Introduction to Real Analysis (1.4.2) - Sequences, Series and the Real Numbers (Nested Interval Theorem, Equivalence Classes, Definition of the real numbers)

At this point the title of this blog has become something of a running gag (for myself and the one odd viewer that occasionally encounters this website).

As much as I would like to actually maintain an average of $\displaystyle \frac{3}{2}$ posts per day, I haven't been able to because of high school work and other obligations.

In any case, I'll try and make a conscious effort to do so, and if it becomes glaringly obvious that such an average isn't, and never will be feasible to maintain, I'll change the title to something like:

$\displaystyle 0 < average \ posts \ per \ day \leq \frac{3}{2}$

$\square$

This post will cover several things. Two important theorems will be proved, and then several definitions pertaining to the real numbers will be provided.

First, we will pretend that the real numbers already exist, and possess all the qualities we expect the real numbers to be endowed with. In other words, through whatever means we may have used in order to construct the set of real numbers, the set of all real numbers is a complete ordered field.

All this means is that non-empty subsets of the real numbers satisfy the 'least upper bound' or 'completeness' property, that all the laws of inequality hold, and that the real numbers are equipped with addition and multiplication and all the good stuff.

The 'completeness property' is formally stated below:

$\displaystyle (Completeness \ Property) -$ Let $S \subset \mathbb{R}$ be non-empty. Furthermore, suppose there exists some number $b \in \mathbb{R}$ such that $a \leq b$ for all $a \in S$. Then there must exist an element $\xi \in \mathbb{R}$ such that $\xi$ is a least upper bound of $S$, meaning that $\xi \geq a$ for all $a \in S$ and for any other upper bounds of $S$ (numbers $b \in \mathbb{R}$ such that $b \geq a$ for all $a \in S$), $\xi \leq b$.

$\square$

In my previous post, I ended on:

 ''Imagine the 'Real Number' line was in front of you. Now mark a point on the line, and another point to the right of this point with a left and right closed bracket respectively. Continue to do this, but make sure that all new left closed brackets you mark are to the right of previously marked left closed brackets, and that all right closed brackets you mark are to the left of previously marked right closed brackets, and that there are no left closed brackets to the right of right closed brackets. What you should begin to notice after a while, is that both closed brackets begin to 'close in' to something.''

$\square$

Lets take this and turn it into a theorem, as promised earlier.

$\displaystyle (Nested \ Interval \ Theorem)$ - Let $I_n = [a_n,b_n]$ (where $a_n < b_n$ and $a_n,b_n \in \mathbb{R}$) denote a closed interval. Suppose $I_{n+1} \subset I_n$ for all $n \in \mathbb{N}$. 

Then we may consider the sequence of such intervals $\{I_n\}_{n=0}^{\infty}$

If all of the criterion above are satisfied, the intersection of all of these intervals is non-empty. 

That is,

 $\bigcap_{i=0}^{\infty} I_n \neq \emptyset$

$\square$

Lets translate what this theorem is saying by comparing it to the description in my previous post. 

Continuing to mark brackets so that left closed brackets are to the right of previously marked left closed brackets and right closed brackets are the the left of previously marked right closed brackets along with the fact that a left closed bracket may never be to the right of a right closed bracket does nothing but ensure that $I_{n+1} \subset I_n$ for all $n \in \mathbb{N}$ . 

Furthermore what these closed brackets (which is just the sequence of nested intervals $\{I_n\}_{n=0}^{\infty}$, mentioned in the aforementioned theorem) are 'closing in' on is simply an element in $\bigcap_{i=0}^{\infty} I_n \neq \emptyset$, which as the theorem asserts, is non-empty.

As it turns out, we can prove the Nested Interval Theorem with nothing but the Completeness property of the real numbers. 

$\square$

$\displaystyle Proof \ (Nested \ Interval \ Theorem)$ - We will denote any sequence $\{k_n\}_{i=0}^{\infty}$ as $\{k_n\}$ from now on. 

