B)\geq 1+t$ , $\delta(Y\cap C)\geq 1+u$ , aaid $Y=(Y\cap A)\oplus_{b}(Y\cap B)\oplus_{c}(Y\cap C)$ . Hence,$1\geq s+t+u$ .
$6(Y)\geq 1+(s+t+u)$ . Therefore, $\delta(X)=\delta(Y/(a=d))=\delta(Y)$, $\delta(Y\cap B)\geq 1$ , $\delta(Y\cap C)\geq 1+u$ ,. Then,Now, consider the case$\delta(Y)\geq
1+(1+u)\geq 1+(s+t+u)$ by$Y=(Y\capA)\oplus(Y\cap
B)\oplus_{c}(Y\cap
C)$. We haveandthe assumption that $s+t\leq 1$ . Therefore, $\delta(X)=\delta(Y/(a=d))=\delta(Y)-1\geq s+t+u$.
We have checked the condition (1) of the definition of $(s+t+u)$ -component.
We turn to check the condition (2) of the definition of $(s+t+u)$ -component.
. Let $X_{1}=X\cap(A\cup B)$
and ,alld $X\neq$ GO. Then
Suppose$A\cupB=$$s+t\leq1$
is a $(1+s +t)$ -component,
and $X_{2}=X\cap C$ . Since
since $d\not\in X$ . Th erefore,.
byby (1), and thus
$a$

$C$

$a$

$D$

$d$

$\overline{D}$$X\subseteq D$$5(\mathrm{X})--$$1$$a_{\dot{l}}d\not\in X$

$a$

$c$

$a\in X\subseteq\overline{D}$$Y\subseteq D$

$a$

$a$

$b$

$\delta(Y\cap$

$c.$

-

$\delta(Y\cap A)$$b\not\in Y$$\geq 1$

$X\subseteq D$$a\not\in X\subseteq\overline{D}$$a$$d\not\in X$$A\oplus_{b}B$$\delta(X_{1})\geq 1$$\delta(X_{1})+\delta(X_{2})-1\geq 1$

SupposeSince$\square$$c\not\in Y$$a\not\in X$$\delta(X)\geq$$\delta(X_{2})\geq 1$.

be such that $X=Y/(a=d)$ and , $d\in Y$ .
$=\delta(Y)-1\geq 1$ .
, $\delta(Y)=\delta(Y\cap(A\cup B))+\delta(Y\cap C)\geq 2$ . Therefore,$c\not\in X$

and $a\in X$ . Let$Y\subseteq D$$a$$\delta(X)$78

Lemma 3.3 Suppose $0\leq s<1/2$ and $a(1+s)$ -component exists. If is rational
there exists suchthen 1-component exists. If is irrational, then for any$>\epsilon’>0$ and $a(1+\epsilon’)$ -component exists.
With Lemma 3. 2, we can choose
that
$s$

$\epsilon>0$

$s$

$a$

$\epsilon’$$a$$\epsilon$$1$-component or $a(1+\epsilon’)$ -component$arb\iota t\uparrow^{\mathrm{Y}}anly$large.

Proof. Suppose $0\leq s<1/2$ and a $(1+s)$ -component exists. Consider the following: Let $s_{0}=s$ . If $s_{i}=0$ then let $s_{i+1}=0$ . If
non-negative decreasing sequence
alld let $s_{i+1}=$
such that$s_{i}>0$ then choose the least positive integer$\{s_{i}\}_{i<\omega}$$m_{i}s_{f}\geq 1$$m_{i}$.
.
By Lemma 3.2 (2), there is a $(1+s_{i})$ -component for each$s_{l}=0$eventually.
$s_{i}>s_{i+1}\geq
0$
thenrational$s_{i}>0$is$s_{0}=s$.
IfthenNote that if
Hence, 1-component exists.
converges to 0. This should be well-known, but we
If $s_{0}=s$ is irrational,isis a decreasing sequence,
give a proof for convenience. Since
converges to 0 because
is unbounded, thenan increasing sequence. If
integer $m$ eventually.
positive$s_{i}<1/(m_{i}-1)$ . If
is bounded, $m_{i}=m$ for someSo, we can assume that $s_{i+1}=ms_{\dot{f}}-1$ for all . Since converges to some $\beta>0$ ,$\beta=m\beta-1$ . Therefore, $\beta=1/(\mathrm{m}-1)$ . But $s_{i}>\beta=1/(m-1)$ contradicts$m_{i}s_{i}-1$$\mathrm{i}<\omega$$s_{i}$$\{m_{i}\}_{i<v}$$\{s_{i}\}_{i<\omega}$($\{m_{i}\}_{i<\omega}$$s_{i}$$\{m_{i}\}_{i<\omega}$$\mathrm{i}$$s_{i}$$(m-1)s_{i}=(m_{i}-1)s_{i}<1$ .

