Minimum Versus Infimum

Attained Lower Edges and Unattained Lower Edges

An infimum describes the greatest lower bound of a set. A minimum describes a lower edge that the set actually contains. As with a maximum and a supremum, the distinction is whether the boundary value is attained: a set can have an infimum without having a minimum.

Recall from “Lower Bounds” that $L$ is a lower bound for $S\subseteq\mathbb{R}$ when $L\leq x$ for every $x\in S$. The infimum is the greatest lower bound. A minimum must itself be an element of the set, as well as being no greater than any other element.

Recall (Minimum). A number $m$ is the minimum of $S\subseteq\mathbb{R}$ if $m\in S$ and $m\leq x$ for every $x\in S$. It is denoted by $\min S$. The infimum $\inf S$, when it exists, is the greatest lower bound of $S$; it need not belong to $S$.

For a nonempty set bounded below, the Existence of the Infimum Theorem guarantees an infimum in $\mathbb{R}$. The important question is whether this greatest lower bound is in the set. The following criterion gives an exact answer.

Theorem (An Infimum Is a Minimum Exactly When It Is Attained). Let $S\subseteq\mathbb{R}$ be nonempty and bounded below. Then $\inf S$ is a minimum of $S$ if and only if $\inf S\in S$.

Proof. Write $\beta=\inf S$. If $\beta$ is a minimum, then by the definition of minimum $\beta\in S$. Conversely, suppose $\beta\in S$. Since $\beta$ is a lower bound, $\beta\leq x$ for every $x\in S$. Thus $\beta$ belongs to $S$ and is no greater than every element of $S$, so it is the minimum. $\square$

This criterion is the lower-edge counterpart of the attainment criterion for a supremum. It does not say that every set bounded below has a minimum. Completeness guarantees the infimum, but the infimum may lie just outside the set.

Worked Example: An Infimum That Is Not a Minimum

Let $A=(0,4]$. Every $x\in A$ satisfies $0<x$, so $0$ is a lower bound for $A$. Given any $\varepsilon>0$, choose $x=\min\{\varepsilon,4\}/2$. This gives $0<x\leq2$, so $x\in A$, and $x<\varepsilon$. Therefore, for every positive $\varepsilon$, $A$ contains an element less than $0+\varepsilon$. By the Approximation Characterization of the Infimum, $\inf A=0$.

But $0\notin A$, so the attainment criterion shows that $A$ has no minimum. The set extends arbitrarily close to its lower edge without reaching it. In particular, the fact that every element is positive does not make any one element the smallest: if $x\in A$, then $x/2$ is also positive, is at most $2$, and is strictly less than $x$.

Worked Example: A Minimum and an Infimum That Agree

Let $B=\{-3,\,0,\,7/2,\,9\}$. The number $-3$ belongs to $B$, and $-3\leq -3$, $-3\leq0$, $-3\leq7/2$, and $-3\leq9$. Hence $-3$ is the minimum of $B$. It is also a lower bound. Any lower bound $L$ for $B$ must satisfy $L\leq-3$, since $-3\in B$. Thus $-3$ is the greatest lower bound, and $\min B=\inf B=-3$.

Here the infimum is attained. Membership matters: the infimum is a minimum precisely because its value is an element of the set.

Recording Approximation with a Sequence

The Approximation Characterization of the Infimum says that a greatest lower bound can be approached from above: for every positive tolerance, the set contains an element less than the infimum plus that tolerance. Choosing tolerances $1/n$ packages these approximations into a sequence. This gives a useful test that parallels the Sequential Approximation to the Supremum.

Theorem (Sequential Approximation to the Infimum). Let $S\subseteq\mathbb{R}$ be nonempty, and let $\beta\in\mathbb{R}$. Then $\beta=\inf S$ if and only if $\beta$ is a lower bound for $S$ and there is a sequence $(s_n)_{n\geq1}$ of elements of $S$ such that $$ \beta\leq s_n<\beta+\frac{1}{n} $$ for every positive integer $n$.

Proof. First suppose $\beta=\inf S$. Then $\beta$ is a lower bound. For each positive integer $n$, apply the Approximation Characterization of the Infimum with $\varepsilon=1/n$. It supplies an $s_n\in S$ such that $s_n<\beta+1/n$. Since $\beta$ is a lower bound, $\beta\leq s_n$, giving the stated inequalities.

Conversely, suppose $\beta$ is a lower bound and there are elements $s_n\in S$ satisfying the displayed inequalities. Let $L$ be any lower bound for $S$. We show that $L\leq\beta$. If instead $\beta<L$, the Archimedean property gives a positive integer $n$ such that $1/n<L-\beta$. Consequently, $$ s_n<\beta+\frac{1}{n}<L. $$ This contradicts the fact that $L$ is a lower bound and $s_n\in S$, which require $L\leq s_n$. Thus every lower bound $L$ satisfies $L\leq\beta$. Since $\beta$ is itself a lower bound, it is the greatest lower bound, so $\beta=\inf S$. $\square$

The left-hand inequality allows the sequence terms to equal the infimum. If a term does, then the infimum belongs to the set and is a minimum. But the right-hand inequality can hold even when every term is strictly greater than the infimum; in that case the sequence describes approach without attainment.

