Notes on Chapter 1: Introductory Ideas

By Susam Pal on 27 Jul 2024 [draft]

Contents

§ 1.1 Foreword for the Student: Is Analysis Necessary?

Example of Non-Differentiable Function

It is not obvious from the graph that \[ f(x) = \begin{cases} x \sin(1/x) & \text{ if } x \neq 0 \\ 0 & \text{ if } x = 0 \end{cases} \] is continuous but not differentiable at \( x = 0. \)

Example of Discontinuous Function

Although the \( \sin \) function is continuous, the function \[ S(x) = \frac{\sin x}{1} - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \dots \] is discontinuous.

§ 1.2 The Concept of Numbers

Irrationality of \( \sqrt{2} \)

Let \( \sqrt{2} = m/n \) where \( m \) and \( n \) be positive integers with no common factor. Then \begin{align} (m / n)^2 = 2 & \implies m^2 = 2n^2 \notag \\ & \implies 2 \mid m^2 \notag \\ & \implies 2 \mid m \label{div1.2.1} \\ & \implies m = 2l \; \text{ for some positive integer } l \notag \\ & \implies m^2 = 4l^2 \notag \\ & \implies 2n^2 = 4l^2 \notag \\ & \implies n^2 = 2l^2 \notag \\ & \implies 2 \mid n^2 \notag \\ & \implies 2 \mid n. \label{div1.2.2} \\ \end{align} But \eqref{div1.2.1} and \eqref{div1.2.2} contradict the assumption that \( m \) and \( n \) have no common factor.

Irrationality of Non-Integer \( \sqrt{n} \)

Let \( n \) be a natural number such that \( n \) is not a perfect square. Then there exists a prime number \( p \) such that \[ n = p^{2k + 1} l \] where \( k \) is a nonnegative integer, \( l \) is a positive integer, and \( p \nmid l. \) Note that the power of \( p \) cannot be even because \( n \) is not a perfect square.

Let \( \sqrt{n} = r/s \) such that \( r \) and \( s \) are positive integers with no common factor. Then \begin{align*} (r / s)^2 = n & \implies r^2 = n s^2 \\ & \implies r^2 = p^{2k + 1} l s^2 \\ & \implies p^{2k + 1} \mid r^2. \end{align*} Since \( r^2 \) is a perfect square, \( p \) must divide it an even number of times. Therefore we get, \begin{align} p^{2k + 1} \mid r^2 & \implies p^{2k + 2} \mid r^2 \notag \\ & \implies p^{k + 1} \mid r \notag \\ & \implies p \mid r. \label{div1.2.3} \end{align} Also \begin{align} p^{2k + 2} \mid r^2 & \implies p^{k + 1} \mid r \notag \\ & \implies r = p^{k + 1} j \; \text{ for some positive integer } j \notag \\ & \implies r^2 = p^{2k + 2} j^2 \notag \\ & \implies n s^2 = p^{2k + 2} j^2 \notag \\ & \implies p^{2k + 1} l s^2 = p^{2k + 2} j^2 \notag \\ & \implies l s^2 = p j^2 \notag \\ & \implies p \mid l s^2 \notag \\ & \implies p \mid s^2 \; \text{ (since } p \nmid l \text{)} \\ & \implies p \mid s. \label{div1.2.4} \end{align} But \eqref{div1.2.3} and \eqref{div1.2.4} contradict the assumption that \( r \) and \( s \) have no common factor.

Irrationality of \( \sqrt{2} + \sqrt{3} \)

If \( x = m / n \) where \( m \) and \( n \) are integers, then \( x^2 = m^2 / n ^2. \) We have shown that if \( x \) is rational, then \( x^2 \) is rational too. Taking its contrapositive, we find that if \( x^2 \) is irrational, then \( x \) is irrational.

Let \( x = \sqrt{2} + \sqrt{3}. \) Then \( x^2 = 5 + 2 \sqrt{6}. \) If \( x^2 = 5 + 2 \sqrt{6} = m / n \) for some integers \( m \) and \( n, \) then \( \sqrt{6} = (m - 5n) / 10n. \) But this is a contradiction since \( \sqrt{6} \) is irrational (by the discussion in the previous section). Thus \( x^2 \) is irrational. Therefore \( x = \sqrt{2} + \sqrt{3} \) is irrational.