$\square$a set of vertices and a set
Lemma 3.4 Let $(V, E)$ be a finite graph where$0<\beta\leq
1$ .
Then a graph which isof edges, and let $\delta(V, E)=|V|-\beta|E|$ utith$0\leq s<1/2$ with respect
that
such
$1+s)$
some
real
number-component
exists{forto this 6.
$E$$V\iota s$

$a$

$s$

Proof. We remark first that ally two different points in each $(1+s)$ -component below
will be its joints.
is a l-component.
. A graph with 4 points and 6 edges
SupposeSuppose $1/2<\alpha_{i}\leq 1$ . Let $s=1-\alpha_{i}$ . Then a graph with 2 points and 1 edge isa $(1+s)$ -component. These 2 points will be joints.
$(K_{4})$$\alpha_{i}=1/2$

Suppose $0<\alpha_{i}<1/2$ . Choose least natural number

Note that$k$$\leq 4$

implies A$-\alpha$$(\begin{array}{l}k^{n}2\end{array})>1$. Therefore,

$k$

$k$

$\geq 5$such that. Consider$k$$-\alpha$$K_{k}.$$(\begin{array}{l}k2\end{array})<1$.

, a complete, $\delta(X)\geq 1$ .
graph with points. Then for any non-empty proper subset $X$ of, remove 1 edge at a time. Then tlle -value of the graph incleases
Starting frombyat a time. Repeat this process until the -value of the graph exceeds 1. Let
be the graph with points obtained by this process. We eventually get $E$ because .
points with no edge has the -value $k\geq 5$ . If $5(\mathrm{E})=1$ , then $E$ is a 1-component. If
not, $1<\delta(E)<1+\alpha_{i}<1+1/2$ since if we put one more edge to $E$ , the -value willbe less than 1.
$K_{k}.$$k$$\delta$$K_{k}$$E$$\delta$$\alpha_{\iota}$

$k$

$k$

$\delta$$\delta$$\square$If we have all -component as a sufficiently large graph, then we can make an-colnponent as a structure with an n-ary relation for any .
$s$

$s$

$n$

79

be a real number such that $0<\beta\leq 1$ . Suppose $(A, E)$ is
Lemma 3.5 Let
such that $|A|-\beta|E|=|A|-\beta|S|$
sufficiently large binary graph. Then there. Here,is the set of edges in $E$
and $|X|-\beta|E_{X}|\leq|X|-\beta|S_{X}|$ for any
connecting vertices in $X$ , and $S_{X}=X^{n}\cap S$ . In particular, if $(A, E)$ is an s-component.
then so$a$$\beta$$\iota s$$S\subseteq A^{n}$$X\subseteq A$$E_{X}$$\mathrm{z}s(A, S)$

are members of $f(\{a,$ &}be a one-tcyone map such that
Proof. Let :is the set of two point subsets of ). We can choose
for any distinct points in (such if $|A|>2n$ . Let $S=f(E)$ (Consider each edge as the set of two end points).
, if $f(e)\in X$ then $e\in E_{X}$ . Hence,
Then $|A|-\beta|E|=|A|-\beta|S|$ . Also, for any$|S_{X}|\leq|E_{X}|$ . Therefore, $|X|-\beta|E_{X}|\leq|X|-\beta|S_{X}|$ .
$f$$[A]^{2}arrow A^{n}$

$A$

$)$

$a_{\dot{t}}b$$A$$[A]^{2}$$f$$X\underline{\subseteq}A$$\square$

As a corollary, we get the following proposition.

