Questions: Extreme Value Theorem (Proof via Compactness)

Question 1

The function f(x) = 1/x is continuous on (0, 1]. Does the Extreme Value Theorem guarantee that f attains a maximum on this interval?

• A. Yes, because f is continuous on the interval (0, 1]
• B. No, because (0, 1] is not compact — it is bounded but not closed, so the EVT does not apply
• C. No, because f is not differentiable at x = 0
• D. Yes, because every continuous function on a bounded interval attains its supremum

Answer: (B)

The EVT requires the domain to be compact (closed and bounded in ℝ). The interval (0, 1] is bounded but not closed — it is missing the endpoint 0. As x → 0⁺, f(x) = 1/x → ∞, so f is not even bounded above, much less attaining a maximum. This counterexample shows continuity alone does not suffice; compactness is an essential hypothesis. Option A is the classic misconception: continuity is necessary but not sufficient. Option D is false — f is unbounded above on (0, 1].

Question 2

What are the two key steps in the rigorous proof of the Extreme Value Theorem?

• A. Step 1: Prove f is bounded above; Step 2: Prove the supremum is attained. Connected by the Bolzano-Weierstrass theorem.
• B. Step 1: Prove the continuous image of a compact set is compact; Step 2: Use Heine-Borel to conclude compact subsets of ℝ are closed, hence contain their supremum.
• C. Step 1: Prove the Intermediate Value Theorem; Step 2: Apply it to f − sup(f) to find a zero.
• D. Step 1: Prove f is uniformly continuous on the compact domain; Step 2: Use uniform continuity to construct a maximizing sequence.

Answer: (B)

The proof has two essential steps. First: if K is compact and f is continuous, then f(K) is compact (compactness is preserved under continuous maps). Second: compact subsets of ℝ are closed and bounded by Heine-Borel; a closed set contains its own supremum and infimum. Together these guarantee that sup f(K) ∈ f(K) — meaning f actually attains its maximum. Bolzano-Weierstrass (option A) is an equivalent characterization of compactness but is not how the standard proof is structured.

Question 3

A continuous function on a bounded open interval (a, b) is highly probable by the Extreme Value Theorem to attain its maximum.

• T. True
• F. False

Answer: False

An open interval is not compact — it is bounded but not closed. The EVT requires the domain to be compact. A continuous function on (a, b) may approach its supremum as x → a or x → b without ever attaining it. For example, f(x) = x on (0, 1) has supremum 1 but attains no maximum on the open interval since x = 1 is not in the domain. The distinction between open and closed intervals is precisely why the closed interval [a, b] — which is compact by Heine-Borel — is the natural domain for optimization in calculus.

Question 4

The Extreme Value Theorem guarantees that a continuous function on a compact set attains its maximum, but says nothing about where that maximum is located.

• T. True
• F. False

Answer: True

This is correct and an important limitation to understand. The EVT is an existence theorem: it guarantees that some point x* exists satisfying f(x*) = max f, but says nothing about how to find x*. The maximum might be at an interior point (where calculus provides the tool: f'(x*) = 0 if f is differentiable) or at a boundary point. Locating the extremum requires additional tools; the EVT only guarantees it exists.

Question 5

Explain why both hypotheses of the Extreme Value Theorem — continuity of f and compactness of the domain — are necessary. Give a counterexample showing what fails when each is dropped.

Model answer: Continuity is necessary: a discontinuous function on [0,1] can fail to attain its supremum (e.g., f(x) = x for x ∈ [0,1) and f(1) = 0 has supremum 1 but never attains it). Compactness is necessary: a continuous function on a non-compact domain can also fail (e.g., f(x) = 1/x on (0,1] is continuous but unbounded, attaining no maximum; g(x) = x on [0, ∞) is continuous but unbounded above). Without compactness, the continuous image need not be compact and so need not contain its own supremum.

Each hypothesis does genuine work in the proof. Continuity ensures that the image of the compact domain is compact. Compactness of the domain ensures the image is compact rather than merely bounded. Compact subsets of ℝ being closed (Heine-Borel) ensures the supremum is actually attained as an element of the image. Remove either hypothesis and one link in the chain 'compact domain + continuity → compact image → closed → contains its supremum' breaks.