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I having problem how to interpret the definition of the cartesian product.

The cartesian product of two sets $A$ and $B$ is the set of all ordered pairs $(a,b)$, where $a\in A$ and $b\in B$. \begin{align} A\times B=\{(a,b): a\in A \text{ and } b\in B\} \end{align}

I understand if $A=\{2\}$ and $B=\{5\}$, we have $A\times B=\{(2,5)\}$. No problem.

But suppose $A=\{3,4\}$ and $B=\{7,8\}$, I interpret the definition as $$A\times B=\{(3,7),(4,8)\}$$ This is wrong, but what have I misunderstood in the definition? How should I analyze the definition word by word?

I know the correct is $A\times B=\{(3,7),(3,8), (4,7), (4,8)\}$. However, I dont' grasp why this is correct based on the definition.

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  • $\begingroup$ you can see it like that: $A\times B = \cup_{a\in A} \{(a,b)\mid b \in B\}$ $\endgroup$ – Surb Aug 7 '17 at 10:03
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    $\begingroup$ Realise that sets have no order: $\{ a, b\} = \{b,a\}$, so this is the only thing the definition could mean. $\endgroup$ – SvanN Aug 7 '17 at 10:04
  • $\begingroup$ There is a possibility of misunderstanding as you do in the finite case such as this one, or in the case where $A,B$ have the same number elements, but how (with your understanding of the definition) would you deal with $\mathbb{R}\times \{0,1\}$, or even more simply $\{0,1\}\times \{2,3,4\}$ ? $\endgroup$ – Max Aug 7 '17 at 10:22
  • $\begingroup$ it's correct based on the definition because it has all the ordered pairs it can have. with one coordinate coming from one set, and the other from the other set. $\endgroup$ – user451844 Aug 7 '17 at 10:29
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here's an example of a Cartesian product for you that's more chess related:

$$\begin {array}{c|c|c|c|c|c|c|c|c|}\hline 8 &(a,8)&(b,8)&(c,8)&(d,8)&(e,8)&(f,8)&(g,8)&(h,8)\\ \hline 7 &(a,7)&(b,7)&(c,7)&(d,7)&(e,7)&(f,7)&(g,7)&(h,7)\\\hline 6 &(a,6)&(b,6)&(c,6)&(d,6)&(e,6)&(f,6)&(g,6)&(h,6)\\\hline 5 &(a,5)&(b,5)&(c,5)&(d,5)&(e,5)&(f,5)&(g,5)&(h,5)\\\hline 4 &(a,4)&(b,4)&(c,4)&(d,4)&(e,4)&(f,4)&(g,4)&(h,4)\\\hline 3 &(a,3)&(b,3)&(c,3)&(d,3)&(e,3)&(f,3)&(g,3)&(h,3)\\\hline 2 &(a,2)&(b,2)&(c,2)&(d,2)&(e,2)&(f,2)&(g,2)&(h,2)\\\hline 1 &(a,1)&(b,1)&(c,1)&(d,1)&(e,1)&(f,1)&(g,1)&(h,1)\\\hline &a&b&c&d&e&f&g&h \end {array}$$

you can now plot relations between the sets on it by coloring it in etc.

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The elements of $A\times B$ are all the pairs that have the first element (of the pair) from $A$ and the second element (also of the pair) from $B$.

A typical example: imagine that you have two T-shirts: red and blue, and three trousers: white, green and black. You can dress in six ways: $$\{\text{red, blue}\}\times\{\text{white, green, black}\}=\\ \{\text{(red, white), (red, green), (red, black), (blue,white), (blue, green), (blue, black)}\}$$

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I'm not sure if it helps, but it seems to me that this is where you misunderstand the thing:

We have $$A\times B=\{(a,b) : a \in A, b \in B \} = \bigcup_{a\in A} \{(a,b): b \in B\}$$ For every element in $A$, we can "attach" any element in $B$ and vice versa.

If $A=\{3,4\}$ and $B =\{7,8\}$, then we have $$ A\times B = \{(3,7),(3,8)\} \cup \{(4,7),(4,8)\}=\{(3,7),(3,8),(4,7),(4,8)\}.$$ Now, it is also true that $$A\times B=\bigcup_{b\in B} \{(a,b): a \in A\}$$ and we get $$ A\times B = \{(3,7),(4,7)\} \cup \{(3,8),(4,8)\}=\{(3,7),(3,8),(4,7),(4,8)\}.$$

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$A=\{3,4\}$

$B=\{7,8\}$

You want to know why $A \times B = \{(3,7),(3,8),(4,7),(4,8)\}$

Look at each element, say $(4,7)$ for example. Is $4 \in A$ and is $7 \in B$?. Yes! Hence $(4,7) \in A \times B$. The same goes for the other three ordered pairs.

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