Proposition 3.6 If a rational number is a member of then 1-component exists.
art
If irrational rvurnber is a member of then for any $>0$ , there are andand $a(1+\epsilon’)$ -compotlent and $a(1-\epsilon’)$ -component exist
such that$a$$\alpha$$\iota n$$\mathrm{K}_{\alpha}$$\epsilon’$$\epsilon’$$\epsilon$$cx_{l}$$\epsilon>\epsilon’>\epsilon’>0$in$\mathrm{K}_{\alpha}$.

positive real
Lemma 3.7 Suppose that $r_{1}+r_{2}+\cdots+r_{k}-n=r$ where thenumbers at most 1, -1 $\leq r\leq 1$ , and a $na$ rural number. Assume further that, . . .,. Then1-component exists and so does $(1+r_{i})$ -component for each$(1+r)$ -component exists.
)$r_{i}s$$a\uparrow’ e$

$n$

$k$

$\mathrm{i}=1$

Proof. We prove the lemma by induction on . The lemma holds for $n=0$ by Lemma
3.2 (1). Suppose $n\geq 1$ . Let be the maximum suffix such that $r_{1}+\cdots+r_{i}<1$ .
Since a 1-component exists, we can assume that $i\geq 2$ .
ifthen must be 1. Therefore, there is a $(1+r)$ -component by Lemma 3.2(3), and we are done.
. Then $r_{1}+\cdots+r_{i}+r_{i+1}\geq 1$ . Let $q=r_{1}+\cdots+r_{i}+r_{i+1}-1\geq 0$ . By
SupposeLemma 3.2 (3), there is a $(1+q)$ -component. We have $q+r_{i+2}+\cdots+r_{k}-(n-1)=r$ .
Therefore, -component exists by the induction hypothesis.
$n$$\mathrm{i}$$\mathrm{i}=k$$n$$\mathrm{i}<k$$\square$$r$

Proposition 3.8 Suppose that a that onal number is a member of .
$\alpha$(1)$A$

1-component exists in$\mathrm{K}_{\alpha}$.

(2) If a reduced fraction $k/m$ is a $membe,$ ? of.
.
in$\alpha$then $a(1+1/m)$ -component$e_{\mathrm{z}}.\tau ists$$\mathrm{K}_{\alpha}$then $a(1+$(3) If $a(1+1/m)$ -component und $a(1+1/m’)$ -component exist in
the least common. Here,-component e.qiisl.s in$m’$ .
$ric$$m$
rrvultof and$\mathrm{K}_{\alpha}$$1/1\mathrm{c}\ln(m, \uparrow\tau x’))$$\cdot$$\mathrm{K}_{\alpha}$$1\mathrm{c}\mathrm{m}(m, m’)?,S$80(4) If $a(1+1/m)$ -component exists in.

$\mathrm{K}_{\alpha}$then $a(1-1/m)$ -component exists in$\mathrm{K}_{\alpha}$

Proof. (1) is in Proposition 3.6.
(2) Suppose a reduced fraction $k/m=\alpha_{i}$ for some . Then a graph with 2 points$=|A|-\alpha_{i}|E|$ .
defined by
and 1 edge is a $(1+(m-k)/m)$ -component for
Since there is a graph which is a 1-component for , we can make a $(1+(m-k)/m)-$component arbitrarily large. By Lemma 3.5, $(1+(m-k)/m)- \mathrm{c}\mathrm{o}$ mponent exists in$i$$\delta_{\alpha_{i}}(A, E)$$\delta_{\alpha_{\dot{x}}}$$\delta_{\alpha_{i}}$$\mathrm{K}_{\alpha}$.

and $m$ are prime each other. Therefore, there are positive
Let $k’=m-k$ .
integers , such that $uf_{\hat{v}}’-vm=1$ . Hence, $u(k’/m)-v$ $=1/m$ . There is aby Lemma 3.7.
$(1+1/m)$ -component in(3) Let be the greatest common divisor of $m$ and $m’$ . Then there are positive
integers $u<m’$ and $v<m$ such that um–vm’ $=d$ . Tl en $u(1/m’)-v(1/m)=$$d/(mm’)$ and thus $u(1/m’)+(m-v)(1/m)-1=d/(mm’)$ . Therefore, there is aby Lemma 3.7.
$(1+d/(mm’))$ -component in(4) Also, by Lemma 3.7.
$k’\wedge$

