数理解析研究所講究録
第 1794 巻 2012 年 62-69

62

On the existence of indiscernible structures
竹内 耕太 (Kota Takeuchi)
筑波大学数理物質科学研究科(Graduate school of Pure and Applied
Sciences, University of Tsukuba)

1

Introduction

There are many types of indiscernible objects, for example, indiscernible sequences, indiscernible arrays, and indiscernible trees. They are useful toanalyze the stability of theories. Therefore, we want to know when they
exist. The existence of indiscernible objects was studied in [2], [4], [3], and[5]. In this paper, I show a general method to prove the existence of indiscernible objects. It consists of two steps, checking amalgamation property
and proving partition theorem.

2

Notations and Preliminaries

Before starting, I remark some elementary facts.

Definition 1. Let be a countable language and $M_{n}(n\in\omega)$ a countablestructure. We say the class
has amalgamation property if: For
any $M_{n_{i}}\ni a_{n_{i}}(i<2)$ such that atp
atp, there are embeddingssuch thatfor some.
$\mathcal{L}$$\mathcal{L}-$$\{M_{n} :n\in\omega\}$$(a_{n_{0}})=$$\sigma_{i}:M_{n_{i}}arrow M_{k}$$(a_{n_{1}})$$k\in\omega$$\sigma_{0}(a_{n_{0}})=\sigma_{1}(a_{n_{1}})$

Proposition 2. Suppose $\{M_{n}:n\in\omega\}$ has amalgamation property. Then
there exists the unique countable -structuresuch that$\mathcal{L}$$\bullet$$M^{*}$

for any $a\in M^{*}$ and for any $b\in M_{k}$ with $atp(a)=atp(b)$ , there is anembedding : $M_{k}arrow M^{*}$ with $\sigma(b)=a$ ,$\sigma$$\bullet$for any $a’,$ $b\in M^{*}$ such that atp $(a)=$ atp $(b)$ there is $b’\in M^{*}$ suchthat atp $(aa’)=$ atp $(bb’)$ .
$a,$$\square$63
Corollary 3.

2. Let$a,$1.

$b\in M^{*}$

is homogeneous.

$M^{*}$. If atp $(a)=$ atp $(b)$ then tp $(a)=$ tp $(b)$ .

口

Example 4. Let $M_{n}=(\omega, <)$ for all $n\in\omega$ . Then $\{M_{n} : n\in\omega\}$ has.
isamalgamation property. The generic model$M^{*}$$(\mathbb{Q}, <)$In what follows, we work in a big modelof complete L-theory . We
having free variables subscripted
often consider a set of L-formulasby. We consider
as an -structure with a countable language .
are denoted by $A,$ $B,$
, and subsets ofSubsets of are denoted by $X,$.
denotes an indiscernible object in some sense, which
depends on the structure on . For example if has the empty structurethen we call an indiscernible set. And, if has a structure of total order,
is theand
we call an indiscernible sequence. For$X\simc^{Y}$
sequence
.
.
means
$T$

$\mathcal{M}$$\Gamma((x_{i})_{i\in \mathcal{I}})$$i\in \mathcal{I}$$\mathcal{L}$$\mathcal{L}$$\mathcal{I}$$Y,$$\mathcal{I}$$\mathcal{M}$$\ldots$$\ldots$$I=(a_{i})_{i\in \mathcal{I}}\subset\Lambda t$$\mathcal{I}$$\mathcal{I}$$I$$\mathcal{I}$$I$$X\subset \mathcal{I}$$I=(a_{i})_{i\in \mathcal{I}},$$a_{X}$$atp_{\mathcal{L}}(X)=atp_{\mathcal{L}}(Y)$$(a_{i})_{i\in X}$

Indiscernibilities and substructure properties

3
Let$\mathcal{I}$and$\mathcal{I}’$be -structures.
$\mathcal{L}$Definition 5. A map-atomic types.

:$\sigma$$\mathcal{I}arrow \mathcal{I}’$is said to be an -embedding if$\mathcal{L}$$\sigma$preserves$\mathcal{L}$Note that :-isomorphism. Let
such that$\sigma$$\mathcal{L}$$\mathcal{I}arrow \mathcal{I}’$is an
is an -embedding if and only if :is the sequence
be a sequence in M.
. The following definitions are in [5] and [3].
$\mathcal{L}$$A=(a_{i})_{i\in \mathcal{I}’}$$B=(b_{i})_{i\in \mathcal{I}}$$b_{i}=a_{\sigma(i)}$Definition 6.
1. We saythere is a realization, i.e.
$\Gamma$$\sigma^{-1}A\models\Gamma$

