The Least Upper Bound Property

Completeness: No Missing Upper Edges

The previous tutorial distinguished an infimum from a minimum by asking whether the lower edge belongs to the set. Here we turn to upper edges. A supremum is the least upper bound, whether or not it is in the set; a maximum must belong to the set. The question now is when a nonempty set that is bounded above is guaranteed to have a supremum in the number system being used.

For the real numbers, the answer is built into their completeness. The least upper bound property is not a consequence of the field and order rules alone: it is a fundamental property that distinguishes the real numbers from the rational numbers. In particular, it rules out gaps in the real line that would prevent some sets from having a least upper bound.

Definition (Least Upper Bound Property). The real numbers have the least upper bound property if every nonempty set $S\subseteq\mathbb{R}$ that is bounded above has a supremum in $\mathbb{R}$. In other words, there is a real number $\alpha$ that is an upper bound for $S$ and is no greater than any other upper bound for $S$.

The existence assertion is important: the property guarantees that the least upper bound is a real number. It does not say that the supremum is an element of the set. As the earlier Approximation Characterization of the Supremum shows, a supremum can be identified by checking that it is an upper bound and that elements of the set occur arbitrarily close below it.

The hypotheses are essential. The empty set is excluded, and an unbounded-above set has no real upper bound to begin with. The property also depends on the number system: a set of rational numbers may be nonempty and bounded above by rational numbers without having a least rational upper bound.

Worked Example: A Supremum Not Attained by a Sequence of Points

Let $S=\{1-1/n:n\text{ is a positive integer}\}$. Each $n$ is positive, so $1/n>0$, and therefore $1-1/n<1$. Thus $1$ is an upper bound for $S$. Given any $\varepsilon>0$, the Archimedean property gives a positive integer $n$ with $1/n<\varepsilon$. For this $n$, $$ 1-\varepsilon<1-\frac{1}{n}<1. $$ The set therefore contains an element greater than $1-\varepsilon$ for every positive $\varepsilon$. By the Approximation Characterization of the Supremum, $\sup S=1$.

The value $1$ is not in $S$, since $1-1/n=1$ would require $1/n=0$. Hence this set has a supremum but no maximum. The least upper bound property guarantees the existence of a supremum; it does not guarantee attainment.

A Consequence: Positive Numbers Have Square Roots

The least upper bound property can do more than assert that suprema exist. It can be used to construct numbers that a set of rational numbers might be unable to supply. We use it to prove that every positive real number has a positive square root. The proof forms a set of nonnegative numbers whose squares fall short of the given number, then identifies its supremum.

Theorem (Existence and Uniqueness of Positive Square Roots). For every real number $a>0$, there is a unique real number $r>0$ such that $r^2=a$.

Proof. Fix $a>0$, and define $$ S=\{x\in\mathbb{R}:x\geq0\text{ and }x^2<a\}. $$ The set is nonempty because $0\in S$. It is bounded above by $a+1$: if $x\geq a+1$, then $x\geq0$ and $$ x^2\geq(a+1)^2=a^2+2a+1>a. $$ Such an $x$ cannot belong to $S$. By the least upper bound property, $\alpha=\sup S$ exists in $\mathbb{R}$. Since $0\in S$, we have $\alpha\geq0$.

We first show that $\alpha^2$ cannot be less than $a$. Suppose $\alpha^2<a$, and write $\delta=a-\alpha^2>0$. Choose $$ h=\min\left\{1,\frac{\delta}{2(2\alpha+1)}\right\}. $$ Then $h>0$, $h\leq1$, and $h(2\alpha+1)\leq\delta/2$. Since $h^2\leq h$, $$ (\alpha+h)^2-\alpha^2=2\alpha h+h^2 \leq h(2\alpha+1) \leq\frac{\delta}{2}<\delta. $$ It follows that $(\alpha+h)^2<a$. Also $\alpha+h\geq0$, so $\alpha+h\in S$. But $\alpha+h>\alpha$, contradicting that $\alpha$ is an upper bound for $S$. Therefore $\alpha^2\not<a$.

