The Greatest Lower Bound Property

Completeness: No Missing Lower Edges

The least upper bound property says that every nonempty real set bounded above has a supremum. Its order-reversed counterpart concerns sets bounded below: such a set has a greatest lower bound, called its infimum. The Existence of the Infimum Theorem, established earlier in this course, gives this existence result. Here we make the greatest lower bound property explicit and examine the useful symmetry between infima and suprema.

Definition (Greatest Lower Bound Property). The real numbers have the greatest lower bound property if every nonempty set $S\subseteq\mathbb{R}$ that is bounded below has an infimum in $\mathbb{R}$. The infimum is the greatest of all lower bounds of $S$: it is a lower bound for $S$, and every other lower bound is less than or equal to it.

A lower bound need not belong to the set. Similarly, an infimum need not be a minimum: it is a minimum only when it is an element of the set. The Approximation Characterization of the Infimum, established in the tutorial “Infimum,” provides a practical test: a lower bound $m$ is the infimum if, for every $\varepsilon>0$, the set contains an element $x$ with $m\le x<m+\varepsilon$.

The hypotheses matter. The set must be nonempty and bounded below. The empty set has no elements to constrain its lower bounds, and an unbounded-below set has no real lower bound at all. When the hypotheses do hold, the greatest lower bound property ensures the infimum is a real number, even if it is not in the set.

Negation Turns Lower Bounds into Upper Bounds

The relationship with the least upper bound property is more than an analogy. Negating every element of a set reverses its order: elements far down in the original set become elements far up in the negated set. This gives a direct correspondence between lower bounds and upper bounds.

Theorem (Order Duality for Supremum and Infimum). Let $S\subseteq\mathbb{R}$ be nonempty, and define $-S=\{-x:x\in S\}$. If $S$ is bounded below, then $-S$ is bounded above and $$ \inf S=-\sup(-S). $$ Conversely, if $S$ is bounded above, then $-S$ is bounded below and $$ \sup S=-\inf(-S). $$

Proof. Suppose first that $S$ is bounded below, and let $L$ be any real number. By the order reversal under negation, $L\leq x$ for every $x\in S$ if and only if $-x\leq -L$ for every $x\in S$. Thus $$ L\text{ is a lower bound for }S \quad\Longleftrightarrow\quad -L\text{ is an upper bound for }-S. $$ In particular, a lower bound for $S$ gives an upper bound for $-S$, so $-S$ is bounded above. It is nonempty because $S$ is nonempty. The least upper bound property therefore gives $\alpha=\sup(-S)$.

Set $\beta=-\alpha$. Since $\alpha$ is an upper bound for $-S$, the correspondence just proved shows that $\beta$ is a lower bound for $S$. If $L$ is any lower bound for $S$, then $-L$ is an upper bound for $-S$. The definition of $\alpha$ gives $\alpha\leq -L$. Negating and reversing the inequality yields $L\leq-\alpha=\beta$. Thus $\beta$ is a lower bound for $S$ that is at least as large as every other lower bound. Hence $\beta=\inf S$, proving $\inf S=-\sup(-S)$.

For the converse, suppose $S$ is bounded above. Negation makes $-S$ nonempty and bounded below. Apply the identity just proved to $-S$: since $-(-S)=S$, it gives $$ \inf(-S)=-\sup(S). $$ Equivalently, $\sup S=-\inf(-S)$, as required. $\square$

This correspondence also explains why the greatest lower bound property and the least upper bound property are two forms of the same completeness principle on the real line. Negation transfers the existence of one kind of boundary to the existence of the other. The result does not require a new argument about how to construct an infimum from scratch; it identifies the infimum with a supremum in a reflected set.

Worked Example: An Infimum That Is Not a Minimum

Let $S=\{2+1/n:n\text{ is a positive integer}\}$. For every positive integer $n$, $1/n>0$, so $2+1/n>2$. Therefore $2$ is a lower bound for $S$, but it is not in $S$: the equation $2+1/n=2$ would require $1/n=0$.

To verify that $2$ is the greatest lower bound, let $\varepsilon>0$. By the Archimedean property, choose a positive integer $n$ with $1/n<\varepsilon$. Then $$ 2<2+\frac{1}{n}<2+\varepsilon. $$ Thus elements of $S$ lie within every positive distance above $2$. By the Approximation Characterization of the Infimum, $\inf S=2$. Since $2\notin S$, this infimum is not a minimum.

Infima Under Scaling and Translation

A change of scale or a translation moves a set’s lower edge in a predictable way. Positive scaling preserves the order of the elements, so the infimum is scaled and translated along with them. Negative scaling reverses the order, so the relevant boundary of the original set is its supremum instead.

Theorem (Infimum Under Affine Transformations). Let $S\subseteq\mathbb{R}$ be nonempty, and let $b\in\mathbb{R}$.

If $S$ is bounded below and $c>0$, then $cS+b=\{cx+b:x\in S\}$ is bounded below and $$ \inf(cS+b)=c\inf S+b. $$
If $S$ is bounded above and $c<0$, then $cS+b=\{cx+b:x\in S\}$ is bounded below and $$ \inf(cS+b)=c\sup S+b. $$

Proof. First suppose $S$ is bounded below, $c>0$, and $\beta=\inf S$. For every $x\in S$, $\beta\leq x$. Multiplying by $c>0$ preserves order, and adding $b$ also preserves order, so $$ c\beta+b\leq cx+b. $$ Thus $c\beta+b$ is a lower bound for $cS+b$.

