🏃🏽‍♀️‍➡️Intro to Mathematical Analysis Unit 4 – Completeness and Supremum/Infimum Properties

Key Concepts and Definitions

• Completeness a fundamental property of the real number system ($\mathbb{R}$) ensures every non-empty subset of $\mathbb{R}$ that is bounded above has a least upper bound (supremum)
• Supremum the smallest upper bound of a set $A$, denoted as $\sup A$, is the smallest real number greater than or equal to every element in $A$
  • Formally, $s = \sup A$ if $s$ is an upper bound of $A$ and for any upper bound $t$ of $A$, $s \leq t$
• Infimum the greatest lower bound of a set $A$, denoted as $\inf A$, is the largest real number less than or equal to every element in $A$
  • Formally, $i = \inf A$ if $i$ is a lower bound of $A$ and for any lower bound $t$ of $A$, $i \geq t$
• Upper bound a real number $b$ is an upper bound of a set $A$ if $b \geq a$ for all $a \in A$
• Lower bound a real number $b$ is a lower bound of a set $A$ if $b \leq a$ for all $a \in A$
• Bounded above a set $A$ is bounded above if there exists a real number $b$ such that $a \leq b$ for all $a \in A$
• Bounded below a set $A$ is bounded below if there exists a real number $b$ such that $a \geq b$ for all $a \in A$

Historical Context and Importance

• The concept of completeness emerged in the 19th century as mathematicians sought to rigorously define the real number system and its properties
• Completeness plays a crucial role in establishing the foundations of real analysis, a branch of mathematics that deals with the properties of real numbers and functions of real variables
• The completeness axiom distinguishes the real number system from other number systems, such as the rational numbers ($\mathbb{Q}$), which are not complete
  • For example, the set of rational numbers between 0 and 2 whose squares are less than 2 does not have a rational supremum, highlighting the incompleteness of $\mathbb{Q}$
• Completeness is essential for proving the existence of limits, continuity, and differentiability of functions, which are fundamental concepts in real analysis
• Many important theorems in real analysis, such as the Intermediate Value Theorem and the Extreme Value Theorem, rely on the completeness of the real number system
• The completeness axiom also has practical implications in various fields, such as physics and engineering, where real numbers are used to model continuous quantities and phenomena

Completeness Axiom Explained

• The completeness axiom states that every non-empty subset of real numbers that is bounded above has a least upper bound (supremum) in the set of real numbers
• Formally, if $A$ is a non-empty subset of $\mathbb{R}$ and $A$ is bounded above, then there exists a real number $s$ such that $s = \sup A$
• The completeness axiom is an essential property that distinguishes the real number system from other number systems, such as the rational numbers
• One consequence of the completeness axiom is the Archimedean property, which states that for any positive real numbers $a$ and $b$, there exists a natural number $n$ such that $na > b$
  • This property ensures that there are no "infinitely large" or "infinitely small" real numbers
• The completeness axiom is closely related to the notion of Dedekind cuts, which provide a way to construct the real numbers from the rational numbers
  • A Dedekind cut is a partition of the rational numbers into two non-empty sets $A$ and $B$, such that every element of $A$ is less than every element of $B$, and $A$ has no greatest element
  • The completeness axiom ensures that every Dedekind cut corresponds to a unique real number

Supremum and Infimum: The Basics

• The supremum (sup) of a set $A$ is the least upper bound of $A$, denoted as $\sup A$
  • If $s = \sup A$, then $s$ is an upper bound of $A$, and for any upper bound $t$ of $A$, $s \leq t$
• The infimum (inf) of a set $A$ is the greatest lower bound of $A$, denoted as $\inf A$
  • If $i = \inf A$, then $i$ is a lower bound of $A$, and for any lower bound $t$ of $A$, $i \geq t$
• A set $A$ may have a supremum or infimum that is not an element of $A$ itself
  • For example, the set $A = (0, 1)$ has $\sup A = 1$ and $\inf A = 0$, neither of which are elements of $A$
• If a set $A$ has a maximum element $m$, then $\sup A = m$; similarly, if $A$ has a minimum element $n$, then $\inf A = n$
• The supremum and infimum of a set $A$ are unique when they exist, as a consequence of the completeness axiom
• For any non-empty set $A$ that is bounded above, $\sup A$ always exists in $\mathbb{R}$; similarly, for any non-empty set $A$ that is bounded below, $\inf A$ always exists in $\mathbb{R}$

Properties and Theorems

• The Supremum Property if $A$ is a non-empty subset of $\mathbb{R}$ that is bounded above, then $\sup A$ exists in $\mathbb{R}$
• The Infimum Property if $A$ is a non-empty subset of $\mathbb{R}$ that is bounded below, then $\inf A$ exists in $\mathbb{R}$
• Theorem if $A$ and $B$ are non-empty subsets of $\mathbb{R}$ with $A \subseteq B$, then $\sup A \leq \sup B$ and $\inf A \geq \inf B$
• Theorem for any real number $c$ and non-empty set $A$, $\sup(c + A) = c + \sup A$ and $\inf(c + A) = c + \inf A$, where $c + A = \{c + a : a \in A\}$
• Theorem for any positive real number $c$ and non-empty set $A$, $\sup(cA) = c \cdot \sup A$ and $\inf(cA) = c \cdot \inf A$, where $cA = \{ca : a \in A\}$
  • If $c$ is negative, then $\sup(cA) = c \cdot \inf A$ and $\inf(cA) = c \cdot \sup A$
• The Nested Interval Property if $\{I_n\}$ is a sequence of closed, bounded, and nested intervals (i.e., $I_{n+1} \subseteq I_n$ for all $n$), then the intersection $\bigcap_{n=1}^{\infty} I_n$ is non-empty
  • This property is a consequence of the completeness axiom and is crucial for proving the existence of certain limits

