The following is a very simple exericse in Rudin.
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 $\alpha \leq \beta$.
The proof goes something like this.
Let $x \in E$ (since $E$ is nonempty). Then $x \geq \alpha$ (definition of lower bound), $x \leq \beta$ (definition of upper bound). Then by transitivity, $\alpha \leq \beta$.
I am trying to understand where the assumption that $E$ is nonempty comes into play in this proof, because this fact surely does not hold when $E$ is empty. For example, I could take $\alpha = -1$, an upper bound, and $\beta = 3$, a lower bound. Could I not reword the above proof like this, for example:
By the definition of lower bound, $\forall x \in E, x \geq \alpha$, and by the definition of upper bound, $\forall x \in E, x \leq \beta$. Hence, $\forall x \in E, \alpha \leq x \leq \beta$, so by transitivity, $\alpha \leq \beta$.
I am trying to figure out why the above proof is wrong, though it very obviously is. Specifically, the first two assertions in the first sentence sound fine even if $E$ is empty. The next assertion sounds incorrect, but I suppose it could be vacuously true because I cannot provide a counterexample to it: there's nothing in $E$. Is the only incorrect step that I cannot invoke transitivity when there is no actual $x$ validating the fact that $\alpha \leq x \leq \beta$?
