数理解析研究所講究録
第 1555 巻 2007 年 1-8

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On a generator of a nonstandard universe
法政大学

村上雅彦 (Masahiko MURAKAMI)

Hosei University

1. Nonstandard Universe
1.1. Superstructure
Given a set $X$ , we define the iterated power set $V_{n}(X)$ by$V_{0}(X)=X$ ,$V_{n+1}(X)=V_{n}(X)\cup\wp(V_{n}(X))$.

The superstructure $V(X)$ is the union$V(X)= \bigcup_{n<w}V_{n}(X)$.

A set $X$ is said to be a base set if$\emptyset\not\in X$ud$\forall x\in Xx\cap V(X)=\emptyset$.

In a superstructure $V(X)$ , the elements of $V(X)\backslash X$ are called sets $re$ lative to. We denote the structure \langle V(X), ) for the language
by the same symbol $V(X)$ .
$V(X)$$\in$$\iota_{\in}=\{\in\}$of set theory2

1.2. Nonstandard Universe
A nonstandard universe is a triple \langle V(X), $V(Y),$$*\rangle$

such that:(1) $X$ and $Y$ are infinite base sets.
(2) (Transfer Principle) The map
into $V(Y)$ : $\star:V(X)arrow V(Y)$ ,iff$V(X)\models\varphi(a,\overline{b})$$\star$is a bounded elementary embedding of $V(X)$$V(Y)\models\varphi(a,\overline{b})$for every$\Delta_{0}$-formula$\varphi(x,\overline{y})$.

(3) $\star x=Y$ .

For $a\in V(\star X)=V(Y)$ ,is standard if
for some $x\in V(X)$ andis intemal if
for some $x\in V(X)$ .
We denote the set of all internal elements $nV(\star X)$ by$a=\star_{X}$$a$$a\in\star_{X}$

$a$

V(X)

$=${ $x\in V(\star X)|x$ is internal}$= \bigcup_{\mathfrak{n}<w}V_{n}(X)$.

The structure $\star V(X)$ is transitive over Y. Then, we can simply denote by single$V(X)$ nonstandard universe.

1.3. Invariants of nonstandard Universe
The norm (of standardness) $nos(a)$ of an internal element$nos(a)=\min\{|x||a\in\star_{x\}}$$a$.

The radius of V(X) is a cardinal defined by
rad $(V(X))= \min\{\kappa|\forall y\in V(X)nos(y)<\kappa\}$ .
Let $E$ be a subset of $\star V(X)$ . We denote$dc1(E)=\{w(s)|w\in V(X), s\in E^{<w}, s\in domw\}$.

The length of V(X) is a cardinal defined bylen (V(X)) $= \min$ { $|E||E\subseteq*V(X)$ and $dc1(E/\star)=V(X)$ }.
’V(X) is monogenic if $1en(*V(X))=1.$$V(X)=dc1(\{a\})$.

$a$

is a generator of monogenic V(X) if

FYom now on, we shall consider “V(X) such that rad $(*V(X)),$ $1en(V(X))<$$|V(X)|$.

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2. Examples of nonstandard universe
2.1. Bounded ultrapowerLet

be an index set. We define$I$$\hat{V}(X)^{I}=$$\varphi(I)$-valued universe by{ $u:Iarrow V(X)|$ ran$w\subseteq V_{n}(X)$for some $n<w$ }with truth values$Iu=vJ=\{i\in I|u(i)=v(i)\}$ ,$I\varphi\wedge\psi\#=\beta\varphi J\cap\beta\phi J$$\beta u\in vJ=\{i\in I|u(i)\in v(i)\}$,$[\urcorner\varphi J=I\backslash I\varphi J$$I\exists x\varphi(x)J=\cup\{[\varphi(u)J|u\in\hat{V}(X)^{I}\}$Let$\mathcal{U}$be an ultradlter over

$I$,,.

