# Notes on Chapter 1: Introductory Ideas

By Susam Pal on 27 Jul 2024

## § 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:

$$(a + b) + c = a + (b + c)$$ where $$a, b, c \in \mathbb{Q}.$$

$$a + b = b + a$$ where $$a, b \in \mathbb{Q}.$$

There exists $$0 \in \mathbb{Q}$$ such that for every $$a \in \mathbb{Q},$$ $$a + 0 = a.$$

There exists an element $$-a \in \mathbb{Q}$$ for every $$a \in F$$ 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 $$a = b.$$

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:

• $$k$$ is an upper bound of $$B;$$
• $$c$$ is an upper bound of $$B$$ $$\implies$$ $$c \ge k.$$

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

• $$k \ge b$$ for all $$b \in B;$$
• $$c$$ satisfies $$c \ge b$$ for all $$b \in B$$ $$\implies$$ $$c \ge k.$$

### 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:

• Axiom of Archimedes: For all $$x > 0 \in \mathbb{R},$$ there exists $$n \in \mathbb{N}$$ such that $$n > x.$$
• Axiom of Completeness: Every non-empty subset of $$\mathbb{R}$$ that is bounded above has a least upper bound in $$\mathbb{R}.$$

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:

• $$c$$ is an upper bound of $$A'$$ (i.e., $$c \ge a'$$ for all $$a' \in A'$$);
• $$d$$ is an upper bound of $$A'$$ $$\implies$$ $$d \ge c$$ (i.e., $$d$$ satisfies $$d \ge a'$$ for all $$a' \in A'$$ $$\implies$$ $$d \ge c$$).

The above properties imply:

• $$-c \le a$$ for all $$a \in A;$$
• $$-d$$ satisfies $$-d \le a$$ for all $$a \in A$$ $$\implies$$ $$-d \le -c.$$

Therefore $$-d$$ 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 $$x < u < y.$$