# Construction of the reals via monotonic sequences

This section discusses the construction of the real numbers from the rationals via the idea of a monotonic nondecreasing sequence. This approach would appear to be perhaps the best one of all from our point of view that the monotone convergence theorem is the most natural formulation of the completeness axiom, however it is not the only possible approach.

In fact, there is an important technical problem concerning this choice of using monotonic nondecreasing sequences as a base for the construction of the reals that makes it more difficult to work with negative numbers directly. This is because the additive inverse or negative of a nondecreasing sequence is most naturally a nonincreasing sequence. For this reason, this web page starts with the construction of the nonnegative reals and then gives the construction of the full real number line from its nonnegative part at the end.

Other approaches to constructions of the reals are discussed in other web pages in this section, as linked from here and from the contents page.

This web page is still under construction. The final version will contain some (but not all) of the proofs omitted here at present.

A bounded monotonic nondecreasing sequence of nonnegative rationals is a sequence $( a n )$ such that: (a) each $a n$ is a nonnegative rational number; (b) $a n ⩽ a n + 1$ for all $n ∈ ℕ$; and (c) there is a bound $B ∈ ℚ$ such that $a n ⩽ B$ for all $n ∈ ℕ$.

The set of nonnegative rationals will be denoted $ℚ +$ and the set of nondecreasing sequences of nonnegative rationals will be denoted on this page by $R +$. This set $R +$ is not the set of (nonnegative) reals, but closely related to it. For example, $R +$ contains sequences converging to what we think of as $2$. The problem is that $R +$ contains several such sequences and we want there to be only one $2$. The solution is to use an equivalence relation to make elements of $R +$ the same.

Definition.

For sequences $( a n )$ and $( b n )$ in $R +$, we write $( a n ) ∼ ( b n )$ to mean

The idea is that two monotonic nondecreasing sequences are equivalent if each contains terms greater than terms in the other. In other words, this will mean that they tend to the same real limit. (Notice that we can't talk about this limit as we haven't constructed the real numbers yet so we have to find a roundabout way of saying that they tend to the same real limit.)

Proposition.

The relation $∼$ is an equivalence relation on $R +$. For each $x , y ∈ ℚ$, the constant sequences with values $x$, $y$ respectively are equivalent if and only if $x = y$.

Proof.

The relation is reflexive because given $( a n ) ∈ R +$ we have so .

It is obviously symmetric because the definition of $∼$ was deliberately symmetric in $( a n ) , ( b n )$.

If and then given arbitrary $n ∈ ℕ$ there is $k ∈ ℕ$ such that $a n ⩽ b k$ and $l ∈ ℕ$ such that $b k ⩽ c l$ hence . The other property is proved similarly.

Definition.

Given a sequence $( a n ) ∈ R +$ we write $[ a n ]$ for the equivalence class of $( a n )$. We define $ℝ +$ to be the set of equivalence classes $[ a n ]$ for $( a n ) ∈ R +$. For $x ∈ ℚ +$ we identify $x$ with the constant sequence with value $x$. This means that we can think of $ℝ +$ as containing (a copy of) $ℚ +$.

There is something to prove here, but it is easy: we must show that two constant sequences with values $x , y ∈ ℚ +$ are inequivalent if $x ≠ y$. In other words, if $x ≠ y$ then the reals identified with $x$ and $y$ are different. But this is easy as if $x < y$, $a n = x$, $b k = y$ then it is not true that .

We also have to define the order $<$ on $ℝ$, as well as the usual operations of addition, multiplication, etc.

Definition.

We write $[ a n ] ⩽ [ b n ]$ to mean

We write $[ a n ] ⩾ [ b n ]$ to mean $[ b n ] ⩽ [ a n ]$; and we write $[ a n ] < [ b n ]$ to mean that $[ a n ] ⩽ [ b n ]$ and these equivalence classes are not equal. We also define $[ a n ] > [ b n ]$ to mean $[ b n ] < [ a n ]$. In particular, we write $[ a n ] ⩾ 0$ to mean

Proposition.

The order relation $⩽$ on the set $ℝ +$ is well-defined and satisfies the following.

Proof.

To make sure that this definition makes sense we should check that if $[ a n ] ⩽ [ b n ]$ and $( c n ) ∼ ( a n )$, $( d n ) ∼ ( b n )$ then $[ c n ] ⩽ [ d n ]$. This argument is the same as the argument for transitivity of $∼$ given above: use , , .

