Let S = {1 − 1/n : n ∈ ℕ} = {0, 1/2, 2/3, 3/4, ...}. What is sup(S)?
A1, because every element of S is strictly less than 1 and we can get arbitrarily close to 1 from within S
BThere is no supremum because 1 is not an element of S
CThe supremum does not exist because the sequence is infinite
D3/4, the largest element listed explicitly in the set description
The supremum is the least upper bound — the smallest value that is ≥ every element of S. Every element of S is less than 1, so 1 is an upper bound. For any ε > 0, choosing n large enough gives 1 − 1/n > 1 − ε, so we can get within ε of 1 from inside S — meaning no number smaller than 1 could be an upper bound. Therefore sup(S) = 1. The fact that 1 ∉ S is irrelevant: the supremum need not be in the set. Option B is the classic confusion between supremum and maximum.
Question 2 Multiple Choice
A student argues: 'The open interval (0, 1) has no supremum because it has no maximum — for any x in (0,1), I can always find something larger in the set.' What is wrong with this reasoning?
ANothing is wrong — if the maximum doesn't exist, neither does the supremum
BThe student confuses supremum with maximum; the supremum is the least upper bound, which equals 1 even though 1 ∉ (0, 1) and no maximum exists
CThe interval does contain its maximum — it is the limit of the sequence 1 − 1/n, which 'reaches' 1
DThe student correctly identifies that no supremum exists, but for the wrong reason — open intervals never have suprema
The supremum and maximum are different concepts. A maximum must be an element of the set that is ≥ all other elements. A supremum is the least upper bound — it must be ≥ all elements of S, but need not belong to S. The open interval (0, 1) has no maximum (you can always find a larger element inside), but its supremum is 1 — the smallest number that every element of (0,1) stays below. The completeness axiom guarantees this supremum exists in ℝ even without a maximum.
Question 3 True / False
The supremum of a bounded nonempty set in ℝ always exists, but may not be an element of the set.
TTrue
FFalse
Answer: True
This is the content of the Least Upper Bound Property (completeness axiom of ℝ): every nonempty subset of ℝ that is bounded above has a supremum in ℝ. The supremum may or may not belong to the set — it is the sup if the set attains it (maximum), and it is outside the set otherwise (as in open intervals or sets like {1 − 1/n : n ∈ ℕ}). Completeness guarantees existence; nothing guarantees membership.
Question 4 True / False
If a set S has no maximum element, then S has no supremum.
TTrue
FFalse
Answer: False
The supremum (least upper bound) and the maximum are distinct. A maximum is an element of S that is ≥ all others — it must belong to S. The supremum is the least upper bound — it need not belong to S. By completeness of ℝ, every bounded nonempty set has a supremum, whether or not it has a maximum. The open interval (0, 1) has no maximum but has supremum 1. The set {1 − 1/n : n ∈ ℕ} has no maximum but has supremum 1.
Question 5 Short Answer
State the epsilon characterization of the supremum and explain why condition (2) — 'for every ε > 0, there exists s ∈ S with s > sup(S) − ε' — is necessary to distinguish the supremum from just any upper bound.
Think about your answer, then reveal below.
Model answer: A value x = sup(S) if and only if: (1) x ≥ s for all s ∈ S (x is an upper bound), and (2) for every ε > 0, there exists s ∈ S with s > x − ε (x is the *least* upper bound). Condition (1) alone would be satisfied by any upper bound — for example, 100 is an upper bound of (0, 1) satisfying condition (1). Condition (2) rules out all upper bounds that are too large: it says you can always find an element of S within ε of x, so no number smaller than x could be an upper bound. Together, the two conditions uniquely characterize the smallest upper bound.
The epsilon structure in condition (2) — 'for every ε > 0, there exists...' — is the same quantifier pattern that appears throughout real analysis in convergence definitions. Recognizing it here builds the template for epsilon-delta arguments in limits and continuity.