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The following is exercise 7.3.1. of Tent and Ziegler's textbook:

Let $T$ be simple, with a monster model $\mathfrak{U}$, and let $\pi(v,w)$ be a collection of $\mathcal{L}_A$-formulas with $b\in\mathfrak{U}$ such that $\pi(v,b)$ is consistent and does not fork over $A$. If $\mathcal{I}$ is an infinite sequence of $A$-indiscernibles containing $b$, then there is a realization $c$ of $\pi(v,b)$ such that $c\perp_A\mathcal{I}$ and such that $\mathcal{I}$ is indiscernible over $A\cup\{c\}$.

Here is my solution: By compactness, it suffices to assume that $\mathcal{I}$ has order type $\omega$, so let $\mathcal{I}=(b_i)_{i\in\omega}$, with $b_0=b$. Now, let $p(v)\in S_n(A\cup\mathcal{I})$ be a non-forking extension of $\pi(v,b)$, with a realization $c$. By construction, we have $\mathfrak{U}\models\pi(c,b)$ and $c\perp_A\mathcal{I}$. We claim that $\mathcal{I}$ is $A\cup\{c\}$-indiscernible. To see this, we want to show $$\varphi(v,b_{i_1},\dots,b_{i_k})\leftrightarrow\varphi(v,b_{j_1},\dots,b_{j_k})\in\text{tp}(c\big/A\cup\mathcal{I})$$ for every $i_1<\dots<i_k\in\omega$ and $j_1<\dots<j_k\in\omega$ and every $\mathcal{L}_A$-formula $\varphi$. To show this, let $n$ be maximal among the $i_l$ and $j_l$. Since $c\perp_A\mathcal{I}$, in particular we have $c\perp_A\{b_0,\dots,b_n\}$. Then $\mathcal{I}$ may be considered as an $A$-indiscernible sequence with first element $(b_0,\dots,b_n)$ in a natural way, and so by corollary 7.1.5. we may find some $c_n\in\mathfrak{U}$ with $c_n\equiv_{A\cup\{b_0,\dots,b_n\}}c$ and such that $\mathcal{I}$ is $A\cup\{c_n\}$-indiscernible. In particular, the desired formula lies in $$\text{tp}(c_n\big/A\cup\{b_0,\dots,b_n\})=\text{tp}(c\big/A\cup\{b_0,\dots,b_n\})\subseteq\text{tp}(c\big/A\cup\mathcal{I}),$$ and so we are done.

I can't see any error in what I've written, but I looked in the back of Tent and Ziegler's book, and they give a hint for a (seemingly) more elaborate solution, which first proves that $\bigcup_{i\in\omega}\pi(v,b_i)$ does not fork over $A$ and then takes a non-forking extension of that collection. Perhaps the ideas are not so different, since the proof of lemma 7.1.5. uses a similar union of pushforward types ranging over an indiscernible sequence, but I'm not sure whether there's a reason Tent and Ziegler didn't just use lemma 7.1.5. in their solution to this exercise. In particular, is there a mistake anywhere in my proof?

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In your proof, you replace the $A$-indiscernible sequence $\mathcal{I}$ with a "thickening": an $A$-indiscernible sequence $\mathcal{I}^n = (b_{(n+1)k},b_{(n+1)k+1},\dots,b_{(n+1)k+n})_{k\in \omega}$. Then its true that you can find $c_n \equiv_{Ab_0,\dots,b_n} c$ such that $\mathcal{I}^n$ is $Ac_n$-indiscernible. But you cannot conclude from this that $\mathcal{I}$ is $Ac_n$-indiscernible.

For example, $b_0$ and $b_1$ may have different types over $Ac_n$. Your construction only ensures that $b_0$, $b_{n+1}$, $b_{2(n+1)}$, etc. have the same type over $Ac_n$.

In general, it does not follow from $\mathfrak{U}\models\pi(c,b)$ and $c\perp_A\mathcal{I}$ that $\mathcal{I}$ is $A\cup\{c\}$-indiscernible. For example, in the theory of the random graph, suppose $\pi(x,y)$ says $xRy$. We can take $c$ to be any element not in $\mathcal{I}$, with $cRb_0$, and we will have $c\perp_A\mathcal{I}$. In particular, $c$ can be connected by an edge or not to the other elements of $\mathcal{I}$. So you have to do something more elaborate to prove this (in particular, you have to use simplicity somewhere, which your solution does not - the statement is not true for arbitrary theories).

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    $\begingroup$ hello! sorry for the bother, but I'm now struggling to complete the proof. I have no issue showing that $\bigcup_{i\in\omega}\pi(v,b_i)$ is consistent, but it's not so clear to me why we can choose a non-forking realization over which $\mathcal{I}$ is still indiscernible. the "obvious" thing to do seems to me to choose a non-forking realization $c$, so that $c\perp_A\mathcal{I}$, and then use the "standard lemma" to get $\mathcal{J}$ a sequence of $A\cup\{c\}$-indiscernibles realizing $\text{EM}(\mathcal{I}\big/A\cup\{c\})$. $\endgroup$ – Atticus Stonestrom Dec 31 '20 at 18:57
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    $\begingroup$ by finding an element of $\text{Aut}(\mathfrak{U}\big/A)$ taking $\mathcal{J}$ to $\mathcal{I}$, we can then find $c'\equiv_A c$ such that $\mathcal{I}$ is $A\cup\{c'\}$-indiscernible, but surely we will in general lose independence this way. $\endgroup$ – Atticus Stonestrom Dec 31 '20 at 18:58
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    $\begingroup$ if, at the beginning of the construction, we replace $\pi(v,b)$ with a non-forking extension in $S(A\cup\{b\})$, then we can ensure $c'\equiv_{A\cup\{b_i\}}c$ for each $i$, since $\pi(c,w)\subseteq\text{EM}(\mathcal{I}\big/A\cup\{c\})$. but I don't see why this allows us to guarantee independence. Tent and Ziegler's hint says it follows from finite character but this isn't totally clear to me... any hint would be appreciated! thank you as always $\endgroup$ – Atticus Stonestrom Dec 31 '20 at 18:58
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    $\begingroup$ @AtticusStonestrom "but surely we will in general lose independence this way" - The neat thing about the standard lemma is that we don't lose independence! It suffices to show $\text{tp}(c/A\mathcal{J})$ does not fork over $A$. If it does, there is some formula $\varphi(x,b'_0,\dots,b'_n)$ in this type which forks over $A$ (where $\mathcal{J} = (b'_i)_{i\in \omega}$, suppressing the parameters from $A$). Since $\models \varphi(c,b'_0,\dots,b'_n)$, also $\models \varphi(c,b_{i_0},\dots,b_{i_n})$ for some $i_0,\dots,i_n\in \omega$, else $\lnot \varphi(c,y_0,\dots,y_n)$ would be in the EM-type. $\endgroup$ – Alex Kruckman Dec 31 '20 at 20:15
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    $\begingroup$ But also $\text{tp}(b_{i_0},\dots,b_{i_n}/A) = \text{tp}(b'_{0},\dots,b'_{n}/A)$ by indiscernibility of $\mathcal{I}$ over $A$, so $\varphi(x,b_{i_0},\dots,b_{i_n})$ forks over $A$, contradicting the fact that $\text{tp}(c/A\mathcal{I})$ does not fork over $A$. $\endgroup$ – Alex Kruckman Dec 31 '20 at 20:18

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