Next, suppose $\alpha^2>a$. Then $\alpha>0$, and $\delta=\alpha^2-a>0$. Choose $$ h=\min\left\{\frac{\alpha}{2},\frac{\delta}{2\alpha}\right\}. $$ We have $0<h<\alpha$ and $2\alpha h\leq\delta$. Consequently, $$ \alpha^2-(\alpha-h)^2=2\alpha h-h^2<2\alpha h\leq\delta, $$ so $(\alpha-h)^2>a$. Set $y=\alpha-h$, which is positive and strictly less than $\alpha$. If $x\in S$ and $x>y$, then $x+y>0$, and $$ x^2-y^2=(x-y)(x+y)>0. $$ Thus $x^2>y^2>a$, contradicting $x\in S$. Therefore every $x\in S$ satisfies $x\leq y$, so $y$ is an upper bound for $S$ smaller than its least upper bound $\alpha$, a contradiction. We conclude that $\alpha^2$ is neither less than nor greater than $a$, and hence $\alpha^2=a$. Since $a>0$, $\alpha\neq0$, so $\alpha>0$.

For uniqueness, suppose $r>0$, $s>0$, and $r^2=s^2=a$. Then $$ (r-s)(r+s)=r^2-s^2=0. $$ Since $r+s>0$, it is nonzero. The zero-product property therefore gives $r-s=0$, so $r=s$. This proves both existence and uniqueness. $\square$

The construction depends on completeness at the precise moment when $\sup S$ is asserted to exist in $\mathbb{R}$. The field and order rules let us reason about the set, but the least upper bound property supplies the boundary value $\alpha$ needed for the argument.

Worked Example: The Upper Edge of a Square-Defined Set

Consider $T=\{x\in\mathbb{R}:x\geq0\text{ and }x^2<10\}$. The positive square root theorem gives a positive number $r$ with $r^2=10$. If $x\in T$ and $x\geq r$, then $x\geq0$ and $$ x^2-r^2=(x-r)(x+r)\geq0, $$ which would imply $x^2\geq10$, a contradiction. Thus every $x\in T$ satisfies $x<r$, so $r$ is an upper bound.

To check that this upper bound is least, let $\varepsilon>0$ and choose $h=\min\{\varepsilon/2,r/2\}$. Then $0<h<\varepsilon$ and $0<r-h<r$. Therefore $$ (r-h)^2<r^2=10, $$ so $r-h\in T$, and $r-h>r-\varepsilon$. The Approximation Characterization of the Supremum gives $\sup T=r=\sqrt{10}$. Since $r^2=10$, $r\notin T$; the supremum is not a maximum.

How Supremum Behaves Under Positive Scaling

A supremum also behaves predictably when every element of a set is multiplied by the same positive number. This is useful because it lets us transfer an upper edge through a simple change of scale without returning to the definition for each individual set.

Theorem (Supremum Under Positive Scaling). Let $S\subseteq\mathbb{R}$ be nonempty and bounded above, and let $c>0$. Define $cS=\{cx:x\in S\}$. Then $cS$ is nonempty and bounded above, and $$ \sup(cS)=c\sup S. $$

Proof. Write $\alpha=\sup S$. For every $x\in S$, $x\leq\alpha$. Multiplication by $c>0$ preserves order, so $cx\leq c\alpha$. Thus $c\alpha$ is an upper bound for $cS$, and $cS$ is nonempty.

Let $U$ be any upper bound for $cS$. For every $x\in S$, $cx\leq U$. Dividing by the positive number $c$ preserves order, giving $x\leq U/c$. Thus $U/c$ is an upper bound for $S$. Since $\alpha$ is the least upper bound of $S$, $\alpha\leq U/c$, and multiplying by $c$ gives $c\alpha\leq U$. Therefore $c\alpha$ is an upper bound for $cS$ that is no greater than any other upper bound. Hence $\sup(cS)=c\alpha=c\sup S$. $\square$

