We have already seen two equivalent forms of the completeness axiom for the reals: the montone convergence theorem and the statement that every Cauchy sequence has a limit. The second of these is useful as it doesn't mention the order relation $<$ and so applies to the complex numbers for instance. (It turns out that the set of complex numbers is also complete in the Cauchy sense.) This web page describes a third approach due to Dedekind. This uses the order relation $<$ of $\mathbb{R}$ but applies to arbitrary subsets of $\mathbb{R}$ rather than sequences. There are a number of places (including results about the radius of convergence for power series, and several results in more advanced analysis) where Dedekind's approach is particularly helpful.

We start with a straightforward definition similar to many others in this course. Read the definitions carefully, and note the use of $\le $ and $\ge $ here rather than $<$ and $>$.

**Definition 2.1**

Let $A\subseteq \mathbb{R}$.

We say that $A$ is bounded above if there is $b\in \mathbb{R}$ such that $\forall x\in A(x\le b)$. The number $b$ is called an upper bound for $A$.

We say that $A$ is bounded below if there is $c\in \mathbb{R}$ such that $\forall x\in A(x\ge c)$. The number $c$ is called an lower bound for $A$.

If we are interested in the best

upper bound
or best

lower bound we need to consider the following.

**Definition 2.2**

Let $A\subseteq \mathbb{R}$.

We say that $b\in \mathbb{R}$ is a least upper bound for $A$ if $b$ is an upper bound for $A$ and no $c<b$ is also an upper bound. In other words if $\forall x\in A(x\le b)$ and $\forall c<b\exists x\in A(xc)$. The least upper bound of a set $A$ may not exist, but if it does it is unique, because if we have two distince upper bounds $c,d$ then one of these must be larger and so it cannot be a least upper bound. When it exists, the least upper bound of a set $A$ is called the supremum of $A$ and denoted $\text{sup}A$.

We say that $c\in \mathbb{R}$ is a greatest lower bound for $A$ if $c$ is a lower bound for $A$ and no $d>c$ is also a lower bound. In other words if $\forall x\in A(x\ge c)$ and $\forall d>c\exists x\in A(xd)$. The greatest lower bound of a set $A$ may not exist, but if it does it is unique, and is called the infimum of $A$ and denoted $\text{inf}A$.

Supremums and infimums are a bit like maximums and minimums for
*infinite* sets $A$.

The problem with maximums and minimums is that they are only
*guaranteed* to exist for *finite* sets $A$.
For example, $A=\left\{x\in \mathbb{Q}:0\le x\le \sqrt{2}\right\}$
doesn't have a maximum element, though it does have a
minimum element, $0$. On the other hand
$B=\left\{x\in \mathbb{Q}:0\le x\le \sqrt{4}\right\}$
has both a maximum (2) and a minumum (0).

In fact, for a nonempty set $A$, the maximum element of $A$ exists if and only if $\text{sup}A$ exists and is an element of $A$. Similarly, the minimum element of $A$ exists if and only if $\text{inf}A$ exists and is an element of $A$.

**Examples 2.1**

Let $A=\left\{1-\frac{1}{n}:n\in \mathbb{N}\right\}$. Then $A$ has no largest element, i.e., $\text{max}A$ doesn't exist, but $\text{sup}A=1$ since $1$ is an upper bound and any $c<1$ is less than some $1-\frac{1}{n}$ by the Archimedean property. Note that $\text{sup}A$ is not an element of $A$.

Let $A=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}$. Then $A$ has no smallest element, i.e., $\text{min}A$ doesn't exist, but $\text{inf}A=0$ since $0$ is a lower bound and any $d>0$ is greater than some $\frac{1}{n}$ by the Archimedean property. Note that $\text{inf}A$ is not an element of $A$.

Let $A=[2,3]$. In this case $A$ does have largest and smallest elements, and $\text{sup}A=3$ and $\text{inf}A=2$.

Let $A$ be the empty set. Then by convention every $b\in \mathbb{R}$ is both an upper bound and a lower bound for $A$. So $A$ does not have least upper bound or greatest lower bound.

Let $A=\mathbb{N}$. Then $A$ does not have any upper bound, by the Archimedean property. But $A$ does have a lower bound, such as $-1$. The greatest lower bound is determined by your convention on the set of natural numbers. If you prefer $0\in \mathbb{N}$ then $\text{inf}\mathbb{N}=0$. Otherwise you will have $\text{inf}\mathbb{N}=1$.

Let $A=\mathbb{Z}$. Then $A$ does not have any upper bound nor any lower bound, by the Archimedean property again.