$u$

$v$

$\mathrm{K}_{\alpha}$$d$$\mathrm{K}_{\alpha}$$\square$

A Property of4$\mathrm{K}_{\alpha}$with which we can show that theIn this section, we show a special property ofis axiomatized by Axioms 1 and 2
elementary theory of the generic structure of[2].
inin the introduction. This will be investigated$\mathrm{K}_{\alpha}$$\mathrm{K}_{a}$Lemma 4.1 Suppose $0\leq s$ , $t\leq 1$ , $C=B\oplus_{a}E$ , and (E., , ) is $a(1+t)$ -component$+s\leq$
and $a\in B-A$ , thenfor any $X$ such that
If 5
$X$
.
for any such that$a$$(A)+s\leq \mathrm{S}(\mathrm{X})$$\delta(A)$$A\subseteq X\underline{\subseteq}B\infty$$A\subseteq Xarrow\subseteq C$$\mathrm{S}(\mathrm{X})$Proof. Suppose$A\subseteq\wedge X\subseteq C$$X=(X\cap B)\oplus(X\cap E)$ .

If

$b$

$(X\cap E)\neq\emptyset$$\delta(A)+s$, thenIf. If$\delta(X\cap E)$$\geq 1$, then $A\subseteq(X\cap B)$ , $a\not\in(X\cap E)$ , and, a1ld thus $\delta(X)\geq\delta(A)+s$ .
, then$a\not\in X$$(X\cap E)=\emptyset$$A\subseteq X\sim\subseteq B$

alld thus $\delta(X)=\delta(X\cap B)+\delta(X\cap E)$ $\geq\delta(A)+1\geq$.

If $a\in X$ , then $X=(X\cap B)\oplus_{a}(X\cap E)$ ,$A\underline{\subseteq}X\cap B$, and$X\cap E\neq\emptyset$. Hence,$\mathrm{S}(\mathrm{X})=\delta(X\cap B)+\delta(X\cap E)-1\geq(\mathrm{S}(\mathrm{A})+s)+(1+t)-1\geq\delta(A)+s$.
$\square$Lemma 4.2 Suppose $0\leq t\leq s\leq 1$ . $C=B\oplus_{a}E$ , and $(E, a, b)$ i.s $a(1-t)-$$+s\leq\delta(X)$and $a\in B-A$ ,component If
for any $X$ such that
$X$ such that$+(s-t)\leq\delta(X)$
.
arvy
then
for
$\delta(A)$$\delta(A)$$A\subseteq\wedge X\subseteq B$$A\subseteq X\subseteq C$

81(; $X\subseteq C$ . If a
, then $A\subseteq(X\cap B)$ , $a\not\in(X\cap E)$ ,Proof. Suppose$(X\capE)=\emptyset$$X=(X\capB)\oplus(X\capE)$, thenand. If, and thus$(X\capE)\neq\emptyset$
5 $(X)\geq\delta(A)$ $+s\geq\delta(A)$ $+s-t$ . If
, then $\delta(X\cap E)\geq 1$ and thus$A$$\not\in X$$A\subseteqarrow X\subseteq B$$\delta(X)=\delta(X\cap B)+\delta(X\cap E)\geq\delta(A)$ $+1\geq\delta(A)$

If

then

$a\in X$ ,$X=(X\cap B)\oplus_{a}(X\cap E)$,$+s-t$ .
, and$A\subseteq Xarrow\cap B$$X\cap E\neq\emptyset$. Hence,.