Worked Example: Scaling a Supremum

For the set $S$ from the first worked example, $\sup S=1$. Let $V=3S=\{3-3/n:n\text{ is a positive integer}\}$. The Supremum Under Positive Scaling Theorem, with $c=3$, gives $$ \sup V=3\sup S=3. $$ Directly, $3-3/n<3$ for every positive integer $n$, so $3$ is an upper bound. Given $\varepsilon>0$, choose $n$ with $3/n<\varepsilon$, using the Archimedean property. Then $3-3/n>3-\varepsilon$, confirming that elements of $V$ occur arbitrarily close below $3$. The value $3$ is not in $V$, since $3-3/n=3$ would require $3/n=0$. Thus $V$ has supremum $3$ but no maximum.

Why the Rational Numbers Do Not Have This Property

The least upper bound property is specific to a complete number system such as $\mathbb{R}$. The corresponding claim fails in $\mathbb{Q}$, even for a simple set bounded above by rational numbers. The failure is not that rational upper bounds are unavailable; rather, among them there is no least one.

Theorem (The Rational Numbers Do Not Have the Least Upper Bound Property). The set $A=\{q\in\mathbb{Q}:q\geq0\text{ and }q^2<2\}$ is nonempty and bounded above in $\mathbb{Q}$, but it has no supremum in $\mathbb{Q}$.

Proof. We have $0\in A$, so $A$ is nonempty, and $2$ is a rational upper bound: if $q\geq2$, then $q^2\geq4>2$, so $q\notin A$. Suppose, for a contradiction, that $u\in\mathbb{Q}$ is the supremum of $A$. Since $1\in A$, we must have $u\geq1$.

Theorem “The Square Root of 2 Is Irrational” shows that no rational number has square equal to $2$. Thus either $u^2<2$ or $u^2>2$. If $u^2<2$, let $g=2-u^2$ and $h=g/[2(u+1)]$. Both $g$ and $h$ are positive rationals, and $g\leq1$ because $u\geq1$. In particular, $h<1$, so $2u+h<2u+2$. It follows that $$ (u+h)^2-u^2=h(2u+h)<2h(u+1)=g. $$ Hence $(u+h)^2<2$, and $u+h\in A$, although $u+h>u$. This contradicts that $u$ is an upper bound.

If $u^2>2$, define $v=(u^2+2)/(2u)$, a positive rational number. Then $$ u-v=\frac{u^2-2}{2u}>0 \qquad\text{and}\qquad v^2-2=\frac{(u^2-2)^2}{4u^2}>0. $$ Thus $v<u$ and $v^2>2$. If some $q\in A$ satisfied $q>v$, then $q,v\geq0$ and $$ q^2-v^2=(q-v)(q+v)>0, $$ so $q^2>v^2>2$, contrary to $q\in A$. Therefore $v$ is an upper bound for $A$ smaller than $u$, contradicting that $u$ is the least upper bound. Both possibilities lead to a contradiction, so $A$ has no supremum in $\mathbb{Q}$. $\square$

In $\mathbb{R}$, the positive square root of $2$ is the supremum of this set, but it is not rational. That explains the gap: rational numbers can approach the boundary from below, while the boundary itself is missing from $\mathbb{Q}$. The least upper bound property guarantees that such a boundary belongs to $\mathbb{R}$, even when it does not belong to the set being bounded.

Key takeaway. Every nonempty subset of $\mathbb{R}$ that is bounded above has a real supremum. That supremum need not be a maximum, and the same guarantee fails in $\mathbb{Q}$. The property supplies the boundary values needed for results such as the existence of positive square roots.

Check Your Understanding

Use the least upper bound property and the proved results to answer each question.

What two conditions on a set are required before the least upper bound property guarantees a supremum?
Why does the supremum of $\{1-1/n:n\text{ is a positive integer}\}$ fail to be a maximum?
In the square root proof, why is the supremum $\alpha$ neither able to satisfy $\alpha^2<a$ nor $\alpha^2>a$?
If $S$ is nonempty and bounded above and $c>0$, what is $\sup(cS)$?
Why does the set of nonnegative rational numbers whose squares are less than $2$ have no rational supremum?

What You'll Learn