Consider the open interval (0,1) covered by the collection Uₙ = (1/n, 1) for each positive integer n. This is an open cover with no finite subcover. What does this prove?
AThat (0,1) is compact, because we found a valid open cover
BThat (0,1) is not compact, because we found one open cover from which no finite subcover can be extracted
CNothing about compactness — we would need to check all open covers to draw any conclusion
DThat (0,1) is compact under certain covers but not others, depending on how it is covered
Compactness requires that EVERY open cover has a finite subcover. To prove a space is NOT compact, it suffices to exhibit just ONE open cover with no finite subcover — which is exactly what this example does. Any finite subcollection {U_{n₁}, ..., U_{nₖ}} only covers (1/N, 1) where N = max(n₁,...,nₖ), leaving (0, 1/N] uncovered. The missing endpoint 0 provides the 'escape route.' Option C is the most tempting wrong answer: it misreads the definition as requiring 'all covers' to be checked for non-compactness, when in fact a single counterexample suffices.
Question 2 Multiple Choice
Which of the following subsets of ℝ is compact, according to the Heine-Borel theorem?
AThe open interval (0,1), because it is bounded
BThe closed ray [0, ∞), because it is closed
CThe entire real line ℝ, because every point has a neighborhood
DThe closed interval [0,1], because it is both closed and bounded
The Heine-Borel theorem for ℝⁿ states that a subset is compact if and only if it is both closed AND bounded. [0,1] satisfies both: it is closed (contains its limit points 0 and 1) and bounded (fits inside the ball of radius 1). The open interval (0,1) is bounded but not closed — its limit points 0 and 1 are missing, providing escape routes. The ray [0,∞) is closed but unbounded — you can cover it with (n-1, n+1) for each integer n ≥ 0 and find no finite subcover. ℝ fails both conditions.
Question 3 True / False
A space is compact if there exists at least one open cover that has a finite subcover.
TTrue
FFalse
Answer: False
This is the most important misconception to avoid. The definition of compactness requires that EVERY open cover has a finite subcover — not just some particularly nice one. Every space has at least one open cover with a finite subcover (cover the whole space with the single open set X itself). The difficulty of compactness lies in the quantifier 'every': you must guarantee that even adversarially constructed open covers — covers specifically designed to be difficult to trim — still yield a finite subcover. The word 'every' is what makes compactness a powerful and non-trivial property.
Question 4 True / False
The closed interval [0,1] is compact in ℝ because the endpoints 0 and 1 prevent open covers from having arbitrarily thin 'escape routes' near the boundary.
TTrue
FFalse
Answer: True
Intuitively, this is exactly right. In (0,1), the missing endpoint 0 allows the open cover Uₙ = (1/n, 1) to keep shifting the cover's left boundary closer and closer to 0 without ever covering 0 itself — a finite subcover would have to stop at some 1/N, leaving (0,