Let $R$ be a relation, $R: A \to A$ where $R$ is both symmetric and transitive. Prove the following:

Question

Let $R$ be a relation, $R: A\rightarrow A$ where $R$ is both symmetric and transitive. Then prove or disprove the following:

($\forall x \in A \ \exists y \in A: \ (xRy)) \rightarrow$ ($R$ is an equivalence relation).

To be an equivalence relation, $R$ must be reflexive, symmetric, and transitive. We are given that $R$ is symmetric and transitive, but do not know if it is reflexive.

Since $R$ is symmetric, we can also say: if ($\forall y \in A \ \exists x \in A$ such that $yRx$) -- right?

And since $R$ is transitive, we can also say: $\forall x, y, z \in A$, if $(xRy \land yRz) \to xRz$.

However, this leaves us with insufficient evidence that R is reflexive, thus we can't prove that it is an equivalence relation.

I'm stuck.

Answer 1

Let $\;a\in A\;$ . Then there exists $\;y\in A\;$ s.t. $\;aRy\;$ , but then also $\;yRa\;$ , and since the relation is transitive

$$aRy\;\;\text{and}\;\;yRa\;\implies aRa$$

and the relation is reflexive.