# Show that $S= \{p:\exists x,y (p=\langle x,y\rangle) \wedge (x \subset y )\}$ is a proper class.

Show that $S= \{p:\exists x,y (p=\langle x,y\rangle) \wedge (x \subset y )\}$ is a proper class.

I believe that $x \subset y$ implies $\{\{x \},\{x,y \} \} = \{\{x \} , \{y \} \} =\{\{y \} \}$

Can one show that $S$ contains every set? And that this contradicts the axiom of extensionality? (I'm new to set theory so please explicitly refer to the ZFC axioms being used.)

• I took the "$\subset$" to mean $\subseteq$, if you had meant $\subsetneq$, do let me know. The answer is still valid, but it would require a minor modification. – Asaf Karagila Nov 22 '14 at 12:34

$S$ is really just the $\subseteq$ relation on the universe of sets.
The easiest way of showing that it is a proper class is to note that there is an obvious surjection from $S$ onto the universe:
$$\langle x,y\rangle\mapsto y$$
Why is this a surjection? Because for any set $y$, we have that $\langle\varnothing,y\rangle\in S$. Now if $S$ were a set, then by the axiom of replacement the image of this function would be a set, and therefore the collection of all sets would be a set. Something which is false, therefore $S$ is a proper class.
Your idea is incorrect since $x\subseteq y$ doesn't mean that $x=y$, therefore $\{x,y\}$ does not necessarily have to equal to $\{y\}$. Your argument would have been true if the definition of $S$ was using $x=y$ rather than $x\subseteq y$.
Just note that $x=\varnothing$ and $y=\{\varnothing\}$ are such that $x\subseteq y$ but $\{x\}\neq\{x,y\}$.