Hi everyone I'd like to know if following is really correct, looks kinda cumbersome, I think, it is for the great quantity of cases to analyze. To be honest I don't know if this is the better way to do it. but I cannot see other approach.
I would appreciate any suggestion.
Definitions: Let $E$ be a subset of the extended real numbers $\overline { \mathbb{R}}$. Then we define the least upper bound of $E$ by the following rule:
1) If $E$ is contained in $\mathbb{R}$ (i.e., $+\infty$ and $- \infty$ are not elements in $E$), then we let $\text{sup } E$ as is define in $\mathbb{R}$.
2) If $E$ is contains $+\infty$, then we set $\text{sup } E:= +\infty$
3) If $E$ does not contain $+\infty$ but does contain $-\infty$, then we set $\text{sup } E:= \text{sup } ( E- \{ -\infty \})$.
We define the infimum by the formula $\text{inf } E:=-\text{sup } (-E)$ where $-E := \{ -x: x\in E\}$
Theorem: Let $E$ be a subset of the extended real numbers. Then the following statements are true:
$(a)$ For all $x\in E$ we have $x\le \text{sup } E$ and $x\ge \text{inf } E$
$(b)$ Suppose $M$ is an upper bound for $E$, i.e., $x \le M$ for all $x\in E$. Then we have $\,\text{sup } E \le M$.
$(c)$ Suppose $M$ is a lower bound for $E$, i.e., $x \ge M$ for all $x\in E$. Then we have $\,\text{inf } E \ge M$.
Proof: Let $S = \text{sup } E$ and let $L= \text{inf } E$
$(a)$
If $S= -\infty$, then $E$ is empty and the result is vacuously true. When $S = +\infty$ by definition we have $x\le S$ for all $x\in \overline { \mathbb{R}}$, in particular for all the $x \in E$. If $S\in \mathbb{R}$, $E$ does not contain $+\infty$ and thus $E$ only contains element in $\mathbb{R}$ or $-\infty$ and in either case the result follows.
So let $-E := \{ -x: x\in E\}$. And for what we've proved above, we have $\,-x\le \text{sup } (-E)$ for all $-x\in -E$, which means $L=-\text{sup } (-E) \le x$ and since $-x\in -E$ iff $x\in E$, we're done.
$(b)$
(1) Suppose $E\subset \mathbb{R}$, i.e., $+\infty$ and $- \infty$ are not elements in $E$. If $S = -\infty$, it follows that $E= \emptyset$ and so $S \le M$ where $M$ can be every element in $\overline { \mathbb{R}}$. If $S = +\infty$, then $E$ is unbounded and then $M\notin \mathbb{R}$ (otherwise we have a contradiction). Thus either $M= -\infty$ or $M= +\infty$; If $M = -\infty$ and by definition of upper bound, for an arbitrary $x\in E$ we have $x \le -\infty$, but this implies $x= -\infty$ which contradicts the fact that $E$ is contained in $\mathbb{R}$, and If $M = +\infty$ then clearly is an upper bound and $S \le M$. Now in the case when $S\in \mathbb{R}$, we have that if $M\in \mathbb{R}$ the result follows from the properties of the lub in $\mathbb{R}$. When $M\notin \mathbb{R}$ by the same argument as before we know that $M\not= -\infty$ so $M = +\infty$ and by definition $S\le M$.
(2) Now suppose $+\infty \in E$. Since $M$ is an upper bound for $E$ and we assume that $+\infty \in E$ thus $+\infty \le M$, which means $M= +\infty$. Thus, $S\le M$.
(3) When $E$ does not contain $+\infty$ but does contain $-\infty$, we thus have $S= \text{sup } (E^{*})$; where $E^*=E- \{ -\infty \}$. Thus falls under case (1).
$(c)$ It follows from $(b)$ and the fact that $\text{inf } E:=-\text{sup } (-E)$ where $-E := \{ -x: x\in E\}$.
Thanks :)
It looks good. The only things that jumped out at me were the fact that
Other than that, it seems pretty reasonable, given the machinery that you're using. (As Hurkyl noted, you can define the extended reals as the order completion of $\mathbb{R}$, in which case everything flows easily. But that's a different branch of mathematics altogether.)