has the -substructure property ifof such that for any -embedding :for every$\mathcal{L}$$\Gamma((x_{i})_{i\in \mathcal{I}})$$A=(a_{i})_{i\in \mathcal{I}}$$\mathcal{I}arrow \mathcal{I},$$\mathcal{I}arrow\sigma(\mathcal{I})$$\sigma$$\sigma^{-1}A$$\Gamma$$\mathcal{L}$$\sigma$$\varphi(x_{i_{1}}, \ldots, x_{i_{n}})\in$$\models\varphi(a_{\sigma(i_{1})}, \ldots, a_{\sigma(i_{n})})$.

2. We saytp.

$I=(a_{i})_{i\in \mathcal{I}}$is -indiscernible if for any$\mathcal{L}$$X\sim c^{Y}\subset \mathcal{I}$, tp $(a_{X})=$$(a_{Y})$

Example 7. Let{$\mathcal{I}$be the structure$x_{j_{n}}):\varphi$$x_{i_{n}})rightarrow\varphi(x_{j_{1}},$$\varphi(x_{i_{1}},$$\ldots,$$\ldots,$$(\omega, <)$and$\Gamma((x_{i})_{i\in \mathcal{I}})$is an L-formula,

the set$i_{n}\}\sim_{<}\{j_{1},$ $\ldots,j_{n}\},$ $n\in\omega$$\{i_{1},$$\ldots,$}.

64
Any realization of is an indiscernible sequence, and vice versa. Since
is -indiscernible, so is any infinite subsequenceof . Hence has-substructure property.
$I$

$I$

$\Gamma$

$I’$

$<$

$I$

$\Gamma$$<$Let$\mathcal{I}$be an -structure and let$\mathcal{L}$$\Gamma((x_{i})_{i\in \mathcal{I}})$

Remark 8. For the following conditions,is realized by an

1.

$\Gamma$2.

$\Gamma^{*}$

3.

$\Gamma’$4.

$\Gamma$$\mathcal{L}$be a set of L-formulas.

$1\Rightarrow 2\Rightarrow 3\Rightarrow 4$

hold:-indiscernible structure .
$I$$:=\{\varphi(x_{Y}):\exists X\sim c^{Y} s.t.

\Gamma\models\varphi(x_{X})\}$$:=\{\varphi(x_{Y}):\exists X\sim c^{Y} s.t.

\Gamma\ni\varphi(x_{X})\}$

is consistent.

is consistent.

has -substructure property.
$\mathcal{L}$In this paper, we see that if the -structure has a nice property thenthe above conditions are equivalent. More precisely, amalgamation properties, and the partition theorems imply
imply.
$\mathcal{L}$$\mathcal{I}$$4\Rightarrow 2$$2\Rightarrow 1$

Example 9. Let
be the set of formulas expressing the unstability, i.e.
of. (If thetheory is unstable) has subsequence property. By Ramsey’s theorem wehave an indiscernible sequence realizing F.
$\Gamma((x_{i})_{i\in\omega})$$\varphi(x, y)$$\Gamma((x_{i})_{i\in\omega})=\{\varphi(x_{i}, x_{j}) :i<j\}\cup\{\neg\varphi(x_{i}, x_{j}) :i\geq j\}$$\Gamma$

The amalgamation property and the substructure property

4

be an

-structure anda set of L-formulas. Suppose that
. Put Endhas -substructure property, witnessed by: is an -embedding.}. If
End
then
{ :
, since. Therefore,. However, there may not besendingEnd
toeven if $X\sim c$ Y.
. The following is an example of such
In generalthat has substructure property but
is inconsistent.
Let$\mathcal{I}$$\Gamma$$\mathcal{L}$$\Gamma((x_{i})_{i\in \mathcal{I}})$$A=(a_{i})_{i\in \mathcal{I}}$$\mathcal{L}$$\mathcal{I}arrow \mathcal{I}$$\sigma$$\sigma\in$$\mathcal{L}$$\sigma$$\Gamma((x_{\sigma(i)})_{i\in \mathcal{I}})$$\sigma^{-1}A\models\Gamma$$\sigma\in$$(\mathcal{I})$$(\mathcal{I})=$$A\models\Gamma((x_{i})_{i\in \mathcal{I}}))\cup$$A \models\bigcup_{\sigma\in End(\mathcal{I})}\Gamma((x_{\sigma(i)})_{i\in \mathcal{I}})$$X\subset \mathcal{I}$$(\mathcal{I})$$Y\subset \mathcal{I}$$\Gamma$$\bigcup_{\sigma\in End(\mathcal{I})}\Gamma((x_{\sigma(i)})_{i\in \mathcal{I}})\subseteq\Gamma^{*}$$\Gamma^{*}$$\Gamma$

