The Foundation of Real Analysis

You may think that constructing the real numbers is the foundation of real analysis. Theoretically, that is true. How can you analyse the real numbers if there aren’t real numbers to begin with?

The practical foundation of real analysis, however, is more properly seen as the notion of limits. Why is that so? The Cauchy construction of real numbers boils down to creating the real numbers using the limits (or lack thereof) of sequences of rational numbers, so we could have started here if we wanted to.

So let’s start by talking about sequences, then limits. For limits, we will need to measure the notion of “closeness”, and that will require us to define the absolute value function. For real analysis, we will adopt the convention $\mathbb N = \mathbb N^+$ so that $0 \notin \mathbb N$ .

Definition 1. Let $K$ be a set. A $K$ -sequence is a function $x : \mathbb N \to K$ . Commonly, we denote $x_n \equiv x(n)$ for each $n$ , and $x \equiv \{x_n\}$ . A real-valued sequence is an $\mathbb R$ -sequence.

When $\mathbb K$ is a field, we inherit many properties of functions into a field.

Theorem 1. Let $\mathbb K$ be a field. For $\mathbb K$ -sequences $x,y$ and any number $k \in \mathbb K$ , $x + y$ , $kx$ are $\mathbb K$ -sequences. In particular, $x-y := x+(-y)$ is a $\mathbb K$ -sequence.

We can also define division of sequences, but care needs to be taken to handle zeroes. We will define the reciprocal sequence in one way for simplicity.

Theorem 2. Let $x$ be a $\mathbb K$ -sequence. Suppose there exists $N \in \mathbb N$ such that for $n > N$ , $x_n \neq 0$ . Then we can define the sequence $y : \mathbb N \to \mathbb K$ by

$y_n := \begin{cases} 0 &\quad \text{if}\ n \leq N, \\ 1/x_n & \quad \text{if}\ n > N,\end{cases}$

such that for $n > N$ , $x_ny_n = 1$ .

We also want to talk about limits. Roughly speaking, we want to write $x_n \to L$ to mean $x_n \approx L$ when $n$ is large. To define $\approx$ in this context requires a notion of closeness (of which one of its most general forms is described using topology; but not in this discussion).

To do that, our tool of choice will be the absolute value function. For the rest of this post, we will let $\mathbb K$ be an ordered field.

Lemma 1. For any $x \in \mathbb K$ , $x^2 \geq 0$ . Furthermore, if $x \neq 0$ , then $x^2 > 0$ . In particular, $1 > 0$ .

Proof. By the antisymmetry of $\leq$ , $x\geq 0$ or $x \leq 0$ . In the former case,

$x^2 = x \cdot x \geq 0 \cdot 0 = 0.$

In the latter case, $x \leq 0$ implies $-x \geq 0$ so that

$x^2 = (-x)^2 \geq 0.$

Finally, if $x \neq 0$ , then

$\displaystyle x^2 = 0\quad \Rightarrow \quad x = \frac 1x \cdot x^2 = \frac 1x \cdot 0 = 0,$

a contradiction. Thus, $x \neq 0 \Rightarrow x^2 \neq 0$ . Combined with $x^2 \geq 0$ , we have $x^2 > 0$ . For the final result, $1 = 1\cdot 1 = 1^2 > 0$ .

Definition 2. The absolute value function $\mathbb | \cdot | : \mathbb K \to \mathbb K_{\geq 0}$ is the function defined by

$|x| := \begin{cases}x &\quad \text{if}\ x \geq 0, \\ -x &\quad \text{if}\ x < 0.\end{cases}$

There are many intriguing properties involving the absolute value function, which we list in the theorem below.

Theorem 3. The absolute value function satisfies the following properties:

For any $x \in \mathbb K$ , $|x| \geq 0$ .
For any $x \in \mathbb K$ , $|x| = 0\Rightarrow x = 0$ .
For any $x,y \in \mathbb K$ , $|xy| = |x||y|$ .
(Triangle Inequality) For any $x,y \in \mathbb K$ , $|x+y| \leq |x| + |y|$ .

Proof. For the first property, $x \geq 0 \Rightarrow |x| = x \geq 0$ and

$x \leq 0 \quad \Rightarrow \quad |x| = -x \geq -0 = 0.$

For the second property, $x^2 = |x|^2 = 0^2 = 0 \Rightarrow x = 0$ .

The third property is an exercise in case-splitting (there are four cases to compute), which we omit. For the fourth property, we first observe that for any $x,y \in \mathbb K$ ,

$-|x| \leq x \leq |x|,\quad -|y| \leq y \leq |y|.$

Adding the inequalities,

$-(|x| + |y|) \leq x + y \leq |x| + |y|\quad \Rightarrow \quad |x+y| \leq |x| + |y|$

by case-splitting.

Corollary 1. We have as a consequence the following useful properties of the absolute value function:

For any $x \in \mathbb K$ , $|{-x}| = |x|$ .
For any $x,y \in \mathbb K$ , $x=y$ if and only if for any $\epsilon \in \mathbb K^+$ , $|x-y| < \epsilon$ .