$6(\mathrm{X})=\delta(X\cap B)+\delta(X\cap E)-1\geq(\delta(A)+s)+(1-t)-1\geq\delta(A)+s-t$$\square$

Theorem 4.3 Consider
where consists of rational numbers only. Suppose ,are -srructure in
and $A\leq B$ . Then for any positive integer , there is an, $B\leq_{n}C$ , $A\leq C$ and.
-structure insuch that$\mathrm{K}_{\alpha}$$B$$\mathcal{L}$$A$$\alpha$$n$$\mathrm{K}_{\alpha}$$C$$\mathcal{L}$$B\subseteq C$$\mathrm{K}_{\alpha}$$5(\mathrm{X})=5\{\mathrm{A})$Proof. We prove the theorem by the induction on $|B-A|$ .
Suppose $|B-A|=0$ . In this case, $A=B$ . Let $C=B=A$ . Then the statement
holds.
andbe a substructure of such that
Suppose $|B-A|>0$ . Let.
for ally set $X$ such that
Note that we have $A\leq B_{0}$ attd $B_{0}\leq B$ .
. We have $|B-B_{0}|<|B-A|$ . By the induction hypothesis,
Suppose$n>0$
is a structure $C\supset B$ such that $B\leq_{n}C$ , $B_{0}\leq C$ and,anyinteger
there
for$A\leqB_{0}$.
, we have the statem
and. Since
Therefore,
Suppose $\delta(A)<\delta(B_{0})$ . If is a point in $B-A$ then$\delta(A)+s\leq\delta(X)$$s=\delta(B_{0})-\delta(A)$$\delta(B_{0}/A)\leq 1$ by the choice of. Then. Let.
such that A
for any substructure $X$ ofis a reduced fraction. Let
where eachLet
. Then $s=\delta(B_{0})-\delta(A)$ is a
$m$ be the least common multiple of, . . .,,
multiple of $1/m$ . Let $s=k/m$ where is a positive integer.
For any positive integer , let $C’$ be an -structure such that$B$

$B_{0}$$\delta(B_{0})\leq\delta(X)$$A\underline{\subseteq}B_{0}\subseteq B$$A\subseteq Xarrow\subseteq B$$\delta(A)=\delta(B_{0})$$\delta(A)$$\delta(C)=5(\mathrm{B}\mathrm{o})$$=\delta(B_{0})$$\mathrm{e}\mathrm{n}\mathrm{t}$$\delta(\{a\}/A)$$a$$\leq 1$

$B_{0}$$B_{0}$$\alpha=$$\leq 1\backslash$$\mathrm{C}Xarrow$$(n_{1}/m_{1}, n_{2}/m_{2}, \ldots, n_{\ell}/m_{l})$$n_{i}/m_{i}$$\mathrm{m}\mathrm{i}$$m_{2}$$m_{l}$$k$$\mathcal{L}$$n$$C’=B\oplus_{a_{0}}E_{0}\oplus_{a_{1}}E_{1}\cdot$..

$\oplus_{a_{n}}E_{n}\oplus_{a_{\iota+1}},F_{1}\oplus_{a_{n+2}}F_{2}\cdots\oplus_{\alpha_{n+\mathrm{L}-}}F_{k}$,is a l-con point for,are pairwise distinct,
where the$(F_{j},a_{n+j},a_{n+g+1})$
for $\dot{J}=1,2$ ,$(1-1/m)$-componellt
is a, , and
, 1!.
each
$C’$
such thatbe a substructure of. . ., . Let$\mathrm{a}\mathrm{f}1\in B_{()}-A$$a_{\iota}$$\mathrm{i}=0$$k$$\ldots$$(E_{i},a_{i}, a_{i+1})$$n$$C_{0}$$C_{0}=B_{0}\oplus_{a_{0}}E_{0}\oplus_{a_{1}}E_{1}\cdots\oplus_{a_{?\mathrm{t}}}E_{n}\oplus_{a_{\iota+1}},F_{1}\oplus_{a_{n+2}}F_{2}\cdot$..

..

$\oplus_{a_{n+k}}F_{k}$.
. Therefore, $C0$
By Lemn as 4.1 aatd 4.2, we have $A\leq C_{0}$ and$X=$$-B_{0}$ and $|X|\leq n$ .
Then
. Suppose
We show that
case,
.
In
eithersome
foror. Sincefor some $j\leq n$ alld.
4.1,
by Lemllla$\delta(A)$$B_{0}\leq_{n}C_{0}$$\{a_{0}, a_{1}, \ldots, a_{n-1}\}$$a_{i}\not\in X$$E_{1}\cdots\oplus_{a_{j}}E_{j})\oplus D$$\delta(B_{0})\leq\delta(X)$$=\delta(C_{0})$$\in \mathrm{K}_{\alpha}$$X\subseteq C_{0}$$\mathrm{i}<n$$X\subseteq(B_{0}\oplus_{a_{0}}E_{0}\oplus_{a1}$$D\in \mathrm{K}_{\alpha}$$B_{0}\leq B_{0}\oplus_{a_{0}}E_{0}\oplus_{a_{1}}E_{1}\cdots\oplus_{a_{j}j}E$

