TITLE: Pinter's Set Theory Exercise 1.7.6: Let G and H be graphs. Prove that if G and H are sets, then G inverse and GoH are sets.
QUESTION [0 upvotes]: I've managed to do a proof of this by using the previous exercise that shows the domain and range of G are both sets. Then {y} is a set as it is a subset of the range for any y in the range, and similarly for any x in the domain. So for all x and y, {y,x} is a set by being a union of {x} and {y}, and then by Axiom 5 {{y}, {y,x}} = (y,x) is a set for all x and y and the union of these is G inverse so this is also a set.
I can also do GoH similarly by breaking the sets down and reconstructing with unions. But... It's so ugly. It's all I've got but I feel like I'm not even in the right ballpark of what the intended proof is. Is there a more attractive path to this result?

REPLY [1 votes]: Hint
I assume that with "$G$ is a graph" you mean $G = \{ (x,y) \mid x \in X \text { and } y \in Y \}$ where $X,Y$ are sets.
Thus, what is $G^{-1}$ ? It is $\{ (y,x) \mid x \in X \text { and } y \in Y \}$ where $X,Y$ are sets.
So, if the condition "$x \in X \text { and } y \in Y$" allows us to use Specification axiom schema to prove that $G$ is a set, the same holds for $G^{-1}$.
With more details: $G^{-1} = \{ z \mid \exists y \exists x ( z= \{ \{ y \} \{ y,x \} \} \text { and } x \in X \text { and } y \in Y \}$.
But browsing on the textbook, it seems that the author follows a difefrent approach.
If so, IMO you can use Axiom A.3: Every subclass of a set is a set, as well as Th.1.51: If $A,B$ are sets, then $A \times B$ is a set.
Thus, if $G \subseteq X \times Y$ is a set, also $G^{-1} \subseteq Y \times X$ will be.
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