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In ZFC, an ordered pair is defined as $\langle a,b\rangle=\{\{a\},\{a,b\}\}$. Can we define it as $\langle a,b\rangle=\{a, \{a,b\}\}$, does $\langle a,b\rangle=\langle x,y\rangle\Rightarrow a=x\wedge b=y$ hold?

The most similar searched results using Google and MathtSackExchange are defined ordered pair as $\langle a,b\rangle=\{a,\{b\}\}$ or $\langle a,b\rangle=\{\{a,\emptyset\},\{b,\emptyset\}\}$, but their answers don't help my problem.

I went to the professor for further consultation, he tell me it was incorrect to define ordered pair as $\langle a,b\rangle=\{a, \{a,b\}\}$ and asked me to find a counterexample to illustrate.

I have been thinking about it for two days and have no idea. Can you give me some help, for example, suggestions on directions or reference materials. : )

A counterexample: $a=\{b,\phi\}$ and $b=\{a,\phi\}$, thank the answer of @bof, and thank the further explanation of @TonyK and @Hope Duncan : )

Further problem: If we don't need the Axiom of Foundation, is there such a set?

xioacd99
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    ZFC itself does not come with Kuratowski's ordered pairs definition. Where did you get that strange idea from? – user10354138 Oct 25 '23 at 07:37
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    Anyway, your version also works in ZFC (but not necessarily in non-well-founded set theories), because ${a,b}$ has a higher rank than $a$. – user10354138 Oct 25 '23 at 07:50
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    If $a={b,\varnothing}$ and $b={a,\varnothing}$ but $a\ne b$ then ${a,{a,\varnothing}}={a,b}={b,{b,\varnothing}}$ is your counterexample. Of course the Axiom of Foundation says that there are no such sets. The Kuratowski ordered pair doesn't need the Axiom of Foundation to work. – bof Oct 25 '23 at 08:07
  • @bof Great solution, thank you for your counterexample, have a nice day. : ) – xioacd99 Oct 25 '23 at 08:17
  • @bof I can't seem to find such a and b. Could you further explain how to ensure that such a and b exist? Or how to find such a and b? : ) – xioacd99 Oct 25 '23 at 09:12
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    @xioacd99: As bof says in their comment, the Axiom of Foundation says that there are no such sets. So you won't find such $a$ and $b$ in ZFC! – TonyK Oct 25 '23 at 13:57
  • @TonyK Thanks for your further explanation, I wonder if we don't consider the Axiom of Foundation, is there such a set for a and b? – xioacd99 Oct 26 '23 at 00:42
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    @xioacd99 you can’t prove that there is and you can’t prove there isn’t. To prove there is you would need to assume some antifoundation axiom. – spaceisdarkgreen Oct 26 '23 at 03:18

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In ZFC, it perfectly works. Indeed, for all $a,b$, $a \neq \{a,b\}$ because $a \notin a$ (it is a consequence of the axiom of foundation). The set you define indeed has two distinct elements. Then, if $\{a,\{a,b\}\} = \{c,\{c,d\}\}$, there are two possibilities,

$*$ $a = c$ and $\{a,b\} = \{c,d\}$. In this case, we have $\{a,b\} = \{a,d\}$ thus $b \in \{a,d\}$. If $b = a$, then $\{a\} = \{a,d\}$ so $d = a = b$, else, $b = d$. In any case, $a = c$ and $b = d$.

$*$ $a = \{c,d\}$ and $\{a,b\} = c$. Let us show that it is actually impossible. Indeed, in this case, we have $a = \{c,d\} = \{\{a,b\},d\}$. Let $x = \{a,\{a,b\}\}$. $x \cap a = x \cap \{\{a,b\},d\} \neq \emptyset$ because it contains $\{a,b\}$ and $x \cap \{a,b\} \neq \emptyset$ because it contains $a$. It contradicts the axiom of foundation.

This way to define pairs work, I don't know why your teacher told you this, unless you work in a version of set theory with no axiom of foundation.

Cactus
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