**Theorem 2.1**** (Completeness of reals, supremum form)**

Let $A\subseteq \mathbb{R}$ be non-empty and bounded above. Then there is a least upper bound of $A$, $\text{sup}A$.

**Proof**

This is proved by a variation of a proof of an earlier result that every real number has a monotonic sequence of rationals converging to it.

We start by defining a sequence $\left({a}_{n}\right)$ using induction on $n$. Start with any ${a}_{0}\in \mathbb{Q}$ with some $x\in A$ such that ${a}_{0}<x$. This just uses the assumption that $A$ is nonempty. Now inductively assume that ${a}_{n}$ is defined and ${a}_{n}<x$ for some $x\in A$, i.e., ${a}_{n}$ is not an upper bound for $A$. If ${a}_{n}+\frac{1}{n}$ is an upper bound for $A$ then let ${a}_{n+1}={a}_{n}$. Otherwise, let ${a}_{n+1}={a}_{n}+\frac{k}{n}$ where $k={K}_{0}-1\in \mathbb{N}$ and ${K}_{0}$ is chosen to be the least natural number such that ${a}_{n}+\frac{{K}_{0}}{n}\ge x$ for all $x\in A$. Such a $K$ exists since $A$ is bounded above by some $b\in \mathbb{R}$ and we need only take $K\in \mathbb{N}$ so that ${a}_{n}+\frac{K}{n}\ge b$, using the Archimedean Property. So since there is some such $K$ there must be a least such number, ${K}_{0}$.

By construction, $\left({a}_{n}\right)$ is a nondecreasing sequence of rationals and bounded above by $b$, an upper bound for $A$. It follows that $\left({a}_{n}\right)$ converges to some $l\in \mathbb{R}$. We shall show that this $l$ is the least upper bound of $A$.

**Subproof**

Suppose $l$ is not an upper bound of $A$.

**Subproof**

Then there is $x\in A$ such that $l<x$. But this gives a contradiction, for if $n\in \mathbb{N}$ is such that $x-l>\frac{1}{n}$ we consider the $n$th stage of the construction, where we chose ${a}_{n+1}={a}_{n}+\frac{k}{n}$ with $k$ greatest so that there is some $y\in A$ with ${a}_{n}+\frac{k}{n}<y$. But ${a}_{n}+\frac{k}{n}\le l<x-\frac{1}{n}$ so ${a}_{n}+\frac{k+1}{n}\le l+\frac{1}{n}<x$ contradicting this choice of $k$.

Thus $l$ is after all an upper bound. To see it is the least upper bound, we again suppose otherwise.

**Subproof**

Suppose $l$ is not the least upper bound of $A$.

**Subproof**

Then there is some $m<l$ such that every $x\in A$ has $x\le m$. This again is impossible. Since ${a}_{n}\to l$, there must be some ${a}_{n}$ with $m<{a}_{n}\le l$. (To see this, put $\epsilon =l-m$ and ${a}_{n}$ with $\left|{a}_{n}-l\right|<\epsilon $.) But by construction of ${a}_{n}$ there is always some $x\in A$ with ${a}_{n}<x$. Therefore $m<x$ and $m$ is not after all an upper bound for $A$.

This completes the proof.

**Theorem 2.2**** (Completeness of reals, infimum form)**

Let $A\subseteq \mathbb{R}$ be non-empty and bounded below. Then there is a greatest lower bound of $A$, $\text{inf}A$.

**Proof**

Let $c\in \mathbb{R}$ be a lower bound for $A$ and let $B=\left\{-x:x\in A\right\}$. Then $b=-c$ is an upper bound for $B$ and hence by the previous result $B$ has a least upper bound, $l$. It follows easily that $-l$ is a greatest lower bound for $A$, for if $x\in A$ then $l\ge -x$ as $-x\in B$ so $-l\le x$, and if $m>-l$ then $-m<l$ so $-m$ is not an upper bound for $B$, so there is $x\in B$ with $x>-m$ hence $-x<m$ and clearly $-x\in A$.

These results are equivalent to the monotone convergence theorem. To see this, suppose $\left({a}_{n}\right)$ is a bounded nondecreasing sequence. Then let $A=\left\{{a}_{n}:n\in \mathbb{N}\right\}$. By the fact that the sequence is bounded and by completeness theorem above, $l=\text{sup}A$ exists, and it is a nice exercise to show that ${a}_{n}\to l$ as $n\to \infty $.

You have seen another form of the completeness axiom for the reals, which is useful in that it doesn't involve any sequences and may be applied to subsets of $\mathbb{R}$.

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