Assume, $R$ is a symmetric, transitive relation on $A$ and the domain of $R$ is $A$.
$Dom(R)=A$ implies $(\forall x \in A)(\exists y \in A)[xRy]$.
Since, $xRy$ is true it follows that $yRx$ is true, by the symmetric property.
By transitivity, $xRy$ and $yRx$ give $xRx$. Therefore, it follows that, $(\forall x \in A)[xRx]$, which means x is reflexive on A.
Please critique or give any advice, thank you!
Best Answer
Apart from the fact that your proof could benefit from some better wording, it is perfectly correct. I would still suggest you try to reword it more clearly as that will help you later on. Make it more clear what you want to do. Something like:
We wish to prove that $R$ is reflexive i.e. that for every $x$, $xRx$ is true:
$$\forall x: xRx$$
Let $x_0\in A$ be an arbitrary element. Then, because we know that
$$\forall x\in A \exists y\in A: xRy,$$
we know there exists some $y_0$ such that....
...
Therefore, we know that $x_0Rx_0$.
Because $x_0$ was chosen arbitrarily, we know that this is true for every $x\in A$, so $R$ is reflexive.
Such a proof is clearer because: