TITLE: Equivalence Relation Proof-Transitive Intersection
QUESTION [0 upvotes]: Does this proof look right?  Thanks!
Suppose $R_1$  and $R_2$ are relations on set $A$.  For each part give either a proof or a counterexample to justify your answer.
If $R_1$  and $R_2$ are transitive, must $R_1 \cap R_2$ be transitive?
Let $x, y, z$ be arbitrary elements of $A$.  Suppose that $(x, y) \in R_1 ∩ R_2$ and $(y, z) \in R_1 ∩ R_2$.  Then $(x, y)$ and $(y, z) \in R_1$ and $(x, y)$ and $(y, z) \in R_2$.  This would mean that $(x, z) \in R_1$ and $R_2$ since $x, y, z$ are arbitrary.  Because $x, y, z$ are arbitrary, $(x, z) \in R_1 ∩ R_2$, meaning $R_1 \cap R_2$ is transitive on $A$.

REPLY [1 votes]: You have the right idea, but some of what you’ve actually said isn’t quite right.

This would mean that $(x,z)\in R_1$ and $R_2$ since $x,y,z$ are arbitrary.

No, it means that $(x,z)\in R_1$ and $(x,z)\in R_2$ because $R_1$ and $R_2$ are transitive; it has nothing to do with whether $x,y$, and $z$ were arbitrary or hand-picked.

Because $x,y,z$ are arbitrary, $(x,z)\in R_1\cap R_2$

No, $(x,z)\in R_1\cap R_2$ because $(x,z)\in R_1$ and $(x,z)\in R_2$; here again $x,y$, and $z$ could have been hand-picked.
It’s at the very last step that the arbitrary nature of the choice of $x,y$, and $z$ comes into play. What you want to say at the end is:

Since $x,y$, and $z$ were arbitrary elements of $A$ such that $(x,y),(y,z)\in R_1\cap R_2$, it follows that $R_1\cap R_2$ is transitive.
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