Greatest Lower Bounds Require Two Arguments
The Supremum Strategy identifies a least upper bound by showing that the set has elements arbitrarily close to the proposed bound from below. The Infimum Strategy is its lower-bound counterpart: to identify the greatest lower bound, show both that every element lies at or above the candidate and that elements can be found arbitrarily close to it from above. This is useful when a set has no smallest element, or when a proof needs a sharp lower estimate.
A lower bound alone does not identify an infimum. If \(l\) is a lower bound, then every number below \(l\) is also a lower bound. What distinguishes the infimum is that it is the greatest of all lower bounds. The useful witness condition says that for every positive error, the set contains an element less than that error above the candidate.
The greatest-lower-bound property of the real numbers guarantees that every nonempty set bounded below has an infimum. Once existence is known, the main task is to verify a proposed candidate. The next characterization converts that task into two direct checks.
Proof. Suppose first that \(l=\inf A\). Then \(l\) is a lower bound, so \(l\leq a\) for every \(a\in A\). Let \(\varepsilon>0\). If no element \(a\in A\) satisfied \(a<l+\varepsilon\), then every element of \(A\) would be at least \(l+\varepsilon\). That would make \(l+\varepsilon\) a lower bound for \(A\), even though \(l+\varepsilon>l\). This contradicts the fact that \(l\) is the greatest lower bound. Thus there is an \(a\in A\) with \(a<l+\varepsilon\).
Conversely, suppose the two stated conditions hold. The first says that \(l\) is a lower bound. Let \(v\) be any lower bound for \(A\). We show that \(v\leq l\). If \(v>l\), choose \(\varepsilon=v-l\), which is positive. The second condition gives an \(a\in A\) such that
But \(v\) is a lower bound, so \(v\leq a\) for every \(a\in A\). This is a contradiction. Therefore every lower bound \(v\) satisfies \(v\leq l\), and \(l=\inf A\). \(\square\)
The proof provides a witness principle: if \(l=\inf A\) and \(c>l\), then some element of \(A\) is less than \(c\). Choose \(\varepsilon=c-l\) in the characterization. This principle is often the decisive step when ruling out a lower bound larger than the proposed infimum.
Finding Elements Near the Candidate
The two parts of the characterization have different jobs. First show that the candidate is a lower bound, checking the inequality for every element. Then take an arbitrary \(\varepsilon>0\) and construct an element of the set that lies below the candidate plus \(\varepsilon\). This element may depend on \(\varepsilon\); that dependence is how the proof expresses arbitrarily close approximation.
Worked Example: A Sequence of Set Elements Approaching Its Infimum
Let \(A=\{3+2/n:n\text{ is a positive integer}\}\). We claim that \(\inf A=3\). For every positive integer \(n\), \(2/n>0\), so \(3<3+2/n\). In particular, \(3\leq a\) for every \(a\in A\), and \(3\) is a lower bound.
Given \(\varepsilon>0\), choose a positive integer \(n\) with \(n>2/\varepsilon\), using the Archimedean property of the real numbers. Then \(2/n<\varepsilon\), so
Thus \(A\) contains an element less than \(3+\varepsilon\) for every positive \(\varepsilon\). The epsilon characterization gives \(\inf A=3\). The infimum is not a minimum: a minimum must belong to the set, but \(3+2/n=3\) would imply \(2/n=0\), which is impossible for a positive integer \(n\).
The comparison with a minimum matters. An infimum is a bound determined by the whole set, whether or not it is attained. The epsilon condition gives elements close to the infimum from above; it does not say that any element equals the infimum. If a proof needs an actual least element, it must establish membership and attainment separately.
Worked Example: A Constrained Set With an Unattained Infimum
Consider \(D=\{x\in\mathbb{R}:x>-1\text{ and }x^2\leq9\}\). We show that \(\inf D=-1\). By the first condition in the definition of \(D\), each \(x\in D\) satisfies \(-1<x\). Hence \(-1\leq x\), so \(-1\) is a lower bound.
Let \(\varepsilon>0\), and set \(\delta=\min(\varepsilon/2,1)\) and \(x=-1+\delta\). Since \(0<\delta\leq1\), we have \(-1<x\leq0\), and therefore \(x^2\leq1\leq9\). Thus \(x\in D\). Also \(\delta<\varepsilon\): if \(\varepsilon\leq2\), then \(\delta=\varepsilon/2<\varepsilon\); if \(\varepsilon>2\), then \(\delta=1<\varepsilon\). It follows that
The epsilon characterization gives \(\inf D=-1\). This choice verifies both constraints defining \(D\); simply choosing a point close to \(-1\) would not suffice without checking that it still belongs to the set. The infimum is not in \(D\), since membership requires \(x>-1\).
How Infima Change Under Positive Affine Transformations
A lower bound can be transported through a positive scaling and a translation. Positivity is essential to the order argument: multiplying an inequality by a positive number preserves its direction. The following result turns that observation into an infimum formula.
Proof. Write \(l=\inf A\). Since \(A\) is nonempty, choose \(a_0\in A\); then \(ca_0+d\in cA+d\), so the transformed set is nonempty. For every \(a\in A\), \(l\leq a\). Because \(c>0\), this gives \(cl+d\leq ca+d\). Thus \(cl+d\) is a lower bound for \(cA+d\).