. We can define Bounded ultrapower$\hat{V}(X)^{I}/\mathcal{U}=\{u/\mathcal{U}|u\in \mathcal{U}\}$

whereis the equivalence class of the relation
For $a\in V(X)$ definebyanda nonstandard universe.

,$\# u=vJ\in \mathcal{U}$$u/\mathcal{U}$$\check{a}\in\hat{V}(X)^{I}$$\check{a}:Iarrow\{a\}$.

$\star_{a}=\check{a}/\mathcal{U}$. Then$\hat{V}(X)^{I}/\mathcal{U}$isTheorem 1. (1) If $|I|<|V(X)|$ then $\hat{V}(X)^{I}/u$ is monogenic.
(2) Monogenic nonstandard universe V(X) is isomorphic to a bounded ultrapower.

Proof. (1) Wlog
(2) Let$a\in\star I$$a$. Define$I$iv a set relative to $V(X)$ . Then $id_{I}/u$ is a generator of

be a generator of “V(X). Let$\mathcal{U}=\{A\subseteq I|a\in\star A\}$$I$$\hat{V}(X)^{I}/\mathcal{U}$.

be a set relative to $V(X)$ such thatthen V(X) is isomorphic to $\hat{V}(X)^{I}/u$ .

$\square$Considering a generator, we have the theorem below.

Theorem 2. If there is a bounded elementary embedding $e:\hat{V}(X)^{I}/uarrow\hat{V}(X)^{J}/V$,then there is $h:Jarrow I$ such that $\mathcal{U}=\{A\subseteq I|h^{-1}" A\in V\}$ and $e(u/u)=(u\circ h)/V$ .

Proof. Let$h/V=e(id_{I}/\mathcal{U})$

.

口

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2.2. Bounded Boolean ultrapower
Let$\langle \mathfrak{B}, \wedge, , \neg, 0,1\rangle$

be a $cBa$ . We define B-valued universe by$\hat{V}(X)^{\langle \mathfrak{B}\rangle}=\{u:V(X)arrow \mathfrak{B}||ranu=1,suppu\in V(X)(x)\wedge u(y)=0forx\neq y,\}$,where supp $u=\{x\in V(X)|u(x)\neq 0\}$ , with truth values$\mathbb{I}u=vJ=\vee\{u(x)\wedge v(x)|x\in V(X)\}$$I\varphi\wedge\psi J=\beta\varphi\#\wedge\#\phi J$,$\# u\in vJ=\vee\{u(x)\wedge v(y)|x\in y\}$$\mathbb{I}\neg\varphi I=\neg I\varphi J$$\beta\exists x\varphi(x)J=\vee\{\beta\varphi(u)J|u\in\hat{V}(X)^{\langle \mathfrak{B}\rangle}\}$,,.

Let be an ultrafilter of B. As bounded ultrapower, we define bounded Booleanultrapower. Bounded ultrapower is a nonstandard universe. If isis not monogenic by Theorem 1.
atomless then$\mathcal{U}$$\hat{V}(X)^{\langle \mathfrak{B}\rangle}/\mathcal{U}$$\mathfrak{B}$$\hat{V}(X)^{\langle \mathfrak{B}\rangle}/\mathcal{U}$2.3. Bounded ultralimit
A set A of subsets of a Ba

(1)$\cup\Lambda=\mathfrak{B}$(2) If$S_{1},$(3) If$S\in\Lambda$

is a locally atomic complete algebra (LA $CA$ ) if.

$S_{2}\in\Lambda$

thenand(4) For every

that$\mathfrak{B}$$S_{1}\cup S_{2}\in\Lambda$$T\subseteq S$$S\in\Lambda$$S\subseteq C\in\Lambda$

then.

$T\in\Lambda$.

, there is an atomic complete regular subalgebra$\overline{\mathfrak{B}(\Lambda)}$$\Lambda$.

such-valued universe bywith thuth value assignment as that of B-valued universe, where$\mathfrak{B}(\Lambda)$$\prime \mathfrak{B}$denoted by $B(A)$ .

