# Supremum and infimum

## 1Introduction

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 $ℝ$ but applies to arbitrary subsets of $ℝ$ 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.

## 2Supremum and infimum

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

Definition 2.1

Let $A ⊆ ℝ$.

We say that $A$ is bounded above if there is $b ∈ ℝ$ such that . The number $b$ is called an upper bound for $A$.

We say that $A$ is bounded below if there is $c ∈ ℝ$ such that . 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 $A ⊆ ℝ$.

We say that $b ∈ ℝ$ 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 and . 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 $sup A$.

We say that $c ∈ ℝ$ 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 and . 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 $inf A$.

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, $A = x ∈ ℚ : 0 ≤ x ≤ 2$ doesn't have a maximum element, though it does have a minimum element, $0$. On the other hand $B = x ∈ ℚ : 0 ≤ x ≤ 4$ 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 $sup A$ exists and is an element of $A$. Similarly, the minimum element of $A$ exists if and only if $inf A$ exists and is an element of $A$.

Examples 2.1

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

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

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

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

Let $A = ℕ$. 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 $0 ∈ ℕ$ then $inf ℕ = 0$. Otherwise you will have $inf ℕ = 1$.

Let $A = ℤ$. 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 $A ⊆ ℝ$ be non-empty and bounded above. Then there is a least upper bound of $A$, $sup A$.

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 $( a n )$ using induction on $n$. Start with any $a 0 ∈ ℚ$ with some $x ∈ A$ 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 $x ∈ A$, i.e., $a n$ is not an upper bound for $A$. If $a n + 1 n$ is an upper bound for $A$ then let $a n + 1 = a n$. Otherwise, let $a n + 1 = a n + k n$ where $k = K 0 - 1 ∈ ℕ$ and $K 0$ is chosen to be the least natural number such that $a n + K 0 n ≥ x$ for all $x ∈ A$. Such a $K$ exists since $A$ is bounded above by some $b ∈ ℝ$ and we need only take $K ∈ ℕ$ so that $a n + K n ≥ b$, using the Archimedean Property. So since there is some such $K$ there must be a least such number, $K 0$.

By construction, $( a n )$ is a nondecreasing sequence of rationals and bounded above by $b$, an upper bound for $A$. It follows that $( a n )$ converges to some $l ∈ ℝ$. 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 $x ∈ A$ such that $l < x$. But this gives a contradiction, for if $n ∈ ℕ$ is such that $x - l > 1 n$ we consider the $n$th stage of the construction, where we chose $a n + 1 = a n + k n$ with $k$ greatest so that there is some $y ∈ A$ with $a n + k n < y$. But $a n + k n ≤ l < x - 1 n$ so $a n + k + 1 n ≤ l + 1 n < x$ 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 $x ∈ A$ has $x ≤ m$. This again is impossible. Since $a n → l$, there must be some $a n$ with $m < a n ≤ l$. (To see this, put $ε = l - m$ and $a n$ with $a n - l < ε$.) But by construction of $a n$ there is always some $x ∈ A$ 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 $A ⊆ ℝ$ be non-empty and bounded below. Then there is a greatest lower bound of $A$, $inf A$.

Proof

Let $c ∈ ℝ$ be a lower bound for $A$ and let $B = - x : x ∈ A$. 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 $x ∈ A$ then $l ≥ - x$ as $- x ∈ B$ so $- l ≤ x$, and if $m > - l$ then $- m < l$ so $- m$ is not an upper bound for $B$, so there is $x ∈ B$ with $x > - m$ hence $- x < m$ and clearly $- x ∈ A$.

These results are equivalent to the monotone convergence theorem. To see this, suppose $( a n )$ is a bounded nondecreasing sequence. Then let $A = a n : n ∈ ℕ$. By the fact that the sequence is bounded and by completeness theorem above, $l = sup A$ exists, and it is a nice exercise to show that $a n → l$ as $n → ∞$.

## 3Summary

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 $ℝ$.

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