Random math: 02

Source of most content: Gunhee Cho

Least upper bound property

This property is equivalent to completeness axiom in real space.

Def. Let $E\subseteq \mathbb{R}$, say $E$ is bounded above (below) if there exists $\beta \in \mathbb{R}$ ($\alpha \in \mathbb{R}$) s.t. for each $x\in E$, $x \leq \beta$ ($x \geq \alpha$)

In this case, $\beta$ is an upper bound and $\alpha$ is a lower bound, and $E$ is bounded if $E$ is bounded above and below.

Def. Let $\emptyset \ne E \subseteq \mathbb{R}$ be bounded above. Say $\beta \in \mathbb{R}$ is the least upper bound of $E$ if $\beta$ satisfies:

  1. $\beta$ is an upper bound of $E$.
  2. If $\alpha < \beta$, then $\alpha$ is not an upper bound of $E$.

Denote $\beta = \sup E$.


  1. If $\sup E$ exists, then $\sup E$ must be unique.

    proof) Let $\alpha, \beta$ be $\sup E$. Then, either $\alpha < \beta \rightarrow \alpha$ is not u.b. or $\alpha > \beta \rightarrow \beta$ is not u.b. or $\alpha=\beta$.

  2. Suppose $\emptyset \ne E \subseteq \mathbb{R}$ is not bouded above. Then, $\forall \beta \in \mathbb{R}, \exists x_\beta \in E$ s.t. $x_\beta > \beta$.
  3. Let $\emptyset \ne E \subseteq \mathbb{R}$ be not bounded above. Then, $\sup E = \infty$
  4. $\sup \emptyset= -\infty$
  5. Let $\emptyset \ne E \subseteq \mathbb{R}$ be bounded above, i.e., there exists $\sup E = \beta \in \mathbb{R}$. Then, for each $\epsilon > 0$, $\beta - \epsilon$ is not an upper bound and by Rmk 2, there exists $x_\epsilon \in E$ s.t. $\beta - \epsilon < x_\epsilon$. Consequently, for each $\epsilon>0$, $\exists x_\epsilon\in E$ s.t. $\beta - \epsilon < x_\epsilon \leq \beta$, which is the equivalent statement to say $\beta = \sup E$.
  6. For each $n \in \mathbb{N}$ and $\epsilon = \frac{1}{n} >0, \exists x_n \in E \text{ s.t. } \beta - \frac{1}{n} < x_n \leq \beta$. This shows that we can find a sequence $(x_n)$ s.t. $\lim_{n\rightarrow \infty}x_n = \beta$.

NB1; In Rmk 3 and 4, since $E$ is not bounded above or $E=\emptyset$, $\sup E = \beta \notin \mathbb{R}$.
NB2; Rmk 5 and 6 is equivalent. (Archimedean property)

M.I. and W.O.

Mathematical induction (M.I.)

\[\text{Let }A \subseteq \mathbb{N},\text{ if (1) }1\in A \text{ (2) if }n \in A, \text{ then }n+1\in A. \text{ Then, }A=\mathbb{N}.\nonumber\]

Well-ordering principle (W.O.)

\[\text{Let } \emptyset \ne A\subseteq \mathbb{N}. \text{ Then, there exists the minimum of }A. \nonumber\]


\[\text{M.I.} \Leftrightarrow \text{W.O.} \nonumber\]


  1. (W.O. $\rightarrow$ M.I.): Suppose M.I. is false, i.e., let $A \subseteq \mathbb{N}$ satisfying (1) and (2) but $A \ne \mathbb{N} \leadsto \mathbb{N} - A \ne \emptyset$, which implies $A$ is proper subset of $\mathbb{N}$. Apply W.O., then there exists the minimum $m_0 \in \mathbb{N}-A$. But by M.I. and $m_0$ is finite, $1 \in A, 2 \in A, \cdots, m_0 \in A$ which is a contradiction. THus, $A=\mathbb{N}$.
  2. (M.I. $\rightarrow$ W.O.): Let $\emptyset \ne A \subseteq \mathbb{N}$. Suppose W.O. is false for $A$, i.e., there is no $\min A$. Define $B := \{ k \in \mathbb{N} \mid k \leq n, \forall n \in A\}$. Since $1 \notin A \leadsto 1 \in B \leadsto B \ne \emptyset$, take a $m \in B$ then $m\notin A$. Thus $m+1 \in B$. (By def. of set $B$ and no minimum in $A$) By, M.I., $B = \mathbb{N} \leadsto A = \emptyset$, which is a contradication.

Archimedean Property

\[\text{Let }x, y \in \mathbb{R}, \text{ with } x >0. \text{ Then, there exists }n \in \mathbb{N} \text{ s.t. }n \cdot x > y. \nonumber\]


If $y \leq 0$, take $n=1$ which can conclude the proof. So, suppose $y>0$.
Claim: there exists $n\in\mathbb{N}$ s.t. $nx >y$.
Suppose not, i.e., for each $n\in \mathbb{N}, nx \leq y$. Define $E := \{ nx \mid n \in \mathbb{N}\}$. Then $\emptyset \ne E \subseteq \mathbb{R}$. Then, by Asm. $E$ is bounded above by $y$, which implies that there exists $\sup E = \beta \in \mathbb{R}$. Since $x > 0, \beta - x < \beta \leadsto \exists n \in \mathbb{N}$ s.t. $\beta - x < nx \leq \beta$ $\leadsto \beta < (n+1)x, n+1 \in \mathbb{N}$ $\leadsto (n+1)x \in E$, which is a contradiction, since $\beta$ is an upper bound of $E$.


  1. Take $x = \epsilon > 0, y =1$. Then, by A.P., there exists $n\in \mathbb{N}$ s.t. $n\epsilon >1 \Leftrightarrow \frac{1}{n} < \epsilon$.
  2. Let $\emptyset \ne E \subseteq \mathbb{R}$ be a bounded above, i.e., $\exists \sup E = \beta \in \mathbb{R}$. $\Leftrightarrow$ For each $\epsilon > 0, \exists x_\epsilon \in E$ s.t. $\beta - \epsilon < x_\epsilon \leq \beta$.
    $\Leftrightarrow$ For each $n\in \mathbb{N}, \exists x_n \in E$ s.t. $\beta - \frac{1}{n} < x_n \leq \beta$. ($\epsilon = \frac{1}{n}$)
  3. Assume $\beta - \frac{1}{n} < x_n \leq \beta$ holds, and let $\epsilon>0$ be fixed. Then by A.P., there exists $n \in \mathbb{N}$ s.t. $\frac{1}{n} < \epsilon \Leftrightarrow - \frac{1}{n} > - \epsilon \Leftrightarrow \beta - \frac{1}{n} > \beta - \epsilon$. Thus for each $\epsilon >0, \exists n \in \mathbb{N}$ s.t. for corresponding $\exists x_n \in E$ satisfying $\beta - \epsilon < \beta - \frac{1}{n} < x_n \leq \beta$. Consequently, we have a sequence $(x_n)^\infty_{n=1}$ s.t. $\lim_{n\rightarrow \infty} x_n = \beta$.

NB1; In Rmk 3, $\epsilon$ has uncountable choice while $\frac{1}{n}$ has countouble choice.
NB2; By Rmk 3, we can hanlde $\beta$ as a seq. that converges to $\beta$ and uncountable choice $\epsilon$ as countable choice $n$.