1.1 The real numbers and completeness

Why the real numbers need more than arithmetic

The rational numbers support the familiar ordered arithmetic, yet the set $\{q\in\mathbb Q:q>0,\ q^2<2\}$ has no rational least upper bound. The real number system adds a completeness property: bounded nonempty sets have least upper bounds. This is the source of many existence results later in analysis, not simply another rule for manipulating inequalities.

Definition

Upper bound and supremum

For a nonempty $A\subset\mathbb R$, $u$ is an upper bound if $a\le u$ for every $a\in A$. A number $s$ is $\sup A$ if it is an upper bound and $s\le u$ for every upper bound $u$. A maximum is an element of $A$ that is also an upper bound. The maximum, when it exists, is the supremum; a supremum need not belong to $A$. Lower bounds, infimum and minimum are defined by reversing inequalities.

Theorem

Completeness axiom

Every nonempty subset of $\mathbb R$ that is bounded above has a real supremum. Equivalently, every nonempty subset bounded below has a real infimum, via $\inf A=-\sup(-A)$.

For example, $\sup(0,1)=1$ although $1$ is not a maximum of that set. Conversely, $\sup[0,1]=1$ and it is a maximum. The empty set and an unbounded-above set do not have a real supremum under this definition.

Theorem

Supremum approximation criterion

An upper bound $s$ of a nonempty set $A$ is $\sup A$ if and only if, for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a\le s$.

Proof

Proving both directions

If $s=\sup A$ and no element exceeds $s-\varepsilon$, then $s-\varepsilon$ is a smaller upper bound, contradiction. Conversely, suppose $s$ is an upper bound and the approximation property holds. If another upper bound $u$ were smaller than $s$, choose $\varepsilon=s-u>0$. The property gives $a>u$, contradicting that $u$ is an upper bound. Thus $s$ is the least upper bound.

Worked example

Use both halves of a supremum proof

Let $A=\{n/(n+1):n\in\mathbb N, n\ge1\}$. Every element is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>1/\varepsilon$ by the Archimedean property below. Then $1-n/(n+1)=1/(n+1)<\varepsilon$. Thus $\sup A=1$. No element equals $1$, so $A$ has no maximum. An upper-bound calculation alone would not prove leastness: $2$ is also an upper bound.

Archimedean property and density

Theorem

Archimedean property

For every real $x$ there exists a positive integer $n>x$. Consequently, for every $\varepsilon>0$ there is $n\in\mathbb N$ such that $1/n<\varepsilon$.

Proof

Completeness forces integers to be unbounded

If the positive integers were bounded above, completeness would give $s=\sup\mathbb N$. Since $s-1$ is not an upper bound, some $n\in\mathbb N$ satisfies $n>s-1$. But $n+1>s$ and $n+1\in\mathbb N$, contradicting that $s$ bounds the set. Apply this to $1/\varepsilon$ to obtain the second statement.

Between any real numbers $x<y$ there is a rational number. Indeed choose $n$ with $n(y-x)>1$; the Archimedean property and the integer-part property yield an integer $m$ with $nx<m<ny$, so $x<m/n<y$. The integer-part property follows from the unboundedness of integers in both directions and the well-ordering of the natural numbers. An irrational also lies between $x$ and $y$: choose a rational $q$ with $x-\sqrt2<q<y-\sqrt2$; then $q+\sqrt2$ is irrational and lies in the interval. Thus density does not say rationals and irrationals are equal sets.

Worked example

An infimum via reflection

For $B=\{2+1/n:n\ge1\}$, every element exceeds $2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$ to obtain $2<2+1/n<2+\varepsilon$. Therefore $2$ is the infimum, but not a minimum. The maximum is $3$, attained at $n=1$.

Proof strategy comparison

To identify a supremum, prove both upper boundedness and approximation from below. To show existence without computing a value, prove the set nonempty and bounded, then invoke completeness. To disprove a maximum, show the candidate supremum is absent; do not confuse that with nonexistence of a supremum. An $\varepsilon$ must be arbitrary and the chosen element may depend on it.

Practice

1. Find the supremum, infimum, maximum and minimum, where they exist, of $A=(-2,3]$.
2. Prove $\sup\{1-2/n:n\ge1\}=1$.
3. If $A$ is nonempty and bounded below, prove $\inf A=-\sup(-A)$.
4. Show there is a rational in $(\sqrt2,\sqrt2+0.01)$ without giving a decimal expansion of $\sqrt2$.

Full solutions

1. $\sup A=3=\max A$ and $\inf A=-2$; there is no minimum because $-2\notin A$.
2. Every term is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>2/\varepsilon$. Then $1-\varepsilon<1-2/n<1$. The supremum criterion gives $1$.
3. Since $A$ is bounded below, $-A$ is nonempty and bounded above. Put $s=\sup(-A)$. For every $a\in A$, $-a\le s$, so $-s\le a$. If $l$ is any lower bound of $A$, then $-l$ is an upper bound of $-A$, hence $s\le-l$, or $l\le-s$. Thus $-s$ is the greatest lower bound.
4. The endpoints are real and ordered, so rational density gives such a rational. Constructively choose $n>100$ and an integer $m$ with $n\sqrt2<m<n\sqrt2+1$; then $m/n$ is in the interval.

Synthesis

Completeness guarantees a least upper bound; approximation shows why it is least. Archimedean estimates allow an integer to be chosen large enough for a desired error, and density provides rationals and irrationals in every nonempty open interval. These ideas will power sequence convergence and continuity proofs.