Take a set $\mathcal{A}\equiv \{a_1,a_2,a_3\}$ of real numbers.
Is there any specific notation in math for the "operator" that transforms $\mathcal{A}$ into the 3-tuple (ordered) $$ (a_1,a_2,a_3) $$ ?
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1$\begingroup$ The operation isn't clear. if the original set is unordered we have $\{a_1,a_2,a_3\}=\{a_3,a_1,a_2\}$ so would the image of your operation be $(a_1,a_2,a_3)$ or $(a_3,a_1,a_2)$? $\endgroup$– luluJan 22, 2019 at 18:20
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$\begingroup$ @lulu Thanks. What I want to formally say is the following: I start my discussion with a set $\mathcal{A}\equiv \{a_1,a_2,a_3\}$. Then I want to tell the reader to fix any ordering of $\mathcal{A}$ and perform some operations with the resulting tuple. $\endgroup$– TEXJan 22, 2019 at 18:28
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$\begingroup$ How can do that formally? $\endgroup$– TEXJan 22, 2019 at 18:28
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2$\begingroup$ I think the way you said it is clear. With finite sets there is no problem with choosing. I think it's always worth making it clear that a choice is involved, even if you are indifferent to the choice being made. $\endgroup$– luluJan 22, 2019 at 18:30
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2$\begingroup$ math.stackexchange.com/questions/851667/… $\endgroup$– EdOverflowJan 22, 2019 at 19:10
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1 Answer
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I'm assuming for this function you want to find, you'd sometimes want a sequence with repeated elements. If this is the case, then the answer to you question is no.
Example:
If there was such a function that did this, say, $f$, then since $\{5\} = \{5,5\}$, you'll have
$f(\{5\}) = (5)$
and
$f(\{5,5\}) = (5,5)$
which doesn't make $f$ a function.
