If $\cal{I}$ is a indiscernible sequence over $A$ then it is indiscernible over $acl(A)$

Question

Let $\cal{I}=(b_i\mid i\in I)$ be an infinite indiscernible sequence over $A$. And let $acl(A)$ be the algebraic closure of $A$. (all in some structure) I am trying to show that $\cal{I}$ is also indiscernible over $acl(A)$.

What i tried so far was to assume towards contradiction that it is not, and by that to conclude that some of the indiscernibles are in the algebraic closure, and thus by the algebraic formula who witnessed that one of them is in $acl(A)$ (that formula is with parameters from $A$.) we can conclude that all of the sequence satisfy that formula. since it is indiscernible over $A$. But we assumed that the sequence is infinite, and that's our contradiction.

The problem is that I'm not sure how to find an algebraic member over $A$ in $\cal I$.

Answer 1

The answer by user27454 is a good "hands-on" way of seeing this. But I'd like to give a more abstract proof, which shows that your proposition is an immediate consequence of a very useful general lemma.

We work in a monster model $\mathcal{U}$. As usual, small means small enough to apply saturation and strong homogeneity of $\mathcal{U}$.

Lemma: Let $A$ be a small set and $I$ an $A$-indiscernible sequence. Then there is a small model $M$ containing $A$ such that $I$ is $M$-indiscernible.

Given the lemma, we just need to note that every model containing $A$ contains $\text{acl}(A)$, and hence $I$ is also $\text{acl}(A)$-indiscernible.

Proof: Let $M'$ be any small model containing $A$. Using Ramsey's theorem and compactness, we build a sequence $I'$ which is $M'$-indiscernible and "based on $I$", in the sense that whenever $\mathcal{U}\models \varphi(\overline{b},\overline{m})$, where $\overline{b}$ is an increasing tuple from $I'$ and $\overline{m}\in M$, there exists $\overline{a}$ an increasing tuple from $I$ such that $\mathcal{U}\models \varphi(\overline{a},\overline{m})$ (this is sometimes called the "Standard Lemma" for indiscernibles). Then since $I$ is actually $A$-indiscernible, we have $\text{tp}(I'/A) = \text{tp}(I/A)$. Let $\sigma\in \text{Aut}(\mathcal{U}/A)$ such that $\sigma(I') = I$, and let $M = \sigma(M')$. Then $M$ is a model containing $A$ and $I$ is $M$-indiscernible.

Answer 2

Suppose $I=\langle a_i:i\in \mathbb{N}\rangle$ is an infinite indiscernible sequence over $A$, and suppose that for some formula $\phi(x,y)$ and $b\in acl(A)$ we have $\models \phi(a_i,b)\wedge \neg\phi(a_j,b)$ for some $i<j$. Let $\psi(y)$ be an algebraic formula over $A$ with $m$ realizations such that $\models\psi(b)$.

Then, $\tau(x_1,x_2)=\exists y(\psi(y)\wedge \phi(x_1,y)\wedge \phi(x_2,y))\in tp(a_i,a_j/A)$ for all $i<j$.

Since $\exists y(\psi(y)\wedge \phi(x,y))\in tp(a_i/A)$, we have by indiscernibility that for every $i<\omega$ there is $b\models \psi(y)$ such that $\phi(a_i,b)$. By piggeonhole principle, there is some $b_1\models \psi$ such that the set $I_1:=\{i<\omega: \phi(a_i,b)\}$ is infinite.

Note that $I_1$ is also an indiscernible sequence. Now, given $a_i,a_j\in I_1$ with $i<j$, $\tau(a_i,a_j)$ implies there is $b\models \psi(y)$ such that $\phi(a_i,b)\wedge \neg\phi(a_j,b)$. It is clear that $b\neq b_1$ by the construction of $I_1$. Again by piggeonhole principle, there is $b_2\neq b_1$ such that the set $$I_2=\{a_i\in I_1: \phi(a_i,b_2)\}=\{a_i\in I:\phi(a_i,b_1)\wedge \phi(a_i,b_2)\}$$ is infinite.

Repeating this argument $m-1$ times, we obtain $b_1,\ldots,b_{m-1}$ realizing $\psi(y)$ and an infinite indiscernible sequence $$I_{m-1}=\{a_i\in I: \models \phi(a_i,b_1)\wedge \cdots \wedge \phi(a_i,b_{m-1})\}.$$

If $a_i,a_j,a_k\in I_{m-1}$ with $i<j<k$, then we have:

1. We have $\models \tau(a_i,a_j)$, and then $\phi(a_i,b_m)\wedge \neg \phi(a_j,b_m)$ for the only element $b_m$ in $\psi(M)\setminus \{b_1,\ldots,b_{m-1}\}$.
2. $\models \tau(a_j,a_k)$, which implies that there is some $b\models \psi(y)$ such that $\phi(a_j,b)\wedge \neg\phi(a_k,b)$. But this is a contradiction because $b\neq b_m$ by (1), and $b\neq b_1,\ldots,b_{m-1}$ since $a_k\in I_{m-1}$.