Example 10. Let $L=$ {on $M$ by$\bullet$$\eta<ini\eta’\Leftrightarrow\eta$$<_{ini},$ $<$lex} and$M=\omega^{<\omega}$

is a proper initial segment of. Consider an$\eta’$,$\mathcal{L}$-structure

65$\eta<$$\bullet$lex

is less than$\eta’\Leftrightarrow\eta$$\eta’$

in the lexicographic order,

for every $\eta’\in M$ . For example, $\{0\}<_{ini}\{0,0\}<$ lex. Let
be also$\{<_{ini}, <lex\}$ .
the structure
and
the
set
Put
lex
$\{x_{\eta}<_{ini}x_{\eta’} : \eta<ini\eta’\}\cup\{x_{\eta}<$: $\eta<$ lex. Immediately has
lex
substructure property in $Th_{L}(M)$ witnessed by $M$ . Let
be the longest
common initial segments of and . ( is definable in $Th_{L}(M).$ ) ThenA
. However,1. Henceis inconsistent in the theory.
$\{0,1\rangle$$\eta,$$(\omega^{<\omega},$$<_{ini},$ $<$$\mathcal{I}$$\mathcal{L}$$)$$\Gamma((x_{\eta})_{\eta\in \mathcal{I}})=$$\Gamma$$\eta’\}$$x_{\eta’}$$\mathcal{L}-$$\eta\cap\eta^{l}$$\eta’$$\eta$$\Gamma\models x_{\langle 0,1\rangle}\cap x_{\langle 1}$$\rangle<_{ini}x_{\langle 0,0\rangle}\cap x_{\langle 0,1\rangle}$$\cap$$x_{(1,0\rangle}\cap x_{\langle 1,1\rangle}\geq_{ini}x_{\langle 0,0\rangle}\cap x_{\langle 1,0\rangle}$$\{0,0\}\{0,1\rangle\langle 1,1\rangle\sim c\langle 0,0\rangle\langle 1,0\}\langle 1,1\}$$\Gamma^{*}$

Proposition 11. Let be an -structure. Suppose thathas amalgamation property, i.e. $\{M_{n}:n\in\omega\}$ has amalgamation property where$(n\in\omega)$ . If
has substructure property thenis consistent.
$\mathcal{I}$$\mathcal{L}$$\{\mathcal{I}\}$$M_{n}=\mathcal{I}$$\Gamma$

Proof. Let$\Gamma^{*}$

be the generic model of. Let
formulas such that
if and only if$\mathcal{I}^{*}$$\Delta((x_{i})_{i\in \mathcal{I}^{*}})$$\{\mathcal{I}\}$be the set of$\triangle((x_{i})_{i\in \mathcal{I}^{*}})\ni\varphi(x_{Y})$$\bullet\Gamma\models\varphi(x_{X})$$\bullet$there is an,$\mathcal{L}$-embedding

In the other words,$\triangle$$\bigcup_{\sigma}\Gamma((x_{\sigma_{k}(i)})_{i\in \mathcal{I}})$

Claim A.

$\triangle((x_{i})_{i\in \mathcal{I}^{*}})$$\sigma$:$\mathcal{I}arrow \mathcal{I}^{*}$

sending $X$ to Y.

is the set( varies over all embeddings from$\sigma$$\mathcal{I}$to$\mathcal{I}^{*}$).

is consistent.

