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I understand the basics of Cartesian products, but I'm not sure how to handle a set inside of a set like $C = \{\{1,2\}\}$. Do I simply include the set as an element, or do I break it down?

If I use it as an element I think it would be something like this:

$$\{(1,\{1,2\}), (2,\{1,2\})\}$$

If I were to break $C = \{\{1,2\}\}$ further, I'm not sure how I would implement that, so I'm guessing what I did above is correct, but I want to make sure.

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    $\begingroup$ This is correct. The cartesian product $B\times C $ contains all the pairs $(b,c)$ where $b\in B$ and $c\in C $. In this case $B $ has two elements, namely $1$ and $2$, and $C $ has one element, namely $\{1,2\}$. The cartesian product should have $|B|\times |C|$ elements, which in this case is two. $\endgroup$
    – Uncountable
    Feb 1, 2016 at 18:01
  • $\begingroup$ Your as-an-element approach is the correct one, since $B\times C := \left\{ \left(b,\,c\right) | b\in B,\,c\in C \right\}$. If you wanted to replace the forms-the-second-coordinate role of $\left\{ 1,\, 2 \right\}$ with $1$ and $2$, the set of second coordinates you'd be taking would be $\bigcup C = B$ and not $C$. Thus that procedure would construct the Cartesian product $B\times \bigcup C = B\times B$. $\endgroup$
    – J.G.
    Feb 1, 2016 at 18:04
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    $\begingroup$ You don't break it down. If you think about it, what sort of sense would that make? A Cartesian product is a set of ordered pairs so why should we break down the pair just because an element is a set? Is there any more reason to do that then to break down an integer if it isn't a prime( {2,3} x {2,4} =(2,2)(2,4)(3,2)(3,4) but we don't say (3,4) = (3,2,2)(3,1,4) so...) $\endgroup$
    – fleablood
    Feb 1, 2016 at 18:13

2 Answers 2

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That's right. Just proceed as you normally would. $B$ has two elements and $C$ has one element. Pair them up (it may help to write "$A$" in place of "$\{1,2\}$" temporarily):

$$A\times B = \{(1,A), (2,A)\} = \{(1,\{1,2\}), (2,\{1,2\})\}$$

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You are correct. Even if a set has sets as elements, you still treat "each element" separately.

mathematica output

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