Proof. The first property follows from

$|{-x}| = |(-1)\cdot x| = |{-1}||x| = -(-1) \cdot |x| = 1 \cdot |x| = |x|.$

For the second property, first assume $y = 0$ . The direction $(\Rightarrow)$ is immediate. For the direction $(\Leftarrow)$ , suppose for any $\epsilon > 0$ that

$|x| < \epsilon \quad \Rightarrow \quad |x| \leq \epsilon = 0 + \epsilon.$

This implies $|x| \leq 0$ . By definition of $|\cdot |$ , $|x| \geq 0$ . Hence,

$0 \leq |x| \leq 0 \quad \Rightarrow \quad |x| = 0 \quad \Rightarrow \quad x = 0.$

For the general case,

$x = y \quad \iff \quad x-y = 0 \quad \iff \quad \forall \epsilon > 0\quad |x-y| < \epsilon.$

With a sufficiently robust idea of the absolute value function, we are now ready to define the limit of a sequence, if it exists.

Definition 3. Let $\{x_n\}$ be a $\mathbb K$ -sequence. For any $L \in \mathbb K$ , we write $x_n \to L$ to mean that

$\forall \epsilon \in \mathbb K^+ \quad \exists N \in \mathbb N:\quad n>N\quad \Rightarrow \quad |x_n - L| <\epsilon.$

We say that $\{x_n\}$ converges in $K \subseteq \mathbb K$ if there exists $L \in K$ such that $x_n \to L$ .

Actually, a very useful perspective to adopt is to view sequences with limits in terms of sequences with limit $0$ .

Theorem 4. Let $x_n$ be a $\mathbb K$ -sequence. Then $x_n \to 0 \iff |x_n| \to 0$ . More generally, $x_n \to L\iff |x_n - L| \to 0$ .

Proof. Use the observation that $|x| \geq 0$ implies $||x|| = |x|$ .

Let’s do arithmetic with sequences, but start with limit $0$ for simplicity (we’ll perform some tricks later on to handle general limits).

Now for sequences with other kinds of limits, how do we know they are unique? For $\mathbb K$ -sequences it turns out that limits must be unique.

Lemma 2. For $k \in \mathbb K$ , define the constant sequence $\{k_n\} \equiv \{k\}$ by $k_n : \mathbb N \to \mathbb K$ , $k_n =k$ . Then $k_n \to 0$ implies $k = 0$ .

Proof. Fix $\epsilon > 0$ . Since $k_n \to 0$ , there exists $N \in \mathbb N$ such that $n > N$ implies

$|k| = |k-0| = |k_n-0| < \epsilon.$

This implies $k = 0$ , as required.

Lemma 3. Let $\{x_n\}$ , $\{y_n\}$ be $\mathbb K$ -sequences such that $x_n \to 0$ and $y_n \to 0$ and $k \in \mathbb K$ . Then

$x_n + y_n \to 0,\quad k x_n \to 0,\quad x_n - y_n \to 0,\quad x_n y_n \to 0.$

Proof. Fix $\epsilon > 0$ . Since $x_n \to 0$ , for any $k_1 > 0$ , there exists $N_1(k_1) \in \mathbb N$ such that

$n > N_1(k_1) \quad \Rightarrow \quad |x_n| < k_1 \epsilon.$

Since $y_n \to 0$ , for any $k_2 > 0$ , there exists $N_2(k_2) \in \mathbb N$ such that

$n > N_2(k_2) \quad \Rightarrow \quad |y_n| < k_2 \epsilon.$

To prove all of these limit laws, the key is to particularise $k_1 >0,k_2 >0$ so that the result is bounded by $\epsilon$ . For the first identity, use the triangle inequality so that for $n > N_1(k_1) + N_2(k_2)$ ,

$|x_n + y_n| \leq |x_n| + |y_n| < k_1 \epsilon + k_2\epsilon = (k_1 + k_2)\epsilon.$

Thus, stipulating $k_1=k_2=1/2$ does the trick. Then there will be corresponding $n > N_1(1/2) + N_2(1/2)$ such that

$|x_n + y_n| < (k_1+k_2)\epsilon = (1/2 + 1/2)\epsilon = \epsilon.$

For the second identity, the result is obvious if $k = 0$ . If $k \neq 0$ , then $|k| > 0$ so that setting $k_1 = 1/|k| >0$ yields

$\displaystyle n > N_1 \quad \Rightarrow \quad|k x_n| = |k||x_n| < |k| \cdot \frac{1}{|k|} \cdot \epsilon = \epsilon.$

For the third identity,

$x_n - y_n = x_n + (-1) \cdot y_n \to 0 + (-1) \cdot 0 = 0.$

For the fourth identity, we aren’t assuming that $\mathbb K = \mathbb R$ , and so will avoid the use of $\sqrt{\cdot}$ , which may or may not be well-defined. Instead, we set $k_1 =1/\epsilon, k_2 = 1$ so that for $n > N_1(1/\epsilon)+ N_2(1)$ ,

$|x_n y_n| = |x_n| |y_n| < ( k_1 \epsilon ) \cdot ( k_2 \epsilon ) = 1 \cdot \epsilon = \epsilon .$

The squeeze theorem is a ridiculously useful limit property.