Let $L$ be any lower bound for $cS+b$. For every $x\in S$, $L\leq cx+b$. Subtract $b$, then divide by the positive number $c$, to obtain $(L-b)/c\leq x$. Hence $(L-b)/c$ is a lower bound for $S$. Since $\beta$ is its greatest lower bound, $$ \frac{L-b}{c}\leq\beta. $$ Multiplication by $c>0$ and addition of $b$ give $L\leq c\beta+b$. Consequently $c\beta+b$ is the greatest lower bound of $cS+b$, proving the first formula.

Now suppose $S$ is bounded above, $c<0$, and $\alpha=\sup S$. Every $x\in S$ satisfies $x\leq\alpha$. Multiplying by $c<0$ reverses the inequality, so $cx\geq c\alpha$. Adding $b$ gives $cx+b\geq c\alpha+b$. Therefore $c\alpha+b$ is a lower bound for $cS+b$.

Let $L$ be any lower bound for $cS+b$. Then $L\leq cx+b$ for every $x\in S$. Subtracting $b$ and dividing by $c<0$ reverses the inequality, giving $$ \frac{L-b}{c}\geq x \qquad\text{for every }x\in S. $$ Thus $(L-b)/c$ is an upper bound for $S$. The least upper bound $\alpha$ satisfies $\alpha\leq(L-b)/c$. Multiplication by $c<0$ reverses the inequality, and then adding $b$ gives $c\alpha+b\geq L$. Therefore $c\alpha+b$ is at least every lower bound of $cS+b$, so it is the infimum. This proves the second formula. $\square$

Worked Example: Reflecting an Interval

Take $S=[-2,5)$, whose supremum is $5$. Consider the set $V=\{-3x+4:x\in S\}$. Since the coefficient $-3$ is negative, the affine transformation theorem gives $$ \inf V=-3\sup S+4=-3(5)+4=-11. $$ The interval endpoints confirm the calculation: as $x$ ranges from $-2$, included, up to $5$, excluded, $-3x+4$ ranges from $10$, included, down toward $-11$, excluded. Thus $V=(-11,10]$. Its infimum is $-11$, but it has no minimum.

Reading the Boundary Correctly

The formulas help prevent a common sign error. Under positive scaling, lower edges remain lower edges. Under negative scaling, lower and upper edges switch roles. For example, if $S$ has a minimum, multiplying by a negative number sends that minimum to a maximum of the image, not its infimum. The infimum of the image comes from the supremum of $S$.

A useful way to check a proposed infimum is to verify both parts of its definition: first, it must be a lower bound; second, no larger number can also be a lower bound. The approximation characterization offers an equivalent check: for every $\varepsilon>0$, the set must contain an element less than $\inf S+\varepsilon$. It is not enough to show only that a number is a lower bound; a set can have many lower bounds.

Worked Example: Infimum of a Half-Open Interval

Let $T=(-4,3]$. Every $x\in T$ satisfies $-4<x$, so $-4$ is a lower bound. It is not an element of $T$, and hence cannot be a minimum. To show it is the infimum, let $\varepsilon>0$ and choose $h=\min\{\varepsilon/2,1\}$. Then $0<h<\varepsilon$, and $x=-4+h$ satisfies $-4<x\leq-3<3$. Thus $x\in T$ and $$ -4\leq x=-4+h<-4+\varepsilon. $$ Elements of $T$ occur arbitrarily close above $-4$, so the Approximation Characterization of the Infimum gives $\inf T=-4$. The endpoint is a greatest lower bound, not a minimum.

Why the Property Matters

The greatest lower bound property ensures that bounded-below sets in $\mathbb{R}$ have genuine boundary values in $\mathbb{R}$. A lower bound alone is not enough: it might sit far below the set, and there may be many such bounds. The infimum identifies the sharpest one. Through negation, this is precisely the same completeness guarantee as the least upper bound property.

The distinction between an infimum and a minimum remains important even when the infimum exists. In the interval $(-4,3]$, the lower edge is approached but excluded; in a set such as $[ -4,3]$, the same lower edge is included and is a minimum. Both sets have infimum $-4$, but only the second attains it. Whether the boundary belongs to the set is a separate question from whether the boundary exists.

Key takeaway. Every nonempty real set bounded below has an infimum. Negating the set converts its infimum into the negative of a supremum, and affine transformations carry the boundary according to whether their scale preserves or reverses order. An infimum need not be a minimum.

Check Your Understanding

Use the definition, order duality, and the affine transformation theorem to answer each question.

What two conditions on a set are required for the greatest lower bound property to guarantee an infimum?
If $S$ is nonempty and bounded below, how can $\inf S$ be expressed using $\sup(-S)$?
Why does a negative scaling use $\sup S$, rather than $\inf S$, to determine the infimum of the transformed set?
What is the infimum of $\{2+1/n:n\text{ is a positive integer}\}$, and is it a minimum?
If $S$ is nonempty and bounded above, what is $\inf(cS+b)$ when $c<0$?