Suprema, infima and the Archimedean property

Try

Consider a_n=2−1/n for positive integers n. The values increase toward 2 but never equal it. The set of values has supremum 2 and no maximum. To show that 2 is least, take a proposed bound b<2 and find n with 1/n<2−b. Then a_n>b. This requires the ability to choose an integer larger than any prescribed real threshold, a fact that will itself be proved from completeness.

Predict

Predict what “arbitrarily close below the supremum” means. It does not mean one distinguished element is infinitesimally close, nor that the supremum belongs to the set. It means that after any positive tolerance ε is chosen, some element lies above s−ε. The element may depend on ε. The order of these choices is the mechanism that will later appear in limit definitions.

Experiment

Build a tolerance table for the worked set.

Tolerance ε	Sufficient integer choice	Result
1/2	n>2	2−ε<a_n<2
1/10	n>10	2−ε<a_n<2
Any ε>0	n>1/ε	Same inequality

Use strict inequalities consistently. If n=1/ε happens to be an integer, the error equals ε rather than being smaller. Produce a counterexample to the claim that every element of a set must lie close to its supremum.

Understand

For a nonempty bounded-above set A, s=sup A exactly when s is an upper bound and every ε>0 admits a∈A with s−ε<a. If the latter property failed, s−ε would be a smaller upper bound. Conversely, any upper bound b<s is defeated by choosing ε=s−b and obtaining a>b. This proves both directions of the useful approximation criterion.

Now prove the Archimedean property. If the natural numbers were bounded above, completeness would give a supremum s. The criterion supplies n>s−1. But the natural number n+1 is greater than s, contradicting that s bounds all natural numbers. Therefore for every real threshold there is a larger natural number. This justifies the choices in the table without circularly assuming 1/n tends to zero.

Apply

Infima are the reversed-order analogue. A lower bound l satisfies l≤a for every a∈A; the infimum is the greatest lower bound. For a nonempty set bounded below, negating elements produces a bounded-above set, and inf A=−sup(−A). Carefully reverse each inequality when verifying this identity.

Repair “a set bounded above must have a minimum.” Upper bounds control only one direction. The interval (−∞,0) is bounded above but has no lower bound or minimum. Even a set bounded on both sides may attain neither endpoint, as an open bounded interval demonstrates.

Squaring a set can reverse which endpoint matters

For E=(−3,1), the supremum of E is one, but the supremum of its squared values is nine. Values close to −3 have the largest magnitudes. Neither −3 nor nine has to belong to the respective set. Predict what changes when an isolated point −3 is added instead of approached.

Does the farthest value belong to the set?

Domain set E Open interval (−3, 1)Two open intervals (−3, −2) and (2, 3)Point −3 together with (−2, 1)Interval (−3, 3] Squared-distance center c -101 Approximating member index n 24816

Map x to (x−c)². Predict the image supremum and whether it is attained, then compare the lower image bound. Open circles exclude endpoints; filled circles include them. Two images may overlap.

inf E=-3, sup E=1. Image infimum 0 (attained minimum); image supremum 9 (not attained).

Domain component	Exact image component
(-3, 1)	[0, 9)

Actual member approaching an extremal endpoint: x=-2.200000, image=4.840000. These displayed decimals are rounded; the endpoint inclusion tests and interval images use exact preset arithmetic. The excluded endpoint is not inserted into E.

Reset

The interval image formulas, not a finite search, establish the bounds. Controls are temporary and award no mastery.

This addresses the transformed-set proof in MIT 18.100B problem set 3, questions 2–3. The epsilon criterion already proved above is the key; no endpoint attainment is assumed.

Prove the shifted-square supremum

Let E be nonempty and bounded, l=inf E, s=sup E, and c any real center. Put R=max(|l−c|,|s−c|). Every x∈E lies in [l,s], so |x−c|≤R and (x−c)²≤R². The image is nonempty and bounded, so completeness supplies its supremum.

Choose an endpoint d, either l or s, with |d−c|=R. The infimum or supremum criterion supplies elements of E arbitrarily close to d, even if d is excluded. For ε>0 choose such an x with |x−d|<ε/(2R+1). Since both |x−c| and |d−c| are at most R, factorization gives

|(x−c)²−(d−c)²| = |x−d| |x+d−2c| ≤ 2R |x−d| < ε.

The previously established upper bound and this approximation prove

sup of (x−c)² over x∈E = max((l−c)², (s−c)²).

If R=0, E is the singleton consisting of c and the result is zero. The formula also follows by choosing aₙ∈E with l≤aₙ<l+1/n and bₙ∈E with s−1/n<bₙ≤s: these converge to l and s by the defining estimates. They are constructed sequences of set members, not necessarily distinct or subsequences of a preexisting sequence. An attained endpoint may be repeated.

Bounds, attained values and holes

The image supremum is a maximum exactly when some member x of E has |x−c|=R. Such a point is a farthest endpoint of [l,s]. If both endpoints tie, inclusion of either one suffices. For E=(−3,3], c=0, the supremum nine is attained at three despite exclusion of −3.

The image infimum does not obey the same endpoint-only rule. The sets (−3,3) and (−3,−2)∪(2,3) have identical infimum and supremum. Their squared images have infima zero and four, respectively. Zero is attained in the first; four is not attained in the second. The gap in the domain matters. For a single interval, the nearest point to c determines the lower image bound, with endpoint inclusion checked separately. For unions, compute each component image and take their union; do not fill gaps merely because the outer endpoints agree.

For the inspector’s shifted center c=1 and two open components, the image components are (9,16) and (1,4). Their overall supremum is sixteen and infimum one, both unattained. The finite approximating member illustrates the bound argument, while the proof establishes what happens for every positive tolerance.

Continue

Optional depth asks how suprema interact with operations on sets. For A+B, an upper bound is sup A+sup B. To prove leastness, choose one element within ε/2 of each supremum and add. The tolerance split is not arbitrary decoration: it ensures the two approximation errors total less than ε.

Write a proof using this pattern without assuming either supremum is attained. Record separately the upper-bound step and the approximation step. In the next lesson, the Archimedean property will place rational points inside any real interval, while suprema will locate common points of nested closed intervals.