Lemma 4. Let $\{x_n\}$ , $\{y_n\}$ , $\{z_n\}$ be $\mathbb K$ -sequences such that $x_n \to 0$ and $y_n \to 0$ . Suppose there exists $N \in \mathbb N$ such that

$n > N \quad \Rightarrow \quad x_n \leq z_n \leq y_n.$

Then $z_n \to 0$ .

Proof. Fix $\epsilon > 0$ . Since $x_n \to 0$ , for any $k_1 > 0$ , there exists $N_1(k_1) \in \mathbb N$ such that

$n > N_1(k_1) \quad \Rightarrow \quad |x_n| < k_1 \epsilon \quad \Rightarrow \quad x_n \geq -|x_n| > -k_1 \epsilon.$

Since $y_n \to 0$ , for any $k_2 > 0$ , there exists $N_2(k_2) \in \mathbb N$ such that

$n > N_2(k_2) \quad \Rightarrow \quad |y_n| < k_2 \epsilon \quad \Rightarrow \quad y_n \leq |y_n| < k_2 \epsilon.$

Setting $k_1 = k_2 = 1$ , for $n > N + N_1(1) + N_2(1)$ ,

$-\epsilon < x_n \leq z_n \leq y_n < \epsilon \quad \Rightarrow \quad -\epsilon < z_n < \epsilon\quad \Rightarrow \quad |z_n| < \epsilon.$

Therefore, $z_n \to 0$ , as required.

Now we want to talk about general limits. Let’s first show that limits in ordered fields must be unique.

Theorem 5. Let $\{x_n\}$ be a $\mathbb K$ -sequence. Suppose $x_n \to L_1$ and $x_n \to L_2$ . Then $L_1 = L_2$ . This allows us to write

$\displaystyle x_n \to L \quad \iff \quad \lim_{n \to \infty}x_n = L$

without ambiguity, whenever such an $L$ exists.

Proof. Observe that $L_1 - x_n = (-1) \cdot (x_n - L) \to 0$ and $x_n - L_2 \to 0$ so that

$L_1 - L_2 = (L_1 - x_n) + (x_n - L_2) \to 0.$

Since $L_1,L_2$ are constants, $L_1 - L_2 = 0$ so that $L_1 = L_2$ .

This brings us to the star of any analysis on limits: the limit laws.

Theorem 6 (Limit Laws). Let $\{x_n\},\{y_n\},\{z_n\}$ be $\mathbb K$ -sequences. Suppose $x_n \to x$ , $y_n \to y$ , and $k \in \mathbb K$ . Then

$x_n \pm y_n \to x \pm y,\quad kx_n \to kx,\quad x_ny_n \to xy.$

Furthermore, if there exists $N \in \mathbb N$ such that

$n > N \quad \Rightarrow \quad x_n \geq 0,$

then $x \geq 0$ . Finally, if $x = y = z$ and there exists $N \in \mathbb N$ such that

$n > N \quad \Rightarrow \quad x_n \leq z_n \leq y_n,$

then $z_n \to z$ .

Proof. The first set of limit laws can be proven algebraically:

$\begin{aligned} (x_n \pm y_n) - (x \pm y) &= (x_n - x) \pm (y_n - y) \to 0 \pm 0 = 0, \\ kx_n - kx &= k (x_n - x) = k \cdot 0 = 0, \\ x_n y_n - xy &= (x_n - x + x) \cdot (y_n - y + y) - xy \\ &= (x_n - x)(y_n - y) + y (x_n - x) + x(y_n - y)\\ &\to 0 + 0 + 0 = 0,\end{aligned}$

as required. The squeeze theorem also works since we can apply the zero-limit case to

$x_n - z \leq z_n - z \leq y_n - z$

so that $x_n - z \to 0, y_n - z \to 0$ implies $z_n - z \to 0$ .

For the monotonicity property, fix $\epsilon > 0$ . Since $x_n \to x$ , find $N_1 \in \mathbb N$ so that for $n > N_1$ ,

$|x_n - x| < \epsilon \quad \Rightarrow \quad x_n < x+\epsilon.$

For $n > N + N_1$ , $0 \leq x_n < x+\epsilon$ . Since $\epsilon$ is arbitrary, $0 \leq x$ , as required.

There is one more question worth asking: if $x_n \to x$ , does it always hold that $1/x_n \to 1/x$ ? Well, if $x = 0$ , then we’ll run into problems, and if $x \neq 0$ , this result is plausibly true. You’ll be pleased to know that this result is true. However, this exploration is worth its own exercise since we won’t be needing it as much in our real-analytic discussion.

—Joel Kindiak, 19 Dec 24, 1508H

KindiakMath

The Foundation of Real Analysis

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The Foundation of Real Analysis

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