$\hat{V}(X)^{\{\langle A\rangle\rangle}=\{u:V(X)arrow \mathfrak{B}(\Lambda)|u(x)\wedgeu(y)=0forx\neq y,$$r\bm{t}u\in\Lambdaranu=1,suppu\inofof.

We say the Boolean algebra U A is base Boolean algebra of

We define$C$$\overline{\mathfrak{B}(\Lambda)}$V(X)$$\}$

is a completion

5Lemma 3. Let$B(\Lambda)$$\varphi$

be a statement of$\hat{V}(X)^{\langle\langle\Lambda\rangle)}$then [$\varphi J\in \mathfrak{B}(\Lambda)$. So$\hat{V}(X)^{\langle(\Lambda\rangle\rangle}$

is-valued.

Let$\mathcal{U}$be an ultrafilter of$B(\Lambda)$. As bounded ultrapower and bounded Boolean

ultrapower, we define bounded Boolean ultralimit. Bounded ultralimit is$\hat{V}(X)^{\langle\langle A\rangle\rangle}/\mathcal{U}$a nonstandard universe.
Theorem 4 (representatioin theorem). For every nonstandard universe V(X), there
is a Bounded ultrvtlimit isomorphic to ‘V(X).

3. Looking for a generator outside “V(X)If V(X) is not monogenic, there is not a generator in‘generator’ outside “V(X).

$V(X)$. We are looking for a

3.1. Ultrasheaf
Let $V^{(B)}$ be a cBa B-valued universe of set theory:$v_{\alpha}^{(\mathfrak{B})}=\{u$:dom $uarrow B|$ dom$V^{(\mathfrak{B})}=\bigcup_{\alpha}V_{\alpha}^{(\mathfrak{B})}$$u\subseteq\cup\{V_{\beta}^{(\mathfrak{B})}|\beta<\alpha\}\}$,,$\check{c};\{\check{x}|x\in c\}arrow\{1\}$with truth values$Iu\in v\#=\vee\{v(x)\wedge\beta x=uI|x\in domv\}$$\mathbb{I}u=vJ=\wedge${,$\beta x\in uI\Leftrightarrow\beta x\in vI|x\in domu\cup$

Inside $v^{(B)}$ , we consider iterated power sets

$V_{\check{n}}(\check{X})$, and let$\hat{V}(X)^{(\mathfrak{B})}=\bigcup_{n<\omega}V_{\hslash}(\check{X})$.

dom$v$}.

6Let$\mathcal{U}$be an ultrafilter of B. As bounded ultrapower and bounded Boolean ultra-power, we define bounded Boolean sheaf
Theorem 5. The map
embedding and “V(X)$\hat{V}(X)^{(B)}/\mathcal{U}$$\star:V(X)arrow\hat{V}(X)^{(\mathfrak{B})}/\mathcal{U}$$ta\eta$$\hat{V}(X)^{(B\rangle}/\mathcal{U}$. If$= \bigcup_{n<\omega}^{\star}V_{n}(X)$is atomless,$B$.

$\star_{a}=\check{a}/\mathcal{U}$is a bounded elemen-

is isomorphic to the Boolean ultrapower$V(X)\neq\hat{V}(X)^{(\mathfrak{B})}/\mathcal{U}$.

Wlog, these inclusions hold:$t^{\star_{x}}|x\in V(X)\}\subseteq V(X)\subseteq\hat{V}(X)^{(\mathfrak{B})}/\mathfrak{U}\subseteq V(\star X)$Suppose$B$is a set relative to $V(X)$ . Canonial generic filter $G\in\hat{V}(X)^{(B)}$ isdefined bydom$G=\check{B}$,$G(\check{b})=b$.

