I think I may be citing the law of transitivity incorrectly in this proof.
Theorem. Let $E$ be a nonempty subset of an ordered set. Suppose $\alpha$ is a lower bound of $E$ and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$.
Proof. Since $\alpha$ is a lower bound of $E$, and $E$ is nonempty, we have that $\forall x \in E, x \geq \alpha$. Similarly, since $\beta$ is an upper bound, we have $\forall x \in E, x \leq \beta$. Stringing these together yields $\forall x \in E, \alpha \leq x \leq \beta$, which by the transitivity property of ordered sets yields $\alpha \leq \beta$.
Here are my questions on this:
(a) Is it a fair assumption that a subset of an ordered set is also ordered? I wasn't sure whether I ought to 'prove' this first, though it seems almost too trivial to prove.
(b) I use transitivity a bit differently from Rudin's definition, based on past formulations I've seen. He seems to define this as $x < y \wedge y < z \implies x < z$, i.e., with strict inequalities. Should I break this into cases, wherein I consider all possible cases of strict inequalities as well as equalities between $\alpha$, $x$, and $\beta$? Or is this a natural extension of this property?