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For any sets $x$, $y$ we write $\langle x,y \rangle$ for the set $\{\{x\},\{x,y\}\}$ Show that this set is unique.

My proof:

By axiom ZF3 of pairs there exists a set (I call it $z:=\langle x,y \rangle$) whose elements are $x$ and $y$. If now $z^{\prime}$ is another such set, then any element of $z^{\prime}$ $x$ or $y$ is also an element of $z$ and vice versa and by the ZF1 (extensionality) axiom we must have that $z=z^{\prime}$.

Am I correct?

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In your construction, what makes your set ordered? In particular, what is the difference between $\langle x,y \rangle$ and $\langle y,x \rangle$? – A Walker May 14 '13 at 17:16
$<x,y>=\{\{x\},\{x,y\}\}$ but $<y,x>=\{\{y\},\{y,x\}\}$ I guess – H.E May 14 '13 at 17:26
What is meant by unique in this context? Since this is the usual set-theoretic definition of an ordered pair it is more common to prove that the following holds: If $\langle x , y \rangle = \langle u , v \rangle$, then $x = u$ and $y = v$. – arjafi May 14 '13 at 19:27

Formally one can write $$\forall z(z\in<x,y>\leftrightarrow (z=\{x\}\lor z=\{x,y\}))$$ (and also the statements $z=\{x\}$ and $z=\{x,y\}$ can be properly expressed as sentences in the language of ZF). Now, if $w$ is another set such that $$\forall z(z\in w\leftrightarrow (z=\{x\}\lor z=\{x,y\})),$$ then obviously the two sets ($w$ and $<x, y>$) are equal by the axiom of extensionality. (I assumed so far you already know (how to prove) that there exists a set whose elements are exactly $\{x\}$ and $\{x,y\}$, given sets $x$ and $y$. If not, it is easy to show that this existence is ensured by the axiom of pair and the separation schema).

I hope this answers your question (I'm not pretty sure I actually understood what you were asking for).

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the qestion was to show that the set defined above is unique – H.E May 14 '13 at 18:00
Then my answer should work... – Marco Vergura May 14 '13 at 19:18

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