Introduction to Real Analysis
$\cdot \mathbb{N}=\{1,2,3,4, \cdots\}$,
$\cdot \mathbb{Z}=\left\{\begin{array}{r}1,2,3,4, \cdots, 0,\\ -1,-2,-3, \cdots\end{array}\right\}$,
$Q=\left\{\frac{p}{q} \mid p, q \in \mathbb{Z}, q \neq 0\right\}$,
$\mathbb{R}=\{$Q$\cup I\}$, here $I=\text{irrational numbers}$.
Least upper bound property (Completeness axiom)
Definition. Let $E \subseteq \mathbb{R}$. We say $E$ is bounded above (below) if there exists $\beta (\alpha) \in \mathbb{R}$ such that for each $x \in E$, $x \leq \beta (x \geq \alpha)$.
In this case, $\beta$ is called an upper bound ($\alpha$ is called a lower bound).
We say $E$ is bounded if $E$ is bounded above and below.
Remark. $E=\varnothing$ is possible.
Ex) $A=\left\{1-\frac{1}{n} \mid n \in \mathbb{N}\right\}$
Show that $A$ has an upper bound and a lower bound (exercise).
ex) $\mathbb{N}$ has a lower bound, but does not hove an upper bound.
ex) $B=\left\{r \in \mathbb{Q} \mid r>0\right.$ and $\left.r^{2}<2\right\}$
Then $B$ has a lower bound. $(\alpha=0)$.
However, $B$ does not have the maximum element.
To justify, it is enough to show if $p \in B$, then there exists $q \in B$ such the $p<q$. To do so, take any $p \in B$.
Take $g=p+\frac{2-p^{2}}{p+2}>p$.
Then $q \in \mathbb{Q}$.
Claim: $q^{2}<2$.
$2-q^{2}=2-\left(p+\frac{2-p^{2}}{p+2}\right)^{2}$
$=2-\left(p^{2}+\frac{2 p\left(2-p^{2}\right)}{p+2}+\frac{\left(2-p^{2}\right)^{2}}{(p+2)^{2}}\right)$
$=2-p^{2}-\frac{2 p\left(2-p^{2}\right)}{p+2}-\frac{\left(2-p^{2}\right)^{2}}{(p+2)^{2}}$
$ =\frac{\left.\left(2-p^{2}\right) p+2\right)^{2}-2 p\left(2-p^{2}\right)(p+2)-\left(2-p^{2}\right)^{2}}{(p+2)^{2}}$
$ =\frac{-\left(p^{2}-2\right)(p+2)^{2}+2 p(p-2)(p+2)-\left(p^{2}-2\right)^{2}}{(p+2)^{2}}$
$ =\frac{1}{(p+2)^{2}}((-1+2-1) p^{4}+(-4+4) p^{3}+(2-4) p^{2} \quad+(8-8) p+(8-4))$
$ =\frac{1}{(p+2)^{2}}\left(4-2 p^{2}\right)=\frac{2\left(2-p^{2}\right)}{(p+2)^{2}}>0$.
Hence $q \in B$.
Definition. Let $\phi \neq E \subseteq \mathbb{R}$ be bounded above We say $\beta \in \mathbb{R}$ is the least upper bound of $E$ if
(1) $\beta$ is an upper bound of $E$.
(2) If $\alpha<\beta$, then $\alpha$ is not an upper bound of $E$.
Denote $\beta=\sup E$ called the Supremum of $E$.
Remark.
(1) If sup E exists then SupE must be unique
Indeed, let $\alpha, \beta$ be supremum of $E$.
Then either $\alpha<\beta \Rightarrow \alpha$ is not an upper bound. It is now esaily deduced that $\alpha=\beta$.
(2) Suppose $\phi \neq E \subseteq \mathbb{R}$ is not bounded above.
$\sim$ [There exists $\beta \in \mathbb{R}$ s.t for each $x \in E, x \leq \beta]$
$\equiv$ For all $\beta \in \mathbb{R}$, there exists $x_{\beta} \in E$ S.t $x_{\beta}>\beta$
(3) Let $\phi \neq E \subseteq \mathbb{R}$, not bounded above,
$\Rightarrow \sup E=\infty(\notin \mathbb{R})$.
(4) $\sup \phi=-\infty$.
(5) Let $\phi \neq E \subseteq \mathbb{R}$ be bounded above.
$\Rightarrow$ there exists sup E $=\beta \in \mathbb{R}$ by definition.
Then for each $\varepsilon>0$, $x-\varepsilon$ is not an upper bound.
Thus by (2), there exists $x_{\varepsilon} \in E$ such that $\beta-\varepsilon<x_{\varepsilon}$.
Consequently, for each $\varepsilon>0$, there exists $x_{\varepsilon} \in E \text { sst } \beta-\varepsilon<x_{\varepsilon} \leq \beta$. Equivalently, $\beta=\sup E$.
(6) For each $n \in \mathbb{N},\left(\varepsilon=\frac{1}{n}>0\right)$.
By (5) there exists $x_{n} \in E$ such that $\beta-\frac{1}{n}<x_{n} \leq \beta$.
In particular, this shows that we have “a sequence” ( $\left.x_{n}\right)$ satisfying $\lim _{n \rightarrow \infty} x_{n}=\beta$
In fact, (6) is equivalent to (5). This is called the Archimedean Property.