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I have a set $\Omega=\{1;2;6\}$ and I want to define another set $A$ consisting of all triples $(a,b,c)$ with $a,b,c\in\Omega$, which contain exactly two 6's.

My first attempt looked like this: $A=\{a^2\times b\vert a=\{6\},b=\{1;2\}\}$

But after looking at the cartesian product's definition, it looks like my set $A$ consists of all triples having two 6's as their first two elements instead of all triples containing two 6's. So here is my second attempt: $A=\{(a^2\times b)\cup(a\times b\times a)\cup(b\times a^2)\vert a=\{6\},b=\{1;2\}\}$

So my questions are, if my last definition of $A$ would be correct (?), if there is a way to shorten the definition of $A$ (?) and if there is something like an "unordered cartesian product"?

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  • $\begingroup$ $\{ f: \{1,2,3\} \to \{1,2,6\} | |f^{-1}(\{6\})| = 2 \}$ $\endgroup$ – copper.hat Jun 4 '15 at 1:02
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There is an "unordered Cartesian product." Given a set $X$, we define $S^k(X)$ to be the quotient of $X\times X\times\cdots \times X$ by the action of the symmetric group on $k$ letters. This is sometimes called the $k$th symmetric power.

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  • $\begingroup$ I nearly didn't find anything about this topic, so I don't know how to rewrite my set with this approach (if it is possible at all) $\endgroup$ – Cubi73 Jun 4 '15 at 17:14
  • $\begingroup$ @Cubinator73 The unordered cartesian product consists of unordered pairs (or more generally $k$-tuples.) (We could use a bracket notation to distinguish unordered from ordered tuples: $[a,b,c]=[b,a,c]=[c,a,b]=...$. Whether you want unordered $3$-tuples is another matter, but you could define your set to simply be $\{[6,6,1],[6,6,2]\}$. $\endgroup$ – Cheerful Parsnip Jun 4 '15 at 18:27
  • $\begingroup$ That's great! Thank you :) $\endgroup$ – Cubi73 Jun 4 '15 at 18:47
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The simplest way that does not involve permutations or functions is to directly express what you want:

  $\{ (a,b,c) : (a,b) = (6,6) \lor (b,c) = (6,6) \lor (c,a) = (6,6) \}$

A slightly shorter alternative based on the fact that $\{6,6\} = \{6\}$ is:

  $\{ (a,b,c) : \{a,b\} = \{6\} \lor \{b,c\} = \{6\} \lor \{c,a\} = \{6\} \}$

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Your idea betrays a minor discomfort with the notation. Literally, $\{(a^2\times b)\cup (a\times b\times a)\cup (b\times a^2)|a=\{6\},b=\{1;2\}\}$ is a set with a single element, namely the set $B=(a^2\times b)\cup (a\times b\times a)\cup (b\times a^2)$, where $a$ is a set with one element, namely $6$, and $b$ is a set with two elements, namely $1$ and $2$. $B$, however, is exactly the set you want.

The notation you choose is more characteristic when $a$ and $b$ are varying elements of a set, rather than fixed sets themselves. For example we could refer to the natural numbers as $\mathbb{N}=\{n| n\geq 0, n\in \mathbb{Z}\}$, where here $n$ actually varies, namely over the integers.

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