§ 1.3 The Language of Set Theory

Notations for Sets

The following notations are introduced in this section:

  1. \( \mathbb{N} \): the set of natural numbers (positive integers)
  2. \( \mathbb{Z} \): the set of all integers
  3. \( \mathbb{Q} \): the set of all rational numbers
  4. \( \mathbb{R} \): the set of all real numbers
  5. \( \mathbb{C} \): the set of all complex numbers

In one of the paragraphs, the book states that \[ X \cup (A \setminus X) = A \] This is not true in general but this is true when \( X \subseteq A. \) as is the assumption made in the book just before stating the above formula. In general though, the following is true instead: \[ X \cup (A \setminus X) = A \cup X. \]

§ 1.4 Real Numbers

Field Axioms

The set \( \mathbb{Q}, \) the addition operation and the multiplication operation on \( \mathbb{Q} \) satisfy the field axioms presented below:

  1. Associative law for addition:
    \( (a + b) + c = a + (b + c) \) where \( a, b, c \in \mathbb{Q}. \)

  2. Commutative law for addition:
    \( a + b = b + a \) where \( a, b \in \mathbb{Q}. \)

  3. Existence of additive identity:
    There exists \( 0 \in \mathbb{Q} \) such that for every \( a \in \mathbb{Q}, \) \( a + 0 = a. \)

  4. Existence of additive inverse:
    There exists an element \( -a \in \mathbb{Q} \) for every \( a \in \mathbb{Q} \) such that \( a + (-a) = 0. \)

  5. Associative law for multiplication:
    \( (ab)c = a(bc) \) where \( a, b, c \in \mathbb{Q}. \)

  6. Commutative law for multiplication:
    \( ab = ba \) where \( a, b \in \mathbb{Q}. \)

  7. Existence of multiplicative identity:
    There exists \( 1 \in \mathbb{Q} \) such that for every \( a \in \mathbb{Q}, \) \(a1 = a. \)

  8. Law of the reciprocal:
    There exists an element \( 1/a \in \mathbb{Q} \) for every \( a \neq 0 \in Q \) such that \( a(1/a) = 1. \)

  9. Distributive law:
    \( a(b + c) = ab + ac \) for every \( a, b, c \in \mathbb{Q}. \)

Order Axioms

The set \( \mathbb{Q}, \) together with the addition and multiplication operations and a relation \( < \) satisfy the order axioms presented below:

  1. Transitive law:
    For all \( a, b, c \in \mathbb{Q}, \) \( a < b \) and \( b < c \) implies \( a < c. \)

  2. Trichotomy law:
    For all \( a, b \in \mathbb{Q}, \) exactly one of the following holds: \( a < b, \) \( b < a, \) or \( a = b. \)

  3. Compatibility with addition:
    For all \( a, b, c \in \mathbb{Q}, \) \( a < b \) implies \( a + c < b + c. \)

  4. Compatibility with multiplication:
    For all \( a, b, c \in \mathbb{Q}, \) \( a < b \) and \( c > 0 \) implies \( ac < bc. \)

Order Relation in \( \mathbb{Q} \)

Let \( m, p \in \mathbb{Z} \) and \( n, q \in \mathbb{N}. \) Then \[ \frac{m}{n} < \frac{p}{q} \iff \frac{mq}{nq} < \frac{np}{nq} \iff mq < np. \]

Density Property

The set \( \mathbb{Q} \) has the property of the density. For all \( q, r \in \mathbb{Q} \) where \( q < r, \) there is \( s \in \mathbb{Q} \) such that \( q < s < r. \) This is trivial to demonstrate by taking \( s = (q + r) / 2. \)

Example: Between \( 5/10 \) and \( 7/10, \) there exists \( 6/10, \) i.e., \( 5/10 < 6/10 < 7/10. \)

There are infinitely many rational numbers between \( q \) and \( r. \) This follows directly from the density property.