It is enough to show that for any embeddings
is consistent. For simplicity, we assume that $n=2$ and is closcd undertaking the conjunction. Take
. Let$(k=1,2)$ and
. (If
then
is clearlyconsistent.) Since $Z_{1}\sim cZ_{2}$ , there are embeddings:such that. By the substructure property,
is consistent.
Therefore,
is consistent. (End of the proof of the claim.)
By the construction of , if
andthen
. Henceis consistent.
a$\sigma_{1},$$\ldots,$$\sigma_{n},$$\bigcup_{k\leq n}\Gamma((x_{\sigma_{k}(i)})_{i\in \mathcal{I}})$$\Gamma$$\varphi_{1}(x_{X_{1}}),$$Z_{k}=\sigma_{k}^{-1}(Y_{1}\cap Y_{2})$$\varphi_{2}(x_{X_{2}})\in\Gamma((x_{i})_{i\in \mathcal{I}})$$Z_{k}=\emptyset$$\sigma_{k}(X_{k})=Y_{k}$$\bigwedge_{k}\varphi_{k}(x_{Y_{k}})$$\mathcal{I}arrow \mathcal{I}$$\tau_{k}$$\tau_{1}(Z_{1})=\tau_{2}(Z_{2})$$\bigcup_{k}\Gamma((x_{\tau_{k}(i)})_{i\in \mathcal{I}})$$\bigwedge_{k}\varphi_{k}(x_{Y_{k}})$$\triangle$$\varphi(x_{Y’})$5$Y\sim cY^{l}\subset \mathcal{I}^{*}$$\triangle\models\varphi(x_{Y})$$\triangle\models$$\Gamma^{*}$

Partition lemmas and the existence of indiscernible structures

We begin with Erd\"os-Rado theorem and indiscernible sequence.

66
Let

$\mathcal{I}=(\omega, <)$. Suppose thathas amalgamation property,

and$\mathcal{L}=\{<\}$

has -subsequence$\mathcal{L}$$\Gamma((x_{i})_{i\in\omega})$

is consistent. By
property. Sincewith sufficiently large . Take acompactness, we may assume. By Erd\"os-Rado, we have a subsequence $I=$
realization of.
of which is indiscernible for n-variable formulas. ClearlyUsing compactness, we have$\{\mathcal{I}\}$$\Gamma$“$((x_{i})_{i\in\omega})$$\mathcal{I}=(\kappa, <)$$J$$\kappa$$\Gamma^{*}((x_{i})_{i\in\kappa})$$J$$(a_{i})_{i\in\omega}$$I\models\Gamma$. If
Fact 12. Letandproperty then it is realized by -indiscernible sequence.
$\mathcal{L}=\{<\}$$\mathcal{I}=(\omega, <)$$\Gamma((x_{i})_{i\in \mathcal{I}})$

has subsequence口$\mathcal{L}$In this section we discuss about three examples of indiscernible structures on an array and a tree. Always, first we show the partition lemma forthe structure. Then we get indiscernible objects. Indiscernible arrays are
discussed in [3]. Indiscernible trees are discussed in [2],[4],[3], and [5]. The
partition theorems are proved in [1] and [4].

Let$\mathcal{L}=\{P_{n}(x),$ $<$$\bullet P_{n}^{\mathcal{I}}=\{n\}\cross\omega$lex$\}_{n\in\omega}$. Let$\mathcal{I}$be the -structure on$\mathcal{L}$$\omega\cross\omega$defined by,$\bullet<_{1ex}^{\mathcal{I}}=\{\{(n, m), (k, l)\}:n<k or n=k\wedge m<l\}$Note that$\lambda$$\omega\cross\lambda$

is an -structure with the natural interpretations, for every$\mathcal{L}$.

Lemma 13 (the partition lemma). Let:
there is an -embedding :$X\sim cY\in(\omega\cross\omega)^{n}$ then $f(\sigma(X))=f(\sigma(Y))$ .

$\kappa<<\lambda$$f$$(\omega\cross\lambda)^{n}arrow\kappa$$\mathcal{L}$$\sigma$and$n\in\omega$. Then for any$\omega\cross\omegaarrow\omega\cross\lambda$

such that if

which is the image of-substructure. We assume $n=2$ , since general cases are similar. For given , we define
a map
by

Proof. We will give an$\mathcal{L}$$\omega\cross\chi\subset\omega\cross\lambda$$f$$\sigma$$g:\lambda^{2}arrow\omega^{2}\kappa$$g(\alpha_{1}, \alpha_{2})=f((-, \alpha_{1}), (-, \alpha_{2})):\omega^{2}arrow\kappa$

By Erd\"os-Rado, we have a homogeneous set$\chi\subset\lambda$.

with respect to .
$g$

Claim A. If $X\sim cY\in(\omega\cross\chi)^{2}$ then $f(X)=f(Y)$ .

