Questions: The Completeness Axiom (Least Upper Bound Property)

Question 1

Consider S = {x ∈ ℚ : x² < 2}. This set is non-empty and bounded above by 2. What does this example reveal about ℚ?

• A. ℚ satisfies the Completeness Axiom because the supremum √2 exists in ℝ
• B. The set has no supremum because it is unbounded in ℚ
• C. ℚ fails the Completeness Axiom — the set is non-empty and bounded above, but has no supremum within ℚ
• D. The set has a supremum in ℚ: the rational number closest to √2

Answer: (C)

√2 is irrational, so there is no rational number that is the *least* upper bound of S in ℚ. For any rational upper bound q > √2, you can find a smaller rational upper bound, so no least upper bound exists in ℚ. This is the exact gap the Completeness Axiom plugs in ℝ. Option A confuses 'the sup exists in ℝ' with 'ℚ satisfies completeness'; option D misunderstands that 'closest rational' doesn't give a least upper bound — there is always a closer one.

Question 2

A student proves that a certain sequence {aₙ} is increasing and bounded above by 5. She concludes the sequence converges. Which property of ℝ does her argument rely on most directly?

• A. The Archimedean property of ℝ
• B. The density of ℚ in ℝ
• C. The ordered field axioms of ℝ
• D. The Completeness Axiom — every non-empty set bounded above has a supremum in ℝ

Answer: (D)

The Monotone Convergence Theorem (an increasing bounded sequence converges) is a direct consequence of the Completeness Axiom. The proof constructs the set of all terms {a₁, a₂, …}, notes it is non-empty and bounded above, invokes completeness to obtain a supremum L ∈ ℝ, then shows the sequence converges to L. None of the other listed properties are sufficient: a bounded increasing sequence over ℚ (like rational approximations to √2) can fail to converge within ℚ.

Question 3

The Completeness Axiom is derivable from the other axioms of an ordered field.

• T. True
• F. False

Answer: False

Completeness is an *additional* axiom that cannot be proved from the ordered field axioms alone. The proof that it is independent is constructive: ℚ is an ordered field that satisfies all other ordered field axioms, yet ℚ is not complete (as the set {x ∈ ℚ : x² < 2} demonstrates). Because ℚ is a model satisfying the ordered field axioms but not completeness, completeness cannot be a logical consequence of those axioms. This is why completeness is listed as a separate axiom characterizing ℝ — and why any complete ordered field is isomorphic to ℝ.

Question 4

The rationals ℚ, despite being an ordered field, fail to satisfy the Completeness Axiom.

• T. True
• F. False

Answer: True

This is exactly what the example S = {x ∈ ℚ : x² < 2} demonstrates. S is non-empty and bounded above in ℚ, but its supremum √2 is not in ℚ. So ℚ contains a non-empty bounded-above set with no least upper bound in ℚ — precisely violating the Completeness Axiom. Real analysis is built on ℝ rather than ℚ for exactly this reason.

Question 5

Why does real analysis require ℝ rather than ℚ as its number system? What would fail if we tried to do analysis over ℚ?

Model answer: ℚ has 'gaps' — bounded sets with no supremum in ℚ. This means existence proofs break down: the standard proof that a bounded increasing sequence converges requires the limit to exist as a number, but in ℚ that number may be irrational and therefore absent. Major theorems — the Monotone Convergence Theorem, Intermediate Value Theorem, Extreme Value Theorem — all rely on suprema (or infima) existing; they all fail over ℚ. For example, f(x) = x² − 2 is continuous on [1, 2] ⊂ ℚ and changes sign, but has no root in ℚ, violating the IVT.

The Completeness Axiom is not a technicality but the foundation of the entire edifice of real analysis. Every major existence theorem — existence of limits, maxima, fixed points — ultimately reduces to invoking completeness. Without it, calculus as we know it simply doesn't work: you can write down all the definitions, but theorems that assert the existence of limits and extrema will have counterexamples over ℚ.