Questions: The Completeness Axiom (Least Upper Bound Property)

Question 1

Consider the set S = {x ∈ ℚ : x² < 2}. Every element of S is rational and S is bounded above. What is true about the supremum of S?

• A. The supremum is √2, which exists as a rational number since it bounds S from above
• B. The supremum is 2, because 2 is the simplest rational upper bound
• C. The supremum exists in ℝ but not in ℚ — it equals √2, which is irrational, revealing a hole in the rationals
• D. S has no supremum because it is an open set with no largest element

Answer: (C)

This example is why the Completeness Axiom matters. S is non-empty and bounded above by any rational greater than √2, so by the axiom it has a supremum in ℝ — and that supremum is √2. But √2 is irrational, so no rational number is the least upper bound of S in ℚ. This demonstrates the 'hole' in ℚ that the axiom fills. Option D confuses supremum with maximum: open sets routinely lack a maximum but still have a supremum.

Question 2

What is the supremum of the open interval (0, 1) as a subset of ℝ, and does the set have a maximum?

• A. The supremum does not exist because the interval is open and no element is the largest
• B. The supremum is 0.999…, which is in the interval and serves as the largest element
• C. The supremum is 1, and 1 is in the interval — it is also the maximum
• D. The supremum is 1, but 1 is not in the interval — the supremum exists but the set has no maximum

Answer: (D)

The supremum is the smallest upper bound, not the largest element. For (0,1), every upper bound is ≥ 1, and 1 itself is an upper bound — so sup = 1. But 1 ∉ (0,1), so the set has no maximum. This is the essential distinction: supremum and maximum coincide only when the supremum is actually achieved by an element of the set. The Completeness Axiom guarantees the supremum exists in ℝ even when the set has no maximum.

Question 3

The set of rational numbers satisfies the Completeness Axiom: most non-empty subset of ℚ that is bounded above has a least upper bound in ℚ.

• T. True
• F. False

Answer: False

This is precisely what the Completeness Axiom denies about ℚ. The set {x ∈ ℚ : x² < 2} is non-empty, bounded above in ℚ, yet has no least upper bound in ℚ — its supremum is √2, which is irrational. The Completeness Axiom is the property that distinguishes ℝ from ℚ. Rationals satisfy all the ordered field axioms but fail completeness, which is why they have 'holes.'

Question 4

A set can have a supremum without having a maximum — the supremum need not be an element of the set.

• T. True
• F. False

Answer: True

The supremum (least upper bound) is defined by two conditions: it is an upper bound, and no smaller number is an upper bound. Neither condition requires the supremum to belong to the set. The open interval (0,1) has supremum 1 but no maximum; the set {1 − 1/n : n ∈ ℕ} = {0, 1/2, 2/3, 3/4, ...} has supremum 1 but 1 is not in the set. The maximum, by contrast, must be an element of the set that is also an upper bound.

Question 5

Why does the Completeness Axiom matter beyond being a technical axiom? Name one major theorem in real analysis that depends on it and explain the dependence.

Model answer: The Completeness Axiom guarantees that ℝ has no holes — every place where a sequence or process converges, the limit actually exists in ℝ. The Monotone Convergence Theorem is a direct application: if a sequence is increasing and bounded above, its range is a non-empty bounded set, and the Completeness Axiom guarantees it has a supremum in ℝ — that supremum is the limit. Without completeness, a bounded increasing sequence of rationals could approach √2 yet have no limit in the field.

The same dependence appears in the Intermediate Value Theorem (if a continuous function changes sign, the zero must exist — but existence requires completeness), the Extreme Value Theorem, and the Bolzano–Weierstrass Theorem. Completeness is not a curiosity; it is the single axiom that makes analysis work over ℝ rather than collapsing over ℚ.