# Proof Problem Rudin PMA 1-4

Rudin's Principles of Mathematical Analysis Chapter 1 #4

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 $\alpha \leq \beta$.

Attempt:

Let E be a nonempty subset of an ordered set S.
Suppose $\alpha$ is a lower bound of E.
Suppose $\beta$ is an upper bound of E.
If $\gamma \gt \alpha$, $\gamma$ is not a lower bound of E.
Hence $\alpha \leq x$ for all x$\in$ E. Therefore $\alpha \in$ E.
If $\gamma \lt \beta$, then $\gamma$ is not an upper bound of E.
Hence $x\leq \beta$ for all x$\in$ E. Therfore $\beta \in$ E.
So $\alpha \leq x \leq \beta$ for all x$\in$ E.

Perhaps my approach is flawed but I'm not sure how to go from this to $\alpha \leq \beta$.

• If γ>α, γ is not a lower bound of E. - This is false. $\gamma$ is not $\inf$, just a lower bound. – Aloizio Macedo Jun 18 '15 at 23:42
• do you mean $\alpha$ instead of $\gamma$ in your second sentence? – smokeypeat Jun 18 '15 at 23:51
• Yes, sorry for that. – Aloizio Macedo Jun 19 '15 at 0:55

You're overthinking this. Since $E$ is nonempty, there is some $x\in E$. Since $\alpha$ is a lower bound for $E$, $\alpha\leqslant x$. Since $\beta$ is an upper bound for $E$, $x\leqslant\beta$. Hence by transitivity, $$\alpha\leqslant\beta.$$
• I have that statement. Are you saying that $\alpha \leq x \leq \beta$ is equivalent to $\alpha \leq \beta$? – smokeypeat Jun 18 '15 at 23:53
• I wasn't precise with my answer; since $\alpha\leqslant x$ and $x\leqslant \beta$ and $\leqslant$ is transitive, it follows that $\alpha\leqslant\beta$. See definition 1.5 in the text. – Math1000 Jun 19 '15 at 0:02