A(0, 1) — an open interval, bounded but not closed
B[0, ∞) — a closed set that extends to infinity
C[0, 1] — a closed and bounded interval
Dℤ (the integers) — a closed set with isolated points
By the Heine-Borel theorem, a subset of ℝ is compact if and only if it is both closed AND bounded. [0, 1] satisfies both conditions. (0, 1) is bounded but not closed — it fails to contain its boundary points 0 and 1, allowing sequences to converge outside the set. [0, ∞) is closed but unbounded — sequences can escape to infinity. ℤ is closed (it contains all its limit points, vacuously — integers have no limit points in ℝ) but unbounded, so not compact. Both conditions are essential.
Question 2 Multiple Choice
Consider the open cover of (0, 1) given by the intervals (1/n, 1) for n = 1, 2, 3, … . Why does this cover prove (0, 1) is not compact?
ABecause infinitely many intervals are needed to cover (0, 1), and compact sets can only be covered by finitely many open sets
BBecause this is a valid open cover of (0, 1) with no finite subcover — every finite subcollection fails to cover points near 0
CBecause the intervals overlap, violating the compactness condition
DBecause (0, 1) has infinitely many points, and compact sets must be finite
Compactness requires that *every* open cover has a finite subcover — not that some covers are finite, but that *all* of them are. The collection {(1/n, 1) : n ∈ ℕ} is a valid open cover of (0, 1): every point x ∈ (0, 1) lies in (1/n, 1) for sufficiently large n. But any finite subcollection {(1/n₁, 1), …, (1/nₖ, 1)} has a minimum 1/nₘₐₓ, and the points in (0, 1/nₘₐₓ) are not covered. Since this one cover has no finite subcover, (0, 1) fails the compactness definition. Option A confuses 'an infinite cover exists' (true for all infinite sets) with 'no finite subcover exists for this cover' (the actual failure).
Question 3 True / False
Nearly every closed subset of ℝ is compact.
TTrue
FFalse
Answer: False
Compactness in ℝ requires *both* closed and bounded (Heine-Borel). A closed set can be unbounded and therefore not compact. For example, ℝ itself is closed (it contains all its limit points) but clearly not compact — the open cover {(−n, n) : n ∈ ℕ} has no finite subcover. Similarly, [0, ∞) is closed but not compact: the cover {[0, n) : n ∈ ℕ} has no finite subcover. Closedness prevents sequences from escaping through the boundary; boundedness prevents them from escaping to infinity. Both are needed.
Question 4 True / False
Every sequence in a compact set K ⊆ ℝ has a subsequence that converges to a point in K.
TTrue
FFalse
Answer: True
This is the sequential characterization of compactness, and it holds in all metric spaces (not just ℝ). The proof uses two properties of compact sets in ℝ: boundedness (guarantees a convergent subsequence exists by Bolzano-Weierstrass) and closedness (guarantees the limit of any convergent sequence in K stays in K). If K were only bounded, the subsequential limit might lie outside K. If K were only closed, Bolzano-Weierstrass might not apply (the sequence could be unbounded). The conjunction is precisely what compact sets provide.
Question 5 Short Answer
Explain in your own words why the extreme value theorem — a continuous function on a compact set attains its maximum — requires compactness rather than just closedness or just boundedness.
Think about your answer, then reveal below.
Model answer: A continuous function on a closed but unbounded set can fail to attain its supremum by 'escaping to infinity' — for example, f(x) = x on [0, ∞) has no maximum. A continuous function on a bounded but open set can fail by approaching but never reaching a boundary value — for example, f(x) = 1/x on (0, 1] approaches infinity as x→0 without attaining it. Compactness (closed AND bounded) rules out both failure modes: boundedness keeps the function from escaping to infinity, and closedness ensures the supremum is actually achieved within the set.
The proof runs: since f is continuous and K is compact, the image f(K) is also compact (continuous images of compact sets are compact). A compact subset of ℝ is closed and bounded, hence contains its supremum. So the supremum of f(K) is in f(K), meaning some point in K achieves it. Each piece of compactness blocks one escape route: boundedness keeps f(K) from going to ±∞, and closedness keeps the sup from being a limit that f never actually hits.