# Ordered Pair Formal Definition

I understand what an ordered pair is, but I'm having trouble the formal Kuratowski definition which says that $\langle a,b \rangle = \{\{a\},\{a,b\}\}$. What about this definition imposes order on the pair?

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Can you write out $\langle 1, 2 \rangle$ and $\langle 2, 1\rangle$? – Calvin Lin Jan 25 '13 at 3:30
Maybe I'm misunderstanding your question, but no, it's an ordered pair because it can only be written $\langle a,b \rangle$. – user58437 Jan 25 '13 at 3:31

Notice: $\langle a,b \rangle = \big\{\{a\},\{a,b\}\big\}$, but $\langle b,a \rangle = \big\{\{b\},\{a,b\}\big\}$. The first elements of these sets are different.

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I guess that's part of my question- why would < b,a, > be written {{b}, {b,a}}? Since the latter isn't ordered, couldn't you reorganize it to say {{a,b},{b}}? – user58437 Jan 25 '13 at 3:33
Oh I think I get it. Thanks. – user58437 Jan 25 '13 at 3:39

The Kuratoiwski definition intends to enforce the one basic notion of an ordered pair, that is $$\langle a,b\rangle=\langle c,d\rangle\iff a=c\land b=d.$$ While one direction is trivial, note that \begin{align}&\langle a,b\rangle=\langle c,d\rangle\\ \implies&\{\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}\\ \implies&\{a\}\in\{\{c\},\{c,d\}\}\\ \implies&\{a\}=\{c\}\lor\{a\}=\{c,d\}\\ \implies&a=c\lor a=c=d\\ \implies&a=c\\ \end{align} and then \begin{align}&\langle a,b\rangle=\langle a,d\rangle\\ \implies&\{\{a\},\{a,b\}\}=\{\{a\},\{a,d\}\}\\ \implies&\{a,b\}\in\{\{a\},\{a,d\}\}\\ \implies&\{a,b\}=\{a\}\lor \{a,b\}=\{a,d\}\\ \implies& b\in\{a\}\lor b\in\{a,d\}\\ \implies & b=a\lor b=d \end{align} and by symmetry also $d=a\lor d=b$. Combined, this yields $(b=a\land d=a)\lor b=d$, i.e. $b=d$. In summary, $$\langle a,b\rangle=\langle c,d\rangle\implies a=c\land b=d.$$

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Rather than thinking about this formally, from which little intuition can be gained, it may be better to see that an ordered pair is coding an order, by listing its initial segments: $\emptyset,\{a\},\{a,b\}$ are the initial segments of the order on the set $\{a,b\}$ where $a$ comes first. Clearly, we do not need to list $\emptyset$, and this list obviously tells us what the elements are, and which one comes first (and the case $a=b$ is coded here as well). – Andrés Caicedo Jan 27 '13 at 23:49

It implies $\langle b,a\rangle=\{{\{{b\}},\{{a,b\}}\}}$, so $\langle a,b\rangle\neq\langle b,a\rangle$.

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What if $a=b$? :-) – Asaf Karagila Jan 25 '13 at 10:00
There's one in every crowd. – Gerry Myerson Jan 25 '13 at 11:35
Well, it's a math crowd. :-) – Asaf Karagila Jan 25 '13 at 11:52
@Asaf: From Gerry’s point of view it may be a maths crowd. :-) – Brian M. Scott Jan 25 '13 at 18:21
@Brian, I live in Oz, but my native tongue is American. – Gerry Myerson Jan 25 '13 at 23:07