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The formal Kuratowski definition of ordered pair is that $\langle a,b\rangle = \{\{a\},\{a,b\}\}$.

While I think I understand the above definition well I wanted to check if below definition also works just fine (and hence is "equivalent" to Kuratowski definition)

$$\langle a,b\rangle = \{a,\{b\}\}.$$

I think that both the definitions are just fine, but maybe I'm missing a subtle point. Also is there any reason to prefer Kuratowski's definition over the later one?

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Unfortunately, with the new definition, both $\langle\{0\},1\rangle$ and $\langle\{1\},0\rangle$ equal $\{\{0\},\{1\}\}$. Thus this definition is not suitable for ordered pairs.

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  • $\begingroup$ yes indeed many thanks $\endgroup$
    – aman_cc
    Jun 28 '20 at 5:03
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With Kuratowski's definition, $\langle a,b\rangle=\langle c,d\rangle$ if and only if $a=c$ and $b=d$, as we'd hope. However, in the proposed approach $\langle a,b\rangle=\{a,\{b\}\}$, observe that $\langle \{1\},2\rangle = \langle \{2\},1\rangle$, so we don't have uniqueness.

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The ordered pair $(a,b)$ is defined in such a way that it satisfies the following:
$(a,b)=(c,d)\Longleftrightarrow a=c$ and $b=d$.
So once your definition of $(a,b)$ satisfies above, it is absolutely fine. But your definition does not satisfy above, so it is not valid. Otherwise you can have any of the following as your definition:
$(a,b)=\{a,\{a,b\}\}$ or, $\{\{b\},\{a,b\}\}$ or, $\{b,\{a,b\}\}$

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  • $\begingroup$ I think {{b},{a,b}} should work just fine $\endgroup$
    – aman_cc
    Jun 28 '20 at 5:17

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