# The Intersection of Ordered Pairs

I've seen that the ordered pair $(a,b)$ is defined as a set that is $(a,b)=\{\{a\},\{a,b\}\}$. Can you explain what do we mean when $(a,b) \cap (b,a) = \{\{a,b\}\}$? I feel that there should be no intersection whenever a is not equal to b.

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The only property you really care about in this context is that $\langle a, b\rangle = \langle c, d\rangle$ should be equivalent to $a=c, b=d$, and this implementation has this property and is technically convenient. The property you mention is an unintended, but harmless, side effect. – Christian Remling Jul 6 '14 at 4:37

If $(a,b)=\{\{a\},\{a,b\}\}$, then $(b,a)=\{\{b\},\{b,a\}\}$. The intersection of two sets $A$ and $B$ is defined as $A\cap B=\{x:x\in A \text{ and }x\in B\}$. So the intersection of $(a,b)$ and $(b,a)$ based on your definition of these sets would be

$$(a,b)\cap (b,a)=\{\{a,b\}\}$$

This is because the set $\{a,b\}$ is the only element that is in both $(a,b)$ and in $(b,a)$. (Note that $\{a,b\}=\{b,a\}$ i.e order of listing elements in sets does not matter)

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Thanks for catching that! I was just in the process of editing that anyway, it slipped by me the first time. – dleggas Jul 6 '14 at 4:41

Given your definition, which is standard, you have $(b,a)=\{\{b\},\{b,a\}\}$ Since the order of elements of a set doesn't matter, you get the result $(a,b)\cap (b,a)=\{\{a\},\{a,b\}\}\cap\{\{b\},\{b,a\}\}=\{\{a,b\}\}$

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Your answer is missing $\{\}$ at the end. – Asaf Karagila Jul 6 '14 at 4:39
@AsafKaragila: I edited while you were commenting, but I don't think my edit dealt with that. I don't understand the comment. – Ross Millikan Jul 6 '14 at 4:42
You should read the very last equality in your answer more carefully then. – Asaf Karagila Jul 6 '14 at 4:46
@AsafKaragila: OK I see. Check this version. Thanks – Ross Millikan Jul 6 '14 at 4:48
Much better now. – Asaf Karagila Jul 6 '14 at 4:48