Worked Example: A Sequence Approaching an Infimum

Let $C=(3,8)$, and define $s_n=3+1/(n+1)$ for each positive integer $n$. Since $n\geq1$, we have $0<1/(n+1)\leq1/2$, so $3<s_n\leq7/2<8$. Hence $s_n\in C$. Also $1/(n+1)<1/n$, so $$ 3\leq s_n=3+\frac{1}{n+1}<3+\frac{1}{n}. $$ The number $3$ is a lower bound for $C$, and the sequence satisfies the conditions of the Sequential Approximation Theorem. Therefore $\inf C=3$. Since $3\notin C$, this infimum is not a minimum.

The sequence does not need to contain the infimum to demonstrate that elements of the set occur arbitrarily close to it. Here every selected term is strictly greater than $3$.

Worked Example: Positive Reciprocals with No Minimum

Consider $D=\{1/n:n\text{ is a positive integer}\}$. Every element is positive, so $0$ is a lower bound. Given $\varepsilon>0$, choose a positive integer $n$ large enough that $1/n<\varepsilon$, using the Archimedean property. Then $1/n\in D$ and $0\leq1/n<0+\varepsilon$. The Approximation Characterization of the Infimum gives $\inf D=0$. Since $0\notin D$, the infimum is not a minimum.

The set has no minimum for a further direct reason. Any element of $D$ has the form $1/k$ for a positive integer $k$. But $1/(k+1)\in D$, and $1/(k+1)<1/k$, since $k<k+1$ and both denominators are positive. Thus every element has a smaller element of $D$.

Deleting an Element When There Is No Minimum

If a set has a minimum, deleting that one element can raise its infimum. If a set has no minimum, no single element serves as its attained lower edge. In that case, deleting any one element leaves the infimum unchanged. The proof uses the approximation property to find an element close to the infimum that is not the one being removed.

Theorem (Deleting One Element from a Set Without a Minimum). Let $S\subseteq\mathbb{R}$ be nonempty and bounded below, and suppose $S$ has no minimum. For every $x\in S$, $$ \inf\bigl(S\setminus\{x\}\bigr)=\inf S. $$

Proof. Write $\beta=\inf S$. The attainment criterion shows that $\beta\notin S$; since $\beta$ is a lower bound, this means $x>\beta$ for every $x\in S$. Fix $x\in S$. The set $S\setminus\{x\}$ is nonempty: otherwise $S=\{x\}$, and $x$ would be its minimum.

Because $S\setminus\{x\}\subseteq S$, $\beta$ is a lower bound for $S\setminus\{x\}$. Now let $\varepsilon>0$, and set $$ \delta=\frac{\min\{\varepsilon,x-\beta\}}{2}. $$ Both numbers in the minimum are positive, so $\delta>0$. Also $\delta<\varepsilon$ and $\delta<x-\beta$, which gives $\beta+\delta<x$. By the Approximation Characterization of the Infimum, there is an $s\in S$ such that $s<\beta+\delta$. Since $\beta$ is a lower bound, $\beta\leq s$; in particular $s<x$, so $s\neq x$ and $s\in S\setminus\{x\}$. Finally, $$ s<\beta+\delta<\beta+\varepsilon. $$ Thus, for every positive $\varepsilon$, the set after deletion contains an element less than $\beta+\varepsilon$. Together with the lower bound $\beta$, the Approximation Characterization gives $\inf(S\setminus\{x\})=\beta$. $\square$

Worked Example: Deleting a Point Without Changing the Infimum

Let $E=(2,7)\setminus\{4\}$. The number $2$ is a lower bound for $E$. For any $\varepsilon>0$, define $\delta=\min\{\varepsilon,2\}$ and take $y=2+\delta/2$. We have $0<\delta\leq2$, so $2<y\leq3<7$, and therefore $y\in(2,7)$. Also $y\neq4$, so $y\in E$. Because $\delta/2<\varepsilon$, we have $y<2+\varepsilon$. The Approximation Characterization of the Infimum now gives $\inf E=2$. As $2\notin E$, there is no minimum.

By contrast, $\{2,5\}$ has minimum and infimum $2$. Deleting $2$ leaves $\{5\}$, whose infimum is $5$. The difference is whether the removed point was the attained lower edge.

How to Check for a Minimum

To show that a number is the minimum of a set, verify both requirements: it belongs to the set and it is no greater than every element. To show that a number is the infimum, verify that it is a lower bound and that elements of the set occur arbitrarily close above it. The Approximation Characterization of the Infimum formalizes the second requirement.

A common pitfall is to treat “positive” as equivalent to “has a smallest positive element.” The set of positive reciprocals in the worked example has infimum $0$, but no minimum. More generally, completeness guarantees an infimum for every nonempty set bounded below; it does not guarantee that this value is attained. When the infimum does belong to the set, the attainment criterion immediately makes it the minimum.

Key takeaway. A minimum is an attained lower edge; an infimum is the greatest lower bound, whether or not the set contains it. A set without a minimum retains its infimum after any one element is deleted.

Check Your Understanding

Use the definitions and results in this tutorial to answer each question.

What two conditions must a number satisfy to be the minimum of a set?
When is the infimum of a nonempty set bounded below also its minimum?
What inequalities characterize sequence terms approximating an infimum?
Why does the set of positive reciprocals have infimum $0$ but no minimum?
Why does deleting one element from a set without a minimum leave its infimum unchanged?

What You'll Learn