Thursday, September 6, 2012

Introduction to the course

Lecture notes for the first MATH 242 class, taught by Ivo Klemes. These lecture notes are student-generated and any errors or omissions should be assumed to be the fault of the notetaker and not of the lecturer. To correct an error, you have to be registered and logged-in; alternatively, you can contact @dellsystem directly.

1Overview of the course¶

The required textbook is Introduction to Real Analysis by Robert G. Bartle and Donald R. Sherbert, fourth edition. The second and third editions are acceptable but page numbers and other references will assume the fourth edition.

You can find the course outline here.

2Introduction to real numbers¶

We begin with chapter 2 of the textbook: real numbers. The reals ($\mathbb{R}$) have all the algebraic properties of fields (associativity, commutativity, distributivity, etc - see 2.1.1 in the textbook). They also have order properties (which are, of course, not unique to the reals - for example, the rational numbers have these properties as well). Finally, there is the supremum, or *completeness, property, which is the last property we need to uniquely characterise $\mathbb{R}$. For instance, $\sqrt{2}$ is contained within $\mathbb{R}$ and not within the rational numbers ($\mathbb{Q}$), and so $\mathbb{Q}$ does not have this property. More on this later.

We will now proceed to build up the real numbers "from scratch", as it were.

2.1Field properties¶

We begin with the existence of two closed¹ binary operations: addition, and multiplication. Both of these operations are commutative and associative. Each also contains a unique identity element. For addition, we call the identity element $0$; for multiplication, we call the identity element $1$ (note that $0 \neq 1$). Formally, we say that $\exists 0 \in \mathbb{R}$ such that $a + 0 = 0 + a = a$, and that $\exists 1 \in \mathbb{R}$ such that $a \times 1 = 1 \times a = a$.

The next property is that every element in $\mathbb{R}$ has an inverse for each operation. (The exception is 0, which has no multiplicative inverse.) The additive inverse of $a \in \mathbb{R}$ is given by $b \in \mathbb{R}$ such that $a + b = 0$, and the multiplicative inverse of $a \in \mathbb{R}$ is given by $b \in \mathbb{R}$ such that $a \times b = 1$. Note that $b$ is unique for a given $a$ for either operation². Consequently, we can define a particular for the additive and multiplicative inverses of $a$, as $-a$ and $\frac{1}{a}$, respectively. Of course, for the multiplicative inverse, the premise is that $a \neq 0$. Note that $-a$ and $\frac{1}{a}$ do not presuppose the existence of subtraction and division; it's simply notation. Also note that we have not yet defined operations with multiple operands: for example, $a + b + c + d$ cannot yet be evaluated.

The last axiom, distributivity, combines the two operations: $a \times (b + c) = (a \times b) + (a \times b)$.

Using these simple properties, we can go on to prove other significant properties of $\mathbb{R}$.

2.1.1Multiplying by zero¶

Prove that $\forall a \in \mathbb{R}, a \times 0 = 0$.

Using the identity property, we know that $0 + 0 = 0$. So we can rewrite $a \times 0$ as $a \times (0 + 0)$. Using the distributivity property, we can rewrite that as $a \times 0 + a \times 0$. So we have:

$$a \times 0 = a \times (0 + 0) = a \times 0 + a \times 0$$

We can then add $-(a \times 0)$ to both sides (the expressions on the far left and the far right), resulting in:

$$a \times 0 + -(a \times 0) = (a \times 0 + a\times 0) + -(a \times 0)$$

Using the fact that $-(a\times 0)$ is the additive inverse of $a \times 0$, which we defined above, we can simplify the left side into 0. For the right side, we use associativity to rewrite the expression as $a \times 0 + (a\times 0 + -(a\times 0))$. The inner expression simplifies to 0, using the same additive inverse property used before, so we have

$$0 = a \times 0 + 0$$

Using the fact that $0$ is the identity element, we have $0 = a \times 0$. $\blacksquare$

2.1.2Subtraction and division¶

Now, we can define two new operations:

• subtraction, defined as $a - b = a + (-b)$ (essentially shorthand notation for adding by the additive inverse)
• division, defined as $a / b = a \times (1/b)$, given $b \neq 0$.

2.1.3The multiplicative inverse of zero¶

Prove that zero does not have a multiplicative inverse - i.e., that there does not exist a $b$ such that $b \times 0 = 1$.³

We can prove this by contradiction. Assume that there does exist a $b$ such that $b \times 0 = 1$. Now, since $0$ is the additive identity, we know that $0 + 0 = 0$ (as in the previous proof). So we can rewrite $b\times 0$ as $b \times (0 + 0) = 1$. Using distributivity, we can rewrite this again as $(b \times 0) + (b \times 0) = 1$. If we subtract $(b \times 0)$ from both sides, we have:

$$((b\times 0) + (b\times 0)) - (b\times 0) = 1 - (b\times 0)$$

Using associativity and additive inverses, we can simplify the left side as follows: $(b\times 0) + ((b\times 0) - (b\times 0)) = (b\times 0)$. So we end up with:

$$(b\times 0) = 1 - (b\times 0)$$

Since our presupposition is that $(b\times 0) = 1$, we can replace $(b\times 0)$ by 1, resulting in:

$$1 = 1 - 1 = 0$$

So we have $1 = 0$, which contradicts the axiom that $1 \neq 0$, and so there does not exist a $b$ such that $b \times 0 = 1$. $\blacksquare$

2.1.4Multiplying by a negative number¶

Prove that $(-a) \times b = -(a \times b)$.

