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I'd like to know if there is already a symbol (notation) for grouping similar elements in a set. For example, suppose $$ S=\{\underbrace{a_1, a_2, \cdots}_{G_1}, \underbrace{b_1,b_2, \cdots}_{G_2},\cdots\}=\{G_1, G_2\} $$ I want to use $G_1$ to represent $a_1, a_2, \cdots$ and $G_2$ to represent $b_1, b_2, \cdots$. So basically $G_1$ is the collection of $a_i$ but without curly bracket. Is there a symbol or notation for such operation?

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    $\begingroup$ I am a set theorist and am telling you it is not, and it doesn't. Please don't add the tag again. $\endgroup$ Commented Jan 13, 2018 at 2:42
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    $\begingroup$ Let me be more explicit. There is no "collection without the curly brackets". Such a thing does not exist as an object. It is not interesting and people do not use it. It may be that notationally sometimes you want to write just, say, $\mathbf a $ rather than $ a_1,a_2,\dots,a_n $. That's fine. It is stuff people do all the time, and there is no standard convention. But $\mathbf a $, or whatever, is not an object. It is just notation. If you want an actual object, you either look at sets, or tuples, or some such. But that is a different thing. $\endgroup$ Commented Jan 13, 2018 at 2:47
  • $\begingroup$ Any notation used for partitions or equivalence classes could be used. $\endgroup$ Commented Jan 13, 2018 at 3:00
  • $\begingroup$ I think @AndrésE.Caicedo has the right idea, you could simply just label such elements as $\textbf{a}$. No reason for a completely new notation. $\endgroup$ Commented Jan 13, 2018 at 3:09
  • $\begingroup$ Ok, if no such symbol exists, we can always invent one as long as there are needs for this operation. $\endgroup$ Commented Jan 13, 2018 at 6:00

2 Answers 2

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Hmmm. What about the union symbol?

Let $G_1 = \{a_1, a_2, \ldots \}$, $G_2 = \{b_1, b_2, \ldots \}$ and $G = G_1 \cup G_2$.

If you have more than two sets, you could use double indexing:

For each $i \in I$, let $G_i = \{a_{i,j} \mid j \in J_i\}$ and let $G = \bigcup_{i \in I} G_i$.

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    $\begingroup$ No, the point is that the $G_i $ are not sets. $\endgroup$ Commented Jan 13, 2018 at 16:46
  • $\begingroup$ @andrés-e-caicedo This is not clear at all from the statement of the question, especially since the OP states that $S$ is a set. $\endgroup$
    – J.-E. Pin
    Commented Jan 13, 2018 at 17:01
  • $\begingroup$ $S=\{G_1,G_2\} $ is stated explicitly in the question. $\endgroup$ Commented Jan 13, 2018 at 17:14
  • $\begingroup$ Close, what I need is $\{G_1\} = \{a_1, a_2, \ldots \}$. $\endgroup$ Commented Jan 13, 2018 at 17:18
  • $\begingroup$ @andrés-e-caicedo Sorry, I was totally confused. When you wrote that the $G_i$ are not sets, I thought you were referring to some set theoretic argument, like Russell's paradox. $\endgroup$
    – J.-E. Pin
    Commented Jan 13, 2018 at 19:00
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In programming this in known as a spread operator, unpacking or splat operator. You could try something like this:

$$G = \{...G_1,\space ...G_2\}$$ $$G = \{*G_1, *G_2\}$$ $$G = \{{}^*G_1, {}^*G_2\}$$

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