# Indiscernible sequences in countable complete theory

Given complete $$T$$ in countable language, prove for any $$\kappa >\omega$$, $$\exists \mathcal{M}\models T$$ such that $$\forall A\subseteq\mathcal{M}, \mathcal{M}$$ realizes at most $$|A|+\omega$$ types n $$S^\mathcal{M}_n(A)$$.

Proof goes like this(in Lecture note I'm using). Take $$\mathcal{M}$$ indiscernible hull of indiscernible $$I$$ cardinality of $$\kappa$$. Here $$\mathcal{M}$$ is model of skolemization of $$T$$ in corresponding language $$\mathcal{L^*}$$. Given $$A\subseteq\mathcal{M}$$, $$a\in A$$ can be written as $$a=t^\mathcal{M}(c_a)$$ where $$c_a \in I$$. take $$J=\cup_{a\in A} c_a$$. then we can show |types over $$A$$| $$\leq$$|types over J| in $$\mathcal{M}$$. So it is enough to show latter one$$\leq |A|+\omega$$. Now define equivalence relation as follows:

$$\forall_{n<\omega}\forall\bar{c},\bar{d}\in[I]^n \bar{c} \sim\bar{d}\Leftrightarrow \forall_{1\leq k\leq n}\forall e\in J, (c_k

where $$\bar{c}=(c_1,...,c_n),\bar{d}=(d_1,...,d_n)$$ increasing sequence in $$[I]^n$$. I'm proving for any term $$t(\bar{x})\in\mathcal{L^*}$$ having n-variables, $$\bar{c}\sim\bar{d}$$ implies $$tp_\mathcal{M}(t^M(\bar{c})/J)=tp_\mathcal{M}(t^M(\bar{d})/J)$$. In my note, it is written that "$$tp_\mathcal{M}(\bar{c}/J)=tp_\mathcal{M}(\bar{d}/J)$$ implies equality above". but I could prove the former one but I don't know how to get the latter one from first one..

This is an almost immediate consequence of indiscernibility. The intuitive meaning of $$\overline{c}\sim\overline{d}$$ is that $$\overline{c}$$ and $$\overline{d}$$ "come in the same order" in $$I$$ relative to the set $$J$$. In particular, if $$\overline{e}$$ is a finite tuple from $$J$$, then $$\overline{c}\overline{e}$$ and $$\overline{d}\overline{e}$$ are both finite tuples from $$I$$ which "come in the same order", so they satisfy the same formulas by indiscernibility.

More precisely:

Suppose $$\varphi(x,\overline{e})\in \text{tp}_\mathcal{M}(t^M(\overline{c})/J)$$ and $$\overline{c}\sim \overline{d}$$. Consider the formula $$\psi(\overline{z},\overline{w})$$ given by $$\varphi(t(\overline{z}),\overline{w})$$.

We have $$\mathcal{M}\models \varphi(t^M(\overline{c}),\overline{e})$$, so $$\mathcal{M}\models \psi(\overline{c},\overline{e})$$. And by indiscernibility, since $$\overline{c}\sim\overline{d}$$, we also have $$\mathcal{M}\models \psi(\overline{d},\overline{e})$$. So $$\mathcal{M}\models \varphi(t(\overline{d}),\overline{e})$$, and $$\varphi(x,\overline{e})\in \text{tp}_\mathcal{M}(t^M(\overline{d})/J)$$. Thus $$\text{tp}_\mathcal{M}(t^M(\overline{c})/J) = \text{tp}_\mathcal{M}(t^M(\overline{d})/J)$$.

• I don't know why $\mathcal{M}\models \varphi(t^{\mathcal{M}}(\bar{c}),e_1,...,e_k)$ and $\bar{c}\sim\bar{d}$ implies $\mathcal{M}\models \varphi(t^{\mathcal{M}}(\bar{d}),e_1,...,e_k)$. it seems that $\bar{c}\sim\bar{d}$ implies $\bar{t(\bar{c})}\sim\bar{t(\bar{d})}$ but I don't know how to prove this. I want to know how $\sim$ affects the argument – fbg Mar 22 at 7:34
• Do you know what "indiscernible" means? @fbg – Alex Kruckman Mar 22 at 13:06
• @fbg I've added a bit more detail to my answer. Let me know if that helps. – Alex Kruckman Mar 22 at 13:58
• Also, your assertion $\overline{c}\sim \overline{d}$ implies $t(\overline{c})\sim t(\overline{d})$ doesn't make any sense, since $\sim$ is only defined on tuples from $I$, and $t(\overline{c})$ might not be an element of $I$. – Alex Kruckman Mar 23 at 5:37
• I couldn't thought about letting $\varphi(t^{\mathcal{M}}(\bar{z}),\bar{w})$ as $\psi(\bar{z},\bar{w})$. Now I see how it works. Thank you – fbg Mar 25 at 4:17