I have to prove that there is no set which contains all ordered pairs $\langle a,b\rangle$. in set theory, $\langle a,b\rangle$ is defined as $\{\{a\},\{a,b\}\}$.
My proof:
Say $S$ is the set with all ordered pairs. Then $S=A\times B$, where $A$ contains all the first coordinates and $B$ contains all the second coordinates. However, there has to be a set that is not contained within $A$, as there can be no set of all sets. Let that set be $r$. Then $\langle r,c\rangle\notin S$, where $c$ is any set of your choice.
Is the proof correct?
Best Answer
No. Take $R_1=\{(1,a),(2,b)\}$.Then $R_1$ cannot be written as $A \times B$. So a set of ordered pair may not be written as cartesian product.
Below I give two proofs independent of perticular definition, say Kuratowski or Weiner or Hausdroff or short def etc.\
Proof1: Let $A$ be a set.Let us construct $B=\{(x,y)$$\in$ $A$$\mid$ $(x,y)$$\notin x$$\}$. We claim $(B,1)\notin A$.\ If possible let $(B,1)\in A$. Then by construction of $B$, $(B,1)\in B$ iff $(B,1)\in A$ and $(B,1)\notin B$. Since we assume $(B,1)\in A$, $(B,1)\in B$ iff $(B,1)\notin B$. So our assumption $(B,1)\in A$ is false. Since $A$ is arbitary, the proof follows. \ Proof2: Let $A$ be the set of all ordered pairs. Since $A$ is a set of ordered pairs, $A$ is relation. Let $B=dom A$.Then $B$ is set$($Since $B\subseteq \bigcup \bigcup A$, by subset axiom, $B$ exists $)$ . There is set $C\notin B$, since there is no set of all sets. Fortiori $(C,1)\notin A$. This contradicts our assumtion. Hence such $A$ does not exists.