Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
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 at 4:37

2 Answers 2

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)

share|improve this answer
    
Thanks for catching that! I was just in the process of editing that anyway, it slipped by me the first time. –  dleggas Jul 6 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 $(a,b)\cap (b,a)=\{\{a\},\{a,b\}\}\cap\{\{b\},\{b,a\}\}=\{\{a,b\}\}$

share|improve this answer
    
Your answer is missing $\{\}$ at the end. –  Asaf Karagila Jul 6 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 at 4:42
    
You should read the very last equality in your answer more carefully then. –  Asaf Karagila Jul 6 at 4:46
    
@AsafKaragila: OK I see. Check this version. Thanks –  Ross Millikan Jul 6 at 4:48
    
Much better now. –  Asaf Karagila Jul 6 at 4:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.