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Can somebody show me how to prove, the statement in the title?

For given set, $A$, exist a set $\{(X,Y): X\subset Y\subset A \}$

Thanks!

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    Let $X=Y=A$, and you just have $A\times A$. If $\subset$ is supposed to denote proper containment, then this is false for a set with 1 or 0 elements.2012-10-26

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To show existence of a set $\{(X,Y): X\subset Y\subset A \}$ for an arbitrary set you can apply the Axiom schema of specification which says that "Every subclass of a set that is defined by a predicate is itself a set." To do this you would like to replace $\phi$ and $A'$ in the following schema:

$ \forall w_1 , \dots, w_n \forall A' \exists B \forall x ( x \in B \iff [x \in A' \land \phi (x,w_1, \dots , w_n, A')])$

Let $A'$ in the formula be the set of all pairs $(X,Y)$ with $X, Y$ sets and let $\phi (x, A') = (x=(X,Y)) \land (X \subset Y \subset A)$. Then $B$ is the set $\{(X,Y): X\subset Y\subset A \}$.

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    Thanks for conf$i$rmi$n$g a$n$d updat$i$ng your answer. The resident ... preferred to sneak out by the back exit :-)2012-10-26