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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?

MJD
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user58437
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    Can you write out $\langle 1, 2 \rangle$ and $\langle 2, 1\rangle$? – Calvin Lin Jan 25 '13 at 03:30
  • 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 03:31
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3 Answers3

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

Hagen von Eitzen
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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 E. Caicedo Jan 27 '13 at 23:49
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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.

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

Gerry Myerson
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