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 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?

share|improve this question
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

3 Answers 3

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, $$\langle a,b\rangle=\langle c,d\rangle\implies a=c\land b=d.$$

share|improve this answer
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). –  Andres Caicedo Jan 27 '13 at 23:49

notice :< a,b > = {{a},{a,b}} but < b,a > = {{b},{a,b}} the first element of these sets are different

share|improve this answer
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

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

share|improve this answer
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

Your Answer


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.