Answer to Question #211421 in Calculus for zass

Question

Your task is to discuss the concept of supremum and infimum for real numbers. Here are some hints on what your presentation can cover.

1. Give the definition of supremum and infimum for a nonempty subset of R.

2. Provide some examples to illustrate the concepts of supremum and infimum.

3. Discuss the Completeness Axiom.

4. Find some standard exercises pertaining to supremum and infimum and discuss its solution. 5. State and prove the Archimedean Property using the Completeness Axiom.

Answer 1

$\textsf{Question 1:}\\ \textbf{Definition 1:} \text{[Supremum] Let $A \subseteq \mathbb{R}$ which is bounded above. An element $\alpha \in \mathbb{R}$ is called} \\ \text{ the supremum of $A\left(\sup(A)\right)$ if it satisfy the following: } \\ (a.) x \le \alpha\,\, \forall x \in A\\ (b.) \text{if $\alpha_0 \in \mathbb{R}$ such that $x\le \alpha_0 $ then $\alpha \le \alpha_0$}$$\textbf{Definition 2:} \text{[Infimum] Let $A \subseteq \mathbb{R}$ which is bounded above. An element $\beta \in \mathbb{R}$ is called} \\ \text{ the infimum of $A\left( \inf(A) \right)$ if it satisfy the following: } \\ (a.) \beta\le x\,\, \forall x \in A\\ (b.) \text{if $\beta_0 \in \mathbb{R}$ such that $\beta_0\le x $ then $\beta_0 \le \beta$}\\ \textsf{Question 2:} \\ \textbf{Examples: } \\ \text{1. consider the interval $A = [1,2] \subset \mathbb{R}$ we observe that every element in $A$ are less than }\\ \text{or equal to every real number greater than or equal to \(2\) with this we can say that \(2\) is}\\ \text{is one of the upper bounds for $A$ , but \(2\) is the smallest of these upper bounds . Hence }\\ \text{we concluede thatthe supremum of $A$ is \(2\). That is, $\sup(A) = 2$}\\ \text{Similarly we also note that $1$ is a lower bound for $A$, but it is greater than the other }\\ \text{lower bounds of $A$ hence we conclude that $\inf(A) = 1$}\\ \boxed{\textcolor{red}{Note:} \text{the $\sup$ or $\inf$ of a set need not be include in the set as in the interval $B=(0,1)$}}\\ \text{which has $\inf(B) = 0$ and $\sup (B)=1$}\\ \textsf{Question 3}\\ \textbf{Theorem[Completeness axiom]. A ordered field $\mathbb{F}$ is complete if every }\\ \text{nonempty subset of $\mathbb{F}$ has a least upper bound that is also contained in $\mathbb{F}$}.\\ \text{This theorem is also true if we use greatest lower bound}.\\ \text{The completeness axiom is used in proving that the set $\mathbb{Q}$ is not complete .}\\ \text{Consider the set $A \subseteq \mathbb{Q}$ defined as $A\{ x \in \mathbb{Q}\mid x^2 \le 2 \}$} \\ \text{It is obvious that $\sup(A) = \sqrt{2}$, but $\sqrt{2}\notin \mathbb{Q}$ which follows by completeness axiom that }\\ \text{$\mathbb{Q}$ is not complete.}\\ \text{For $\mathbb{R}$ the reverse is the case as for every nonempty subset of $\mathbb{R}$ has their least upper bound }\\ \text{contained in $\mathbb{R}$, which shows the completeness of $\mathbb{R}$}\\ \textsf{Question 4}\\ \textbf{Example: }\text{Find the $\inf$ and $\sup$ of the following set if the exist}\\ 1. \text{ $A = \{ x\in \mathbb{R} \mid 2x + 5 > 0 \} $}\\ 2. \text{ $B = \{ x\in \mathbb{R} \mid x<\frac{1}{x} \} $}\\ \textcolor{teal}{Solution:}\\ 1. \text{ Note that solving the inequality in $A$ we have $A = \{ x\in \mathbb{R} \mid x > \frac{-5}{2} \} $}\\ \implies \frac{-5}{2} < x\,\, \forall x \in A \implies A = (\frac{-5}{2},\infty) \implies \inf(A) = \frac{-5}{2}\\ \text{it is obovious that $A$ is not bounded above hence it follows that $\sup(A)$ does not exist.}\\ 2.\text{ SImilarly as in 1 above the set $B$ is equivalent to $B = \{ x \in \mathbb{R} \mid -1 < x < 1 \}$}\\ \implies B = (-1,1) \implies \inf(B) = -1 \text{ and } \sup(B) = 1\\ \textsf{Question 5:}\\ \boxed{\textbf{Theorem [Archimedean Property].} \text{ if $x\in \mathbb{R}$, then there exists $n_x \in \mathbb{N}$ such that $x \leq n_x$}}\\ \texttt{Proof:}\\ \text{If the assertion is false, then $n \leq x$ for all $n \in \mathbb{N}$; therefore, $x$ is an upper bound of $\mathbb{N}$.}\\ \text{Therefore, by theCompleteness axiom,the nonemptyset $\mathbb{N}$ has a supremum $u \in \mathbb{R}$.}\\ \text{Subtracting $1$ from $u$ gives a number $u - 1$, which is smaller than the supremum $u$ of $\mathbb{N}$.}\\ \text{Therefore $u-1$ is not an upper bound of $\mathbb{N}$, so there exists $m \in \mathbb{N}$ with $u-1 < m$.}\\ \text{Adding $1$ gives $u < m+1$, and since $m+1 \in \mathbb{N}$, this inequality contradicts the fact that $u$}\\ \text{is an upper bound of $\mathbb{N}$.}$