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Let $A$ be an uncountable set with arbitrary element $a\in A$. I want to define the Cartesian product of all its elements. In other words, all the vectors belonging to

\begin{gather} \underbrace{A\times\dots\times A}_{A} \end{gather}

Suppose $A=\{a_1,a_2\}$. Then, I am looking for the notation that generalises the following expression to the case in which the set $A$ is uncountable:

\begin{gather} A\times A\equiv\{(a_1,a_1),(a_1,a_2),(a_2,a_1),(a_2,a_2)\} \end{gather}

I have thought of two possibilities, but both seem unsatisfactory:

  1. To define $\prod_{a\in A}i(a)$, where $i:A\to A$ is the identity function (i.e., $i(a)=a$ for all $a\in A$);
  2. To simply define $\prod_{a\in A}a$.

Any help will be much appreciated.

Thank you all.

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    $\begingroup$ Your example with two elements doesn't clarify it for me. Let's say $A$ has three elements. Do you want a set with all 9 possible ordered pairs, or a set will all possible 27 ordered triples? If your set were $\mathbb{N}$, do you want the set with all possible sequences of natural numbers? $\endgroup$ Commented Nov 4, 2021 at 17:57
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    $\begingroup$ Then what you want is $\prod_{a\in A} A$, or equivalently $A^A$, the set of all functions from $A$ to $A$. $\endgroup$ Commented Nov 4, 2021 at 18:06
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    $\begingroup$ $\prod_{a\in A}a$ would given tuples in which the $a$th entry is an *element* of $a$, not an element of $A$. By definition, $\prod_{i\in I} S_i$ is the set of all functions $f\colon I\to \cup S_i$ such that $f(i)\in S_i$. If $A=\{ \varnothing, \{1\}\}$, then $\prod_{a\in A}A$ is $\{(\varnothing,\varnothing), (\varnothing,\{1\}), (\{1\},\varnothing),(\{1\},\{1\})$, whereas $\prod_{a\in A}a$ is $\varnothing\times \{1\}$, which is empty. $\endgroup$ Commented Nov 4, 2021 at 18:10
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    $\begingroup$ It is true that $\prod_{a\in A}a = a_1\times a_2$. It is false that $a_1\times a_2 = (a_1,a_2)$. After all, $a_1\times a_2 = \{ (x,y)\mid x\in a_1, y\in a_2\}$. $\endgroup$ Commented Nov 4, 2021 at 20:48
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    $\begingroup$ First, in standard set theory, there are no sets that satisfy $x=\{x\}$. If you are working in a non-standard set theory where that is possible, then $a_1\times a_2 = \{ (a_1,a_2)\}$ the set whose only element is the ordered pair $(a_1,a_2)$. Second, depending on your definition of the ordered pair, it could be that $\{(a_1,a_2)\} = (a_1,a_2)$, but most of the time they won't be equal. I would say you do not want to live in a world where $x=\{x\}$ is even possible right now, at least until you actually want to start thinking about non-standard set theories. $\endgroup$ Commented Nov 4, 2021 at 20:53

2 Answers 2

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I’m not exactly sure what you want. From your text I suppose you want $$ \prod_{a\in A} a$$ from your formulae I get you want $$ \prod_{A}A = A^A $$ The latter can be understood as all functions from $A$ to $A$. The first one can be understood as all maps $f$ from $A$ to $\bigcup A$ so that for $a\in A$ we have $f(a)\in a$.

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  • $\begingroup$ Thank you for your answer. Please, help me fully understand it. Suppose that $A\equiv\{a_1,a_2\}$. Then, does $\prod_{a\in A}a=\{(a_1,a_1),(a_1,a_2),(a_2,a_1),(a_2,a_2)\}$ or does $\prod_{A}A= \{(a_1,a_1),(a_1,a_2),(a_2,a_1),(a_2,a_2)\} $? Or neither / both? $\endgroup$ Commented Nov 4, 2021 at 17:42
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    $\begingroup$ $$\prod_A A$$ could be understood as that set, by bijectively identifiying $a_1,a_2$ with $1,2$. $$\prod_{a\in A} A$$ would be $a_1\times a_2$. $\endgroup$ Commented Nov 4, 2021 at 17:52
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    $\begingroup$ Sorry, the second one should have been $\prod_{a\in A} a$. $\endgroup$ Commented Nov 4, 2021 at 17:58
  • $\begingroup$ Thank you for your clarifying comments. Can I write $\prod_{A}A$ as $\prod_{a\in A}A$ instead? In other words, is it true that $\prod_{A}A\equiv\prod_{a\in A}A$? $\endgroup$ Commented Nov 4, 2021 at 19:18
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    $\begingroup$ Yes, it is. I wrote $\prod_A$ as shorthand for $\prod_{a\in A}$, as we do not depend on $a$. As I said, we can also simply write $A^A$, the space of functions from $A$ to $A$. $\endgroup$ Commented Nov 4, 2021 at 20:37
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Since I'm not able to comment yet, I'll post as an answer. It is $\prod_{A}A$ - and not $\prod_{a\in A}a$ - the set you're pointing out. Cartesian products are defined between sets, not elements of a set. Note that in your example $A$ is in fact finite. When $A$ is uncountable, all the "$A$-indexed vectors", with coordinates in $A$, would be denoted by $$\prod_{A}A$$

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    $\begingroup$ Note in set theory the elements are also sets. So the second one does also make sense. Also $\prod_A A\neq \emptyset$ is not equivalent to the axiom of choice. $\endgroup$ Commented Nov 4, 2021 at 17:56
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    $\begingroup$ That $\prod_A A$ is empty if and only if $A$ is empty is not equivalent to the Axiom of Choice; it is a theorem of ZF. $\endgroup$ Commented Nov 4, 2021 at 18:04
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    $\begingroup$ Elaborating on the other comments: consider the function $f(a) = a$. Then $f \in \prod\limits_A A$. No choice needed. $\endgroup$ Commented Nov 4, 2021 at 18:06
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    $\begingroup$ Thanks for the correction, how can I edit my answer? $\endgroup$ Commented Nov 4, 2021 at 18:12
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    $\begingroup$ You click on the "Edit" button and make the changes. $\endgroup$ Commented Nov 4, 2021 at 18:15

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