Suppose$X\sim c^{Y}$ ,$X=\{(x_{1}, \alpha_{1}), (x_{2}, \alpha_{2})\}$$\bullet x_{k}=y_{k}(k=1,2)$,

and$Y=\{(y_{1}, \beta_{1}), (y_{2}, \beta_{2})\}$. Because

67$\bullet$$(\alpha_{1}, \alpha_{2})$andBy the homogeneity ofandhave the same order type.

$(\beta_{1}, \beta_{2})$. Hence, $f((-, \alpha_{1}), (-, \alpha_{2}))$are the same function.
,
Since
口
.

$\chi,$$f((-, \beta_{1}), (-, \beta_{2}))$$g(\alpha_{1}, \alpha_{2})=g(\beta_{1}, \beta_{2})$$=$$x_{k}$$y_{k}$$f((x_{1}, \alpha_{1}), (x_{2}, \alpha_{2}))=f((y_{1}, \beta_{1}), (y_{2}, \beta_{2}))$

Fact 14. Letandlexof L-formulas. Suppose thathas
it is realized by an -indiscernible array.
$\mathcal{L}=\{P_{n},$ $<$be a set-substructure property. Then$\mathcal{I}=(\omega\cross\omega;\mathcal{L})$$\}_{n}$$\Gamma((x_{i})_{i\in \mathcal{I}})$$\mathcal{L}$, and let$\Gamma((x_{i})_{i\in \mathcal{I}})$$\mathcal{L}$

Proof. Notice thathas the amalgamation property. Henceis consistent. By compactness, we may assume
lex with sufficiently large . Using the partition lemma and compactness, we have arealization of which is -indiscernible.
口$\Gamma^{*}$$\{\mathcal{I}\}$$\mathcal{I}=$$(\omega\cross\lambda;(P_{n})_{n},$

$<$

$)$

$\lambda$$\Gamma$$\mathcal{L}$Note that there are many other structures on
which give the same indiscernibility. For example,, defined by:$m<k\}$ , is one of such structures.
$\omega\cross\omega$$<_{n}^{\mathcal{I}}=\{\{(n, m), (n, l)\}$$\mathcal{L}=\{P_{n}, <_{n}\}_{n}$Next we consider another structure on
Let be the -structure on
defined by$\omega\cross\omega$$\mathcal{I}$$\mathcal{L}$. Let $\mathcal{L}=\{E(x,$ $y),$ $<$ lex .
$\}$$\omega\cross\omega$$\bullet E^{\mathcal{I}}=\{\langle(n, m), (k, l)\rangle:n=k\}$,$\bullet<_{1ex}^{\mathcal{I}}=\{\{(n, m), (k, l)\rangle:n<k or n=k\wedge m<l\}$Lemma 15 (the partition lemma). Let:
there is an -embedding :then $f(\sigma(X))=f(\sigma(Y))$ .

$\kappa<<\lambda$$f$$(\lambda\cross\lambda)^{n}arrow\kappa$$\mathcal{L}$$\sigma$.

and$n\in\omega$. Then for any
such that if$\omega\cross\omegaarrow\lambda\cross\lambda$$X\sim c^{Y}\in(\omega\cross\omega)^{n}$

Proof. We will give an -substructure

which is the imageof . Assume that $\kappa<<\lambda’<<\lambda$ . By the same argument of the proofof the previous partition lemma, we have an infinite subset
such$(k_{n},m_{n})\}=y\in\lambda’\cross\chi’$that ifthen $f(x)=f(y)$ . By Erd\"os-Rado, there is an infinite setsuch$(m_{n}, k_{n})\rangle=y\in\chi\cross\chi’$ thenthat if$f(x)=f(y)$ . Suppose.
Then we have$\mathcal{L}$$\chi\cross\chi’\subset\lambda\cross\lambda$$\sigma$$\chi^{l}\subset\lambda$$x=\langle(k_{1}, l_{1}),$$(k_{n}, l_{n})\rangle\sim c\langle(k_{1}, m_{1}),$$\ldots,$$\ldots,$$\chi\subset\lambda’$$x=\langle(l_{1}, k_{1}),$$\ldots,$$(l_{n}, k_{n}))\sim c\langle(m_{1}, k_{1}),$$\ldots,$$(k_{n}, l_{n})\rangle\sim c\langle(j_{1}, m_{1}),$$\{(k_{1}, l_{1}),$$\ldots,$$\bullet\langle(k_{1}, l_{1}),$$\ldots,$$(k_{n}, l_{n})\rangle\sim c\{(k_{1}, m_{1}), \ldots, (k_{n}, m_{n})\}$$(j_{n}, m_{n})\rangle\in\chi\cross\chi^{l}$$\ldots,$,$\bullet\{(k_{1}, m_{1}), \ldots, (k_{n}, m_{n})\}\sim c\{(j_{1}, m_{1}), \ldots, (j_{n}, m_{n})\}$

.

