# $E$ is a nonempty subset of an ordered set. If $\alpha$ is a lower bound of $E$ and $\beta$ is an upper bound, then $\alpha \leq \beta$.

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.

• It seems good. Regarding your second question, it is a natural extension. – Dog_69 Aug 18 '18 at 17:15

For your first question, your ordering actually relies on the ordering on the larger set, since the upper and lower bounds need not be members of the subset $E$ itself! For instance, if the original set is $\mathbb{R}$ and the subset is $(0,1)$, your lower bound could be $0$. So really you rely on the fact that the ordering is defined on the original set.

For your second question, to be completely rigorous, the transitivity of $\leq$ follows from transitivity of $<$ and $=$. Personally, I think it is fine to simply leave it as you have, but perhaps your professor wants you to add in additional detail if this is a first course in proofs or something.

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

• They can be considered as strict orders. In fact, some books define orders as $<$ instead of $\leq$. – Dog_69 Aug 18 '18 at 17:28
• Would you mind explaining why they don't satisfy antisymmetry? It seems that $a < b$, for example, cannot coexist with $b < a$. Surely $<$ wouldn't be reflexive, though, whereas $\leq$ would be. – Matt.P Aug 18 '18 at 17:33
• @Matt.P Oh, very good point. I think he/s wanted to say reflexivity. – Dog_69 Aug 18 '18 at 17:47
• Yes, my mistake I wanted to say reflexivity. I fixed it, thx – Edouard Berthe Aug 20 '18 at 5:48