Theorem 6. For every $a\in\star V(X)$ , there is $w\in V(X)$ such thatand $a=\cup\cap M$ “$G/u.\hat{V}(X)^{(\mathfrak{B})}/u$

contains $V(X)\cup\{G/u\}$ .

$w:\mathfrak{B}arrow V(X)\backslash X$

is the least transitive substructure of $V(\star X)$ that

Theorem 7. If there is a bounded $elementa\eta$ embedding$e:\hat{V}(X)^{(A)}/\mathcal{U}arrow\hat{V}(X)^{(B)}/V$,then there is a cBa homomorphism $h:Aarrow B$ such that$(h\circ u)/V$$\mathcal{U}=h^{-1}$ “V and$e(u/\mathcal{U})=$.

Proof. Since$G/\mathcal{U}$trafilter $H$ of$\check{\mathcal{A}}$is $V(A)$ -complete ultrafilter of $A$ , there is a
inside $V(X)^{(B)}$ such that $H/V=e(G/u)$ .

homomorphism $h(a)=\mathbb{I}\delta\in HJ$ .

Compare Theorem 2 with Theorem 7.

$\wp(A)^{\sqrt{}}$-complete ul-Then, we have the$\square$7

3.2. Generator of bounded ultralimitLet

be a bounded ultralimit. Suppose$\hat{V}(X)^{\langle(\Lambda\rangle\rangle}/u$

generator$\Gamma$of$\hat{V}(X)^{\langle\langle\Lambda\rangle\rangle}/u${$\Gamma=$$\prime B(\Lambda)$

is a cBa in $V(X)$ . Define

by$id_{P}/u|P\in\Lambda$

is a partition of $unity$ } $\in V(\star X)$ .

Theorem 8. The genemtor
is -complete ultrafilter of $B(A)$ .
of, there is $w\in V(X)$ such that $w:B(\Lambda)arrow V(X)\backslash X$ andFor every$\Gamma$$\hat{V}(X)^{((\Lambda)\rangle}/\mathcal{U}$$\star\Lambda$$a\in\hat{V}(X)^{(\langle A)\rangle}/\mathcal{U}$$a=\cup\cap bv$ “

F. If$\Lambda$

is the largest LACA on a cBa$\prime \mathfrak{B}$then$\Gamma=G/\mathcal{U}$.

Lemma 9. Let A be an LA $CA$ . There is the least LA $CA$ A such that$\mathfrak{B}(\overline{\Lambda})=\overline{\mathfrak{B}(\Lambda)}$. If$\mathcal{U}\subseteq\overline{\mathcal{U}}$then$\Lambda\subseteq$

A and.

$\hat{V}(X)^{((A)\rangle}/\mathcal{U}\cong\hat{V}(X)^{\langle(\overline{\Lambda}\rangle\rangle}/\overline{\mathcal{U}}$4. Questions
Let$\hat{V}(X)^{((\Lambda\rangle)}/\mathcal{U}$Supposewithbe the least transitive substructure which contains $V(X)\cup\{\Gamma\}$ .

$\hat{V}(X)^{\langle\langle\Lambda\rangle\rangle}/\mathcal{U}_{1}=V(X)=\hat{V}(X)^{\langle(\Lambda\rangle\rangle}/\mathcal{U}_{2}$$\hat{V}(X)^{((\Lambda))}/\mathcal{U}_{2}$?

. Dose$\hat{V}(X)^{((\Lambda))}/u_{1}$coincide8References[1] C. CHANG and J. KEISLER, Model Theory, 3rd ed, North-Holland, Amsterdam, (1990).

[2] M. MUHAKAMI, Standardization rinciple of Nonstandard universes, Joumal$p$of Symbolic Logsc, 64, 4(1999), 1645-1655.
[3] M. MURAKAMI, Nonstandard universe, 数瑠廓析研究所鵠究蝕 1213 Model

Theory and Its A pplications, 39-49 (2001).