Consider the sequence $\{I_n\}$ of nested intervals. As $I_{n+1} \subset I_n$ it is clear that the left endpoints of $\{I_n\}$ form a non-strictly increasing sequence, we will call this sequence $\{a_n\}$, and the right endpoints of $\{I_n\}$ form a non-strictly decreasing sequence, we will call this sequence $\{b_n\}$.

That is, $a_0 \leq a_1 \leq ...$ and $b_0 \geq b_1 \geq b_2 \geq ...$

We will show that $a_i \leq b_j$ for any natural numbers $i$ and $j$.

Suppose $a_i > b_j$, if $i>j$ we know that $b_i>a_i>b_j$ this implies $b_i > b_j$ , yielding a contradiction, as $\{b_n\}$ is non-strictly decreasing.

If $i<j$ then $a_j < b_j < a_i$, again yielding a contradiction, as $\{a_n\}$ is non-strictly increasing.

Therefore $i = j$. But if $i = j$ then $a_i < b_i$ as $I_i$ is a non-empty closed interval, this yields a final contradiction.

Thus $a_i \leq b_j$ for any natural $i$ and $j$.

This means that the set of all terms of the sequence $\{a_n\}$ is bounded above, which implies (by the completeness/least upper bound property) that there exists a least upper bound of $\{a_n\}$, we will call this least upper bound $\xi$.

So we know that $\xi \geq a_n$ for all natural $n$.

We will now show that $\xi \leq b_n$ for all natural $n$.

Suppose $\xi > b_j$ for some natural number $j$.

But $a_n$ is less than $b_j$ for all natural $n$, as we proved that $a_i \leq b_j$ for any natural numbers $i$ and $j$.

Therefore $b_j$ is an upper bound of $\{a_n\}$.

Then $\xi$ is not the least upper bound of $\{a_n\}$. This yields a contradiction.

Therefore $\xi \leq b_n$ for all natural $n$.

This implies that $\xi \in [a_n,b_n]$ for all natural $n$, and so $\bigcap_{i=0}^{\infty} I_n \neq \emptyset$

$\blacksquare$

So we have proved that $\bigcap_{i=0}^{\infty} I_n \neq \emptyset$ by showing that the least upper bound of the sequence of left endpoints of our intervals is contained within the intersection of all nested intervals. We also know that this least upper bound must exist as per the completeness property of the real numbers.

This motivated me to pose the following question 'We know that working with numbers that possess the completeness property ensures the Nested Interval Theorem is true. Is it then possible to start with the Nested Interval Theorem and prove the completeness property?'

Read as stated above, the question is crude and isn't coherent. What I was precisely wondering was, since we can 'narrow down' on real numbers using the Nested Interval Theorem, can we just 'define' real numbers as collections of Nested Intervals that get smaller and smaller and thus seemingly 'narrow down' on the same \ fixed point?

The word 'collection' in the paragraph above is also crude, which is why we will now introduce the notion of 'equivalence classes' in order to make this precise.

Before we introduce equivalence classes, we will define what is called an 'equivalence relation' on a non-empty set $S$.

$Definition \ (Equivalence \ Relation ) -$ A relation $R$ on a set $S$ is defined as any non-empty subset of the set $SXS = \{ (a,b) | a \in S , b \in S \}$

If $(a,b) \in R$ we write $aRb$. That is $aRb$ if and only if $(a,b) \in R$.

An equivalence relation on $S$ is a relation on $S$ that satisfies the following properties:

(Reflexive) $aRa$ for all $a \in S$

(Symmetric) If $aRb$ then $bRa$ for any $a , b \in S$

(Transitive) If $aRb$ and $bRc$ then $aRc$ for any $a, b, c \in S$

$\square$

Now lets get back to defining equivalence classes:

$Definition \ (Equivalence \ Class) -$ An equivalence class for an equivalence relation $R$ on a non-empty set $S$ is the set $[a] = \{ x \in S | xRa \}$

We will also present an important theorem about equivalence relations and equivalence classes:

$Theorem (Equivalence \ relation \ creates \ equivalence \ class \ partition ) -$ An equivalence relation $R$ on the non-empty set $S$ partitions $S$ into disjoint equivalence classes

The proof is also presented below:

$Proof \  ( aforementioned \ theorem ) -$ Consider the equivalence classes $[a]$ and $[b]$ . Clearly these classes are non-empty, as $aRa$ and $bRb$.