We also need to verify at this stage that this order relation on $ℝ +$ is compatible with the usual one on $ℚ +$ via the embedding defined above. This is easy too: if $x ⩽ y ∈ ℚ +$ and $a n = x$, $b n = y$ then it is easy to see that $[ a n ] ⩽ [ b n ]$.

The other axioms are left as exercises for now.

Note that the last part of the above proposition is a version of the Archimedean property for $ℝ +$.

Definition.

The real number $[ a n ] + [ b n ]$ is defined to be the equivalence class of the sequence $c n$ defined by $c n = a n + b n$.

The real number $[ a n ] · [ b n ]$ is defined to be the equivalence class of the sequence $c n = a n · b n$.

Proposition.

The addition and multiplication operations $+,·$ just given are well-defined on $ℝ +$. In particular, the definitions above do not depend on the choice of representative of equvalence class.

Proof.

Exercise.

This defines the nonnegative reals, $ℝ +$ as a structure extending $ℚ +$ with order, $<$, and addition and multiplication stucture. Because we have not defined $-$ (indeed, we cannot define this of $ℝ +$) we are not able to prove the field axioms in $ℝ +$. But most of the obvious algebraic properties are summed up in the next proposition.

Proposition.

The set $ℝ +$ with operations $+,·$ and order relation $<$ satisfy, for all $x , y , z ∈ ℝ +$,

• $( x + y ) + z = x + ( y + z )$
• $x + y = y + x$
• $x + 0 = x$
• $( x y ) z = x ( y z )$
• $x y = y x$
• $x 0 = 0$
• $x ≠ 0 ⇒ x 1 = x$
• $x ( y + z ) = x y + x z$
• $x ⩽ y ⇒ x + z ⩽ y + z$
• $x ⩽ y ⇒ x z ⩽ y z$

Proof.

Exercise.

Although this only defines the nonnegative reals, this is enough to make the idea of a sequence of nonnegative real numbers to make sense, and in particular to interpret the idea of limit in $ℝ +$. When we do this we get,

Proposition.

The set $ℝ +$ has the completeness property, that every monotonic nondecreasing bounded sequence $( a n )$ of numbers $a n ∈ ℝ +$ has a limit $l ∈ ℝ +$.

Proof.

Exercise.

To pass from the nonnegative reals to the full set of reals, one just needs to repeat a construction given earlier when we constructed the $ℤ$ from the $ℕ$ .

Definition.

Let $∼$ be the following relation defined on $ℝ + × ℝ +$: we define $( x 1 , y 1 ) ∼ ( x 2 , y 2 )$ to mean $x 1 + y 2 = x 2 + y 1$.

Proposition.

$∼$ is an equivalence relation on $ℝ + × ℝ +$.

Proof.

Exercise.

Definition.

We write $[ a , b ]$ for the equivalence class of $( a , b ) ∈ ℝ + × ℝ +$. The set $ℝ$ is the set $ℝ + × ℝ + / ∼$ of equivalence classes.

Proposition.

The function $i$ defined by $i ( n ) = [ n , 0 ]$ is a one-to-one function mapping $ℝ +$ into $ℝ$.

Proof.

Let $x , y ∈ ℝ +$ with $i ( x ) = i ( y )$. Then $[ x , 0 ] = [ y , 0 ]$ hence $( x , 0 ) ∼ ( y , 0 )$ hence $x + 0 = y + 0$ hence $x = y$.

The elements $x$ of $ℝ +$ are identified with their images $i ( x ) ∈ ℝ$ in the usual way.

Definition.

We define the arithmetic structure on reals as follows.

1. $[ x 1 , x 2 ] + [ y 1 , y 2 ] = [ x 1 + y 1 , x 2 + y 2 ]$
2. $[ x 1 , x 2 ] - [ y 1 , y 2 ] = [ x 1 + y 2 , x 2 + y 1 ]$
3. $[ x 1 , x 2 ] · [ y 1 , y 2 ] = [ x 1 y 1 + x 2 y 2 , x 2 y 1 + x 1 y 2 ]$
4. $[ x 1 , x 2 ] ⩽ [ y 1 , y 2 ] ↔ x 1 + y 2 ⩽ x 2 + y 1$

Theorem.

These definitions are well-defined. They make $ℝ$ into an Archimedean ordered field satisfying the completeness axiom.

Proof.

Exercise.