82

Co. Hence, Co $\leq C’$ and
Now we have $C’=B\oplus_{B_{0}}C_{0}$ , $B_{0}\leq C$ , andsuch that $C\supset C’$ ,$B\leq_{n}C’$ . Since $|C’-C_{0}|=|B-B_{0}|<|B-A|$ , we have$C’\leq_{n}C$ , $C_{0}\leq C$ and(Cq). Therefore, $B\leq_{n}C$ , $A\leq C$ , and $\delta(C)=\delta(A)$ .
$B_{0}\leq_{n}$$C\in \mathrm{K}_{\alpha}$$\mathrm{S}(\mathrm{C})=6$

口C-str uctures in
Theorem 4.4 Suppose ,$>0$positive
integer , there
any
and fornumber$A\leqC$$\delta(C)<\delta(A)$$B\leq_{n}C$
.
,
and
,
that
$A$

$Ba\uparrow e$$\mathrm{K}_{\alpha}$$n$$\epsilon$$B\subseteq C$$?S$

and $A\leq B$ . Then for any realsuch
an -srructure in$C$$\mathcal{L}$$\mathrm{K}_{\alpha}$$+\epsilon$

Proof. If a consists of ratinal numbers only, then the statement holds by Theorem
4.3.
irrational number. The proof for this case is similar
Assume that a contains
as in the proof ofSo,give only a sketch. Choose
weto that of Theorem 4.3.
$>0$
be an arbitrary (small)
Theorem 4.3. Assume that $s=\delta(B_{0})-\delta(A)>0$ . Letreal number. Let$\partial \mathrm{J}^{\cdot}1$$B_{0}$$\epsilon$$C’=B\oplus_{a_{0}}E_{0}\oplus_{a_{1}}E_{1}$..

$\oplus_{a_{n}}E_{n}\oplus_{a_{n+1}}F_{1}\oplus_{a_{n+2}}F_{2}\cdot$

as in the proof of Theorem 4.3. except that, . . ., so thatsufficiently small for each$(E_{i}, a_{i}, a_{i+1})$$\mathrm{i}=0,1$

$t$

..

$\oplus_{a_{n+k}}F_{k}$is a $(1+t)$ -component with$n$$\delta(B_{0}\oplus_{a_{0}}B_{0}^{\gamma}\oplus_{a_{1}}E_{1}\cdots\oplus_{a_{n}}E_{n})<\delta(B_{0})+\epsilon/4$,, ,and $(F_{j}, a_{n+j}, a_{n+j+1})$ is a $(1-t’)$ -component with $0<t’<\epsilon/4$ for $j=1$ , 2,
$A\leqC_{0}$, andwhere A is the largest integer such that $kt’<s$ . Thenfor
as in the proof of Theorem 4.3. Then we have $C_{0}\leq C’$ and $B\leq_{n}C’$ .
such that $C’\leq_{n}C$ , $C_{0}\leq C$ ,By the induction hypothesis, we can choose
and $\delta(C)<\delta(C_{0})+\epsilon/2$ . Therefore, we have the theorem.
$\ldots$$k$$\delta(C\mathrm{o})<\delta(A)+F./2$$C_{0}$$C\in \mathrm{K}_{\alpha}$$\square$

References[1] J.T. Baldwin, and N. Shi, Stable generic structures, Ann. Pure Appl. Logic 79(1996) 1-35.
[2] K. Ikeda, H. Kikyo, and A. Tsuboi, On generic structures with a strong amalgamation property, in preparation.

[3] N. Peatfield and B. Zilber, Analytic Zariski structures and the Hrushovski construction, Ann. Pure Appl. Logic 132 (2005) 127-180