Archimedean Property

For all \( q > 0 \) in \( \mathbb{Q}, \) there exists \( n \in \mathbb{N} \) such that \( n > q. \)

Example: For \( q = 7/3, \) there exists \( n = 5 > 7/3. \) In fact, there also exists \( n = 3 > 7/3. \)

Upper Bound

A subset \( B \) of \( \mathbb{Q} \) is said to be bounded above if there exists an upper bound \( K \in \mathbb{Q} \) such that \( K \ge b \) for all \( b \in B. \)

Example: The set \( \{ a \in Q : a < 4 \} \) is bounded above by \( 4. \) It is also bounded above by \( 5, \) by \( 6, \) by \( 100, \) etc. The least upper bound is \( 4. \)

Example: The set \( \mathbb{N} \subset \mathbb{Q} \) is not bounded above.

Least Upper Bound

If \( B \subset \mathbb{Q} \) and \( B \ne \emptyset \) and \( B \) is bounded above then a least upper bound of \( B \) is a number \( k \in \mathbb{Q} \) such that:

Here is an equivalent definition of least upper bound purely in symbols without relying on the definition of "upper bound":

Uniqueness of Least Upper Bound

Let \( k_1 \) and \( k_2 \) be the least upper bounds of \( B. \)

Since \( k_1 \) is a least upper bound and \( k_2 \) is an upper bound, we get \( k_1 \le k_2. \)

Since \( k_2 \) is a least upper bound and \( k_1 \) is an upper bound, we get \( k_2 \le k_1. \)

Thus \( k_1 = k_2. \)

Properties of Real Numbers

The set \( \mathbb{R} \) is an ordered field satisfying the following axioms:

Note that the set \( \mathbb{Q} \) does not satisfy the axiom of completeness. For example, the set \[ C = \{ a \in \mathbb{Q} : a > 0 \text{ and } a^2 < 2 \} \] is a non-empty subset of \( \mathbb{Q} \) that is bounded above but its least upper bound is \( \sqrt{2} \not\in \mathbb{Q}. \)

Greatest Lower Bound

Theorem. If the set \( A \subset \mathbb{R} \) and \( A \ne \emptyset \) and \( A \) is bounded below, then \( A \) has a greatest lower bound in \( \mathbb{R}. \)

Proof. First let \( A' = \{ -a : a \in A \}. \) We will use this set in a moment.

Since \( A \) is bounded below, there exists a lower bound \( k \in \mathbb{R} \) such that \( k \le a \) for all \( a \in A. \) Thus \( -k \ge -a \) all \( a \in A. \) Therefore \( -k \ge a' \) for all \( a' \in A'. \) Thus \( -k \) is an upper bound of \( A'. \)

We have shown that the set \( A' \) is bounded above. Therefore by the axiom of completeness, \( A' \) has a least upper bound \( c \in \mathbb{R}. \) This means the following:

The above properties imply:

Therefore \( -c \) is the greatest lower bound of \( A. \)

Supremum and Infimum

A non-empty set \( A \in \mathbb{R} \) that is bounded above has a least upper bound in \( \mathbb{R} \) by the axiom of completeness. If \( A \) is bounded below, it has a greatest lower bound in \( \mathbb{R} \) as demonstrated earlier.

The least upper bound of \( A \) (when it is bounded above) is known as the supremum of A. This is denoted as \( \sup A. \)

The greatest lower bound of \( A \) (when it is bounded below) is known as the infimum. This is denoted as \( \inf A. \)

For open and closed intervals, we have \begin{align*} \sup (a, b) & = \sup [a, b] = b, \\ \inf (a, b) & = \inf [a, b] = a. \end{align*}

Irrationality of \( x + y \) for \( x \in \mathbb{Q} \) and \( y \not\in \mathbb{Q} \)

If \( x \) is rational and \( y \) is irrational, then \( x + y \) is irrational. To show this first let \( x = m/n \in \mathbb{Q}. \) Now suppose \( x + y = p/q \in \mathbb{Q}. \) Then \[ y = \frac{p}{q} - \frac{m}{n} = \frac{pn - qm}{qn} \in \mathbb{Q}. \] This is a contradiction since \( y \) is irrational. Therefore \( x + y \not\in \mathbb{Q}. \)

Irrational Between Rationals

If \( x, y \in \mathbb{Q} \) and \( x < y, \) then there exists an irrational number \( u \) such that \( x < u < y. \)