Discrete Mathematics Proof through Equivalence Relations

Question

Suppose $\alpha$ and $\beta$ are equivalence relations on the set $S$. Suppose further that the relation $\gamma$ is defined as follows: For $x,y \in S$, $x\gamma y$ means $x\alpha y$ and $x\beta y$. Prove that $\gamma$ is also an equivalence relation on $S$.

I'm not quite sure where I should start on this, but I think it has something to do with both $\alpha$ and $\beta$ being equivalence relations, and then proving that $\gamma$ is also an equivalence relation through the use of reflexive, symmetric, and transitive properties. Any ideas?

Answer 1

First note that since $x\alpha x$ and $x \beta x$ for all $x$ in your domain (by reflexivity of these equivalence relations), we have $x \gamma x$.

For symmetry, note that $$x \gamma y \iff x \alpha y \ \wedge \ x \beta y \iff y \alpha x \ \wedge \ y \beta x \iff y \gamma x $$ (by symmetry of these relations).

Finally for transitivity, suppose that $x \gamma y$ and $y \gamma z$ for some $x,y,z \in S$; from here you should be able to continue on your own using the fact that $\alpha$ and $\beta$ are transitive relations, so just "unwrap" the definitions and everything should fall out nicely.

Answer 2

For example, to prove reflexivity, for every $x\in S$ we know $x\alpha x$ and $x \beta x,$ because $\alpha,\beta$ are reflexive. But definition of $\gamma,$ we have $x\gamma x.$ But that means that $\gamma$ is reflexive.

Prove the other properties the same way.