Now, suppose $x \in [a]$ and $x \in [b]$. Then $xRa$ and $xRb$. This implies that $aRx$ (by symmetry), which implies that $aRb$ (by transitivity).

Therefore, for any $y \in [a]$, that is, for any $yRa$, $yRb$ and $y \in [b]$. Thus given $[a] \cap [b] \neq \emptyset$ , $[a] \subset [b]$.

We can prove (symmetrically) that $[b] \subset [a]$. Therefore, if $[a] \cap [b] \neq \emptyset$, it follows that $[a] = [b]$.

The theorem thus follows.

$\blacksquare$

Finally, we will end this post by presenting a definition of 'real number'.

$\square$

Before we begin to define real numbers, a distinction between the 'limit' we are used to and the 'limit' we will be talking about in this post and the posts to come will need to be made.

In the context of this construction of the real numbers, $lim_{n \to \infty} a_n = L$ where $L \in \mathbb{Q}$ means that for any positive $\epsilon \in \mathbb{Q}$, there exists a $M \in \mathbb{N}$ such that if $n>M$, $|a_n - L| < \epsilon$. We also assume that $a_n \in \mathbb{Q}$ for all $n$, as we have not constructed the set of all real numbers yet.

From now on, we will pretend the real numbers do not exist. Any mention of 'numbers' will immediately be assumed to be rational.

Now we will move on to a definition:

$Definition \ ( Constructive \ Interval \ Sequence ) -$ We call the sequence of closed intervals with rational endpoints, $I_n = [a_n,b_n]$ for $n \in \mathbb{N}$ a constructive interval sequence if it satisfies the following conditions:

(1) $I_{n+1} \subset I_n$ for all $n \in \mathbb{N}$

(2) $\displaystyle lim_{n \to \infty} (b_n - a_n) = 0$

$\square$

We will also define a relation on the set of all constructive interval sequences. Instead of denoting this relation $R$, we will denote it $=$.

$Definition \ (Equality \ of \ Constructive \ Interval \ Sequences ) -$ Two constructive interval sequences, 
$\{I_n\}$ and $\{K_n\}$, where $I_n = [a_n,b_n]$ and $K_n = [c_n, d_n]$ are said to be 'equal' that is $\{I_n\} = \{K_n\}$ if and only if :

$\displaystyle lim_{n \to \infty} (c_n - a_n) = 0$.

$\square$

We will show that this is indeed an equivalence relation.

First we will show that this relation is reflexive. Clearly, $a_n - a_n = 0$. Thus this follows.

Now, Suppose $\{I_n\} = \{K_n\}$. Then for all rational $\epsilon > 0$ there exists a natural number $M$ such that if $n>M$, $|c_n - a_n| < \epsilon$

As $|a_n - c_n| = |c_n - a_n|$ this implies $|c_n - a_n| < \epsilon$.

Thus $\displaystyle lim_{n \to \infty} (a_n - c_n) = 0$, which implies symmetry.

Now, suppose $\{I_n\} = \{K_n\}$ and $\{K_n\} = \{P_n\}$ where $\{P_n\} = [e_n,f_n]$ is another constructive interval sequence.

Then choose $M$ such that $|c_n - a_n| < \frac{\epsilon}{2}$ and $|e_n - c_n| < \frac{\epsilon}{2}$

We know that $|e_n - c_n| = |e_n - c_n + a_n - a_n| \leq |c_n - a_n| + |e_n - c_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

It follows that $\{I_n\} = \{P_n\}$ which implies transitivity.

Thus '$=$' is an equivalence relation on the set of all constructive interval sequences.

This allows us to take advantage of the fact that an equivalence relation on a set creates a partition of equivalence classes. Indeed it is these equivalence classes that we will define as 'real numbers'.

$\square$

$Definition \ (Real \ Number) -$ A 'real number' $\alpha$ is an equivalence class of constructive interval sequences.


$\blacksquare$

I will follow up on this definition in the next post, where the properties of the real numbers will be explored, and operations on the real numbers will be defined.