To verify that it is the greatest lower bound, let \(\varepsilon>0\). Apply the epsilon characterization to \(A\) with the positive error \(\varepsilon/c\). There is an \(a\in A\) such that
Multiplying by \(c>0\) and adding \(d\) yields \(ca+d<cl+d+\varepsilon\). Since \(ca+d\in cA+d\), the epsilon characterization shows that \(\inf(cA+d)=cl+d\). In particular, the set is bounded below by this finite number. \(\square\)
Worked Example: Scaling and Translating a Set
Let \(A=\{5+1/n:n\text{ is a positive integer}\}\). Every element is greater than \(5\), and for each \(\varepsilon>0\), choosing \(n>1/\varepsilon\) gives \(5<5+1/n<5+\varepsilon\). Thus \(\inf A=5\). Consider the transformed set \(C=\{4a-7:a\in A\}\). The positive affine transformation theorem gives
The elements can also be checked directly: for each positive integer \(n\),
Given \(\varepsilon>0\), choose a positive integer \(n>4/\varepsilon\). Then \(4/n<\varepsilon\), so \(13<13+4/n<13+\varepsilon\). This confirms both the lower-bound and approximation conditions for the transformed set.
The hypothesis \(c>0\) should not be dropped casually. If \(c=0\), then \(cA+d\) consists only of \(d\), so its infimum is \(d\). If \(c<0\), multiplication reverses order, and the infimum of the transformed set is governed by the supremum of \(A\), when that supremum exists. The theorem above is deliberately stated only for positive scaling.
Combining Infimum Arguments
For a sum set \(A+B=\{a+b:a\in A,\ b\in B\}\), lower bounds for the two sets add to give a lower bound for every sum. To prove that the resulting bound is greatest, approximate each infimum with a suitably chosen error. This is the lower-bound counterpart of the Supremum of a Sum Set Theorem.
Proof. Write \(\alpha=\inf A\) and \(\beta=\inf B\). Since \(A\) and \(B\) are nonempty, choose \(a_0\in A\) and \(b_0\in B\). Then \(a_0+b_0\in A+B\), so the sum set is nonempty. For any \(a\in A\) and \(b\in B\), the lower-bound properties give \(\alpha\leq a\) and \(\beta\leq b\). Adding yields \(\alpha+\beta\leq a+b\). Thus \(\alpha+\beta\) is a lower bound for \(A+B\).
Let \(\varepsilon>0\). By the epsilon characterization for \(A\), there is an \(a\in A\) with \(a<\alpha+\varepsilon/2\); for \(B\), there is a \(b\in B\) with \(b<\beta+\varepsilon/2\). Adding these inequalities gives
The element \(a+b\) belongs to \(A+B\), so the epsilon characterization for the lower bound \(\alpha+\beta\) proves that \(\inf(A+B)=\alpha+\beta\). The sum set is bounded below because this finite number is a lower bound. \(\square\)
Worked Example: The Infimum of a Sum Set
Let \(A=\{-2+1/n:n\text{ is a positive integer}\}\) and \(B=\{6+2/m:m\text{ is a positive integer}\}\). For \(A\), every element is greater than \(-2\), and choosing \(n>1/\varepsilon\) gives \(-2<-2+1/n<-2+\varepsilon\). Hence \(\inf A=-2\). For \(B\), every element is greater than \(6\), and choosing \(m>2/\varepsilon\) gives \(6<6+2/m<6+\varepsilon\). Hence \(\inf B=6\). The infimum of a sum set theorem gives
Indeed, each sum has the form
For a direct approximation check, let \(\varepsilon>0\). Choose positive integers \(n>2/\varepsilon\) and \(m>4/\varepsilon\). Then \(1/n<\varepsilon/2\) and \(2/m<\varepsilon/2\), so
Thus the sums approach \(4\) from above. The infimum is not attained, because attaining \(4\) would require \(1/n+2/m=0\), whereas both terms are positive.
A Reliable Infimum Workflow
The Infimum Strategy is easiest to use when the lower-bound and approximation arguments are kept separate. In a set defined by constraints, first establish a bound that works for every permitted element. Then design an element whose distance above the proposed infimum is controlled by the given error. In a theorem involving several sets, divide the error among the separate approximations so that their total remains within the target.
Verify that the set is nonempty and bounded below, so its infimum is defined.
Show that the proposed number is less than or equal to every element, including elements at any permitted boundary.
For arbitrary \(\varepsilon>0\), find an element strictly below the candidate plus \(\varepsilon\).
Apply the epsilon characterization, or contradict any proposed lower bound strictly larger than the candidate.
A common pitfall is to show only that a candidate is a lower bound and then call it the infimum. There may be larger lower bounds, so the approximation condition is necessary. Another pitfall is to confuse an infimum with a minimum: closeness does not imply attainment. If a proof needs the infimum to be an element of the set, it must verify that separately. In applications involving sums or transformations, also check that the hypotheses make the operations legitimate; in particular, positive scaling preserves the order used in the affine formula.
Check Your Understanding
Use the epsilon characterization and the arguments in this tutorial to answer these questions.
- What two conditions identify \(l\) as the infimum of a nonempty set bounded below?
- If \(l=\inf A\) and \(c>l\), how does the witness principle produce an element of \(A\) less than \(c\)?
- Why does \(\inf\{3+2/n:n\geq1\}=3\) not imply that \(3\) belongs to the set?
- Why does the affine transformation formula require a positive scaling factor?
- In the infimum of a sum set theorem, why is an error of \(\varepsilon/2\) used for each set?
- What additional property must be checked before calling an infimum a minimum?