Applications in Real Analysis

• The completeness axiom is essential for proving the existence of limits of sequences and functions
  • For example, the Monotone Convergence Theorem states that if a sequence is monotone (either non-decreasing or non-increasing) and bounded, then it converges to a limit in $\mathbb{R}$
• Completeness is crucial for establishing the continuity and differentiability of functions
  • The Intermediate Value Theorem, which relies on completeness, states that if $f$ is a continuous function on $[a, b]$ and $y$ is between $f(a)$ and $f(b)$, then there exists a $c \in (a, b)$ such that $f(c) = y$
• The Extreme Value Theorem, another consequence of completeness, states that a continuous function on a closed and bounded interval attains its maximum and minimum values
• Completeness is essential for defining the Riemann integral, which is used to calculate the area under a curve
  • The Riemann integral of a function $f$ on $[a, b]$ is defined as the limit of Riemann sums, which rely on the completeness of $\mathbb{R}$
• The Heine-Borel Theorem, which states that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, is a consequence of the completeness axiom
  • This theorem has important applications in topology and functional analysis

Common Pitfalls and Misconceptions

• Confusing completeness with other properties, such as density or continuity
  • Density means that between any two real numbers, there exists another real number, which is true for both $\mathbb{Q}$ and $\mathbb{R}$
  • Continuity is a property of functions, not number systems
• Assuming that every set has a supremum or infimum
  • Sets that are not bounded above or below may not have a supremum or infimum (e.g., $\mathbb{R}$ itself)
  • Empty sets do not have a supremum or infimum
• Thinking that the supremum or infimum of a set must be an element of the set
  • The supremum or infimum may not be contained in the set itself (e.g., $\sup (0, 1) = 1 \notin (0, 1)$)
• Misunderstanding the relationship between supremum, maximum, infimum, and minimum
  • A set may have a supremum or infimum without having a maximum or minimum element
  • If a set has a maximum or minimum element, it will be equal to the supremum or infimum, respectively
• Incorrectly applying properties of supremum and infimum
  • When multiplying a set by a negative number, the supremum and infimum are swapped (i.e., $\sup(cA) = c \cdot \inf A$ and $\inf(cA) = c \cdot \sup A$ for $c < 0$)
• Forgetting that the completeness axiom is specific to the real number system
  • Other number systems, such as the rational numbers or the hyperreal numbers, may not satisfy the completeness axiom

Practice Problems and Solutions

1. Find the supremum and infimum of the set $A = \{x \in \mathbb{R} : x^2 < 5\}$. Solution $\sup A = \sqrt{5}$ and $\inf A = -\sqrt{5}$. The set $A$ consists of all real numbers whose squares are less than 5. The largest such number is $\sqrt{5}$, and the smallest is $-\sqrt{5}$. Note that $\sqrt{5} \notin A$ and $-\sqrt{5} \notin A$.

2. Prove that if $A$ and $B$ are non-empty subsets of $\mathbb{R}$ with $A \subseteq B$, then $\sup A \leq \sup B$. Solution Let $s_A = \sup A$ and $s_B = \sup B$. Since $A \subseteq B$, every element of $A$ is also an element of $B$. Therefore, $s_B$ is an upper bound for $A$. By the definition of supremum, $s_A$ is the least upper bound of $A$, so $s_A \leq s_B$.

3. Let $A = \{1 - \frac{1}{n} : n \in \mathbb{N}\}$. Find $\sup A$ and $\inf A$. Solution $\sup A = 1$ and $\inf A = 0$. The sequence $\{1 - \frac{1}{n}\}$ is increasing and bounded above by 1, so its supremum is 1. The infimum is 0 because $1 - \frac{1}{n} > 0$ for all $n \in \mathbb{N}$, and as $n$ approaches infinity, $1 - \frac{1}{n}$ gets arbitrarily close to 0.

4. Prove that if $a < b$, then there exists a rational number $r$ such that $a < r < b$. Solution Let $a, b \in \mathbb{R}$ with $a < b$. Consider the set $A = \{r \in \mathbb{Q} : r \leq a\}$. Since $a$ is an upper bound for $A$ and $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists an $r \in \mathbb{Q}$ such that $a < r < b$.

5. Prove that the set $\{x \in \mathbb{Q} : x^2 < 2\}$ does not have a supremum in $\mathbb{Q}$. Solution Suppose, for the sake of contradiction, that the set $A = \{x \in \mathbb{Q} : x^2 < 2\}$ has a supremum $s$ in $\mathbb{Q}$. Then $s^2 \leq 2$. If $s^2 = 2$, then $s = \sqrt{2} \notin \mathbb{Q}$, a contradiction. If $s^2 < 2$, then there exists a rational number $r$ such that $s < r < \sqrt{2}$, which means $r \in A$ and $r > s$, contradicting the fact that $s$ is the supremum of $A$. Therefore, $A$ does not have a supremum in $\mathbb{Q}$.