68
Hence
$=f(\langle(k_{1}, m_{1}), \ldots, (k_{n}, m_{n})\rangle)$$f(\langle(k_{1}, l_{1}), \ldots, (k_{n}, l_{n})\rangle)$$=f(\{(j_{1},$ $m_{1}),$$(j_{n},$$\ldots,$$m_{n})\})$.
口, and letbe a set
andFact 16. Letlex
has -substructure property then itof L-formulas. Suppose thatis realized by an -indiscernible array.
$\mathcal{L}=\{E,$ $<$$\mathcal{I}=(\omega\cross\omega, \mathcal{L})$$\}_{n}$$\Gamma((x_{i})_{i\in \mathcal{I}})$$\mathcal{L}$$\Gamma((x_{i})_{i\in \mathcal{I}})$$\mathcal{L}$

Proof. Because$\{\mathcal{I}\}$

has amalgamation property, the proof is similar to theproof of Fact 14.

口

Next we consider a structure on the tree
Definition 17. Let
lowing structure on

1.

2.
3.

4.

$\eta<_{ini}\nu\Leftrightarrow\eta$$\eta<$$\nu\Leftrightarrow\eta$.

$\mathcal{L}=\{<_{ini}, <lex, \cap, <len, (P_{n})_{n\in\omega}\}$$\omega^{<\omega}$: For$\eta,$$\nu\in\omega^{<\omega}$$\eta<\iota_{en}\nu\Leftrightarrow$. We consider the fol-,is a proper initial segment of ;$\nu$

is less than$\nu$

in the lexicographic order;

the longest common initial segment of

ノ $=$$\eta\cap\iota$$\eta$

5.

lex$\omega^{<\omega}$

len $(\eta)<$ len$(\nu)$, where len$(\eta)$$\eta$and l ノ;is the length of the sequence;$P_{n}(\eta)\Leftrightarrow$the length of$\eta$

is .
$n$. Note that
Let be the structure
has amalgamation property.
The following lemma is in [1, p.662] and [4]. We omit the proof.
$(\omega^{<\omega};<ini, <_{1ex}, \cap, <len, (P_{n})_{n\in\omega})$$\mathcal{I}$$\{\mathcal{I}\}$be a tree, and : $O^{k}arrow\mu$ a k-palace
Lemma 18 (Shelah). Letfunction. If is sufficiently large (depending only on ), then there is an-embedding :such that $f(\sigma(X))=f(\sigma(Y))$ for any k-tuples
口$X,$
with $X\sim c$ Y.
$O=\lambda^{<n}$$f$$\lambda$$\mu$$\mathcal{L}$$\sigma$$\omega^{<n}arrow\lambda^{<n}$$Y\subset\omega^{<n}$By similar way to the above proofs of the existence of indiscernible array,
we also have
Fact 19. Let
property, thenbe a set of L-formulas. If
is realized by an -indiscernible tree.

$\Gamma((x_{\eta})_{\eta\in\omega^{<\omega}})$$\Gamma$$\mathcal{L}$$\Gamma$

has the$\mathcal{L}$-subtree$\square$69

References[1] S. Shelah, Classification Theory, Second Edition. (North Holland, Amsterdam, 1990).

[2] B. Kim and H. Kim, Notions around the tree property 1, Annals of Pureand Applied Logic Volume 162 (2011), No. 9, 698-709.
[3] A. Tsuboi, On Coheir Sequences-Indiscernible Array, Three and Forest-, in New developments of independence notions in model theory (RIMS
Kokyuroku), 2010.

[4] B. Kim, H. Kim and L. Scow, Tree indiscernibilities, revisited, arXiv:
1111. $0915v$ .
[5] K. Takeuchi and A. Tsuboi, On the existence of indiscernible trees,

submitted.