(Note that we can't just use the fact that $(-a) \times (-b) = ab$, we haven't proven this yet.)

Here's the proof:

$\begin{align*} (a\times b) + ((-a) \times b) & = (b\times a) + (b \times (-a)) \tag{by commutativity} \\ & = b \times ((-a) + a) \tag{by distributivity} \\ & = b \times 0 \tag{by the properties of additive inverses} \\ & = 0 \, \blacksquare \tag{by the exercise in section 2.1.1: Multiplying by zero} \end{align*}$

2.2Order properties¶

Now, we will take a look at order properties. I don't really know what else to say about this or how to define "order property", because Google only gives me stuff about lattices and CSS. However, it seems that if we define the set of positive numbers, it will miraculously give birth to the greater-than $>$ relation⁴. So let's define this set by making it an axiom.

2.2.1Defining the positive numbers¶

Axiom. The field $\mathbb{R}$ has the following extra property: there is a subset, $\mathcal{P}$, such that:

• $\forall a, b \in \mathcal{P}$, $(a + b) \in \mathcal{P}$ (i.e., it is closed under addition); and
• $\forall a, b \in \mathcal{P}$, $(a \times b) \in \mathcal{P}$ (i.e., it is closed under multiplication).

Axiom. $\forall \alpha \in \mathbb{R}$, exactly one of the following cases is true:

• $\alpha \in \mathcal{P}$; or
• $-\alpha \in \mathcal{P}$; or
• $\alpha = 0$.

We call the elements of $\mathcal{P}$ positive numbers.

2.2.2The greater-than relation¶

We can now define the greater-than ($>$) relation in terms of $\mathcal{P}$:

If $a, b \in \mathbb{R}$, $a > b$ means that $a - b \in \mathcal{P}$.

In detail: $a > 0$ means $a - 0 \in \mathcal{P}$, and so $a \in \mathcal{P}$.

So $a > b$ means that $a - b \in \mathcal{P}$, and so $(a - b) > 0 \iff (a - b) \in \mathcal{P}$.

2.2.2.1Adding to both sides of the inequality¶

Prove that if $a > b$ and $c \in \mathbb{R}$, then $(a + c) > (b + c)$. In other words, prove that adding the same number to both sides of a greater-than inequality preserves the properties of the inequality (i.e., it doesn't change its veracity).

Essentially, what we have to prove is that $(a + c) - (b + c) \in \mathcal{P}$. Through some basic algebraic manipulations⁵, we can simplify that to $(a -b)$. But we know that $(a - b) \in \mathcal{P}$ because we know that $a > b$. So that's it. $\blacksquare$

2.2.2.2Multiplying both sides of the inequality¶

Prove that if $a > b$ and $c \in \mathcal{P}$, then $ac > bc$.

This "operating on both sides of the inequality" thing doesn't hold as strongly for multiplication as it does for addition - we need the additional assumption that we're multiplying by a positive number. To prove that it holds in this case, we need to show that $(ac - bc) \in \mathcal{P}$. Now, using distributivity, we can rewrite $(ac - bc)$ as $c(a-b)$. So we need to show that $c(a-b) \in \mathcal{P}$. But we know that both $c$ and $(a-b) \in \mathcal{P}$ (from the premises), and since $\mathcal{P}$ is closed under multiplication, then it must be that $c(a-b) \in \mathcal{P}$. $\blacksquare$

2.2.2.3Proving that one is greater than zero¶

Prove that $1 > 0$.⁶

From the second axiom in section 2.2.1, we have that exactly one of the following cases is true:

1. $1 \in \mathcal{P}$; or
2. $-1 \in \mathcal{P}$; or
3. $1 = 0$.

Now, if the first case is true, then it's trivial and our proof is finished. So we move on to the second case. If $-1 \in \mathcal{P}$, then, by closure under multiplication, we must have that $(-1) \times (-1) \in \mathcal{P}$, i.e., $1 \in \mathcal{P}$. It cannot be true that both $1$ and $-1$ are members of $\mathcal{P}$, by the aforementioned axiom; consequently, case 2 is not possible. The last case is trivially false (as $1 \neq 0$ is an axiom, mentioned in section 2.1.

As cases 2 and 3 are not possible, we conclude that case 1 must be true and so $1 > 0$. $\blacksquare$