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Suppose $\bar{b}_0, \bar{b}_1, \dots$ is a sequence of elements of a model $\mathcal{M}$ such that for every finite $X,Y\subset \mathbb{N}$ with the same size we have $\bigcup_{i\in X} \bar{b}_i \equiv_A \bigcup_{i\in Y}\bar{b}_i$.

Then, can we conclude that the sequence $\bar{b}_0, \bar{b}_1, \dots$ is an $A$-indiscernible set? Why?

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    The answer to this question depends on what exactly is meant by the notation. Is $\bigcup_{i\in X} \overline{b_i}$ a tuple or a set? If it's a tuple, in what order are the $b_i$ listed (since $X$ is a set, not a tuple)? If it's a set, does $B\equiv_A C$ (with $B$ and $C$ sets) just mean that *some* enumerations of $B$ and $C$ as tuples have the same type over $A$? – 2017-02-05
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    @AlexKruchman Thank you Alex for your answer. About your question I should say $\bigcup_{i\in X} \bar{b}_i $ is a finite set of tuples. Can we conclude that sequence is an indiscernible set? – 2017-02-06
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    That still doesn't answer the question of what $B \equiv_ A C$ means when $B$ and $C$ are sets of tuples. Remember that to write down the type of a set, you need to name the elements of the set by variables, and the answer to your questions depends on how you do this for $B$ and for $C$. – 2017-02-06
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    If you mean "for any labeling of these sets by variables..." then the sequences are indiscernible sets (but to me, this is the least natural reading of the notation $\equiv_A$). If you mean "when we label these sets by variables according to the order of the elements of $X$ and $Y$ in $\mathbb{N}$", then the sequences are indiscernible sequences. If you just mean "there is some way of labeling these sets by variables...", then the sequences aren't necessarily indiscernible at all. – 2017-02-06
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    @Alex, I think it doesn't matter, since we have $\bigcup_{i\in X} \bar{b}_i \equiv_A \bigcup_{i\in Y}\bar{b}_i$ for **every** finite set $X$ and $Y$. Am I right? – 2017-02-06
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    The question doesn't even mean anything until you *define the notation*. – 2017-02-06
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    Maybe I can illustrate my concern with an example. Consider the structure $(\mathbb{Q},<)\models \text{DLO}$. Let's say we have a sequence $1, 5, \frac{1}{2}, 2, \frac{7}{15},\dots$ of distinct elements (no repeats). Let $X = \{0,1,3\}$, and let $Y = \{2, 1, 4\}$. Then $\bigcup_{i\in X} b_i = \{1, 5, 2\}$, and $\bigcup_{i\in Y} b_i = \{\frac{1}{2}, 5, \frac{7}{15}\}$. What does $\{1, 5, 2\}\equiv_\emptyset \{\frac{1}{2}, 5, \frac{7}{15}\}$ mean? Sets don't come with an order, and it's true that $\text{tp}(1,2,5) = \text{tp}(\frac{7}{15},\frac{1}{2},5)$. Is that sufficient? – 2017-02-06
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    On the other hand, maybe you just meant the following condition: for all finite sequences $x_1,\dots,x_k$ and $y_1,\dots,y_k$ from $\mathbb{N}$, $\overline{b_{x_1}},\dots,\overline{b_{x_k}}\equiv_A \overline{b_{y_1}},\dots,\overline{b_{y_k}}$. In this case, the meaning is totally clear, and yes, this is essentially the *definition* of being an $A$-indiscernible set. – 2017-02-06
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    @Alex Thank you for your explanation, I got it. In the problem it seems the index sets which are subsets of natural numbers induce an ordering. – 2017-02-06

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