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?
Thanks.
Best Answer
You're proof is almost correct.
The fact that E is nonempty is useless to state that ∀x∈E, x≥α (it would also be the case if E were empty!). Same for ∀x∈E, x≤β. These two relations are the definition of lower and upper bounds, respectively.
However, you need the fact that E is nonempty to prove that α≤β, because ∀x∈E,α≤x≤β would be also true if E were empty, but you couldn't argue that α≤β (because in this case α≤x≤β would never happen). A correct statement would be "E is nonempty so ∃x∈E. So α≤x and x≤β so α≤β."
(a) Yes, you can prove it easily.
(b) "<" and ">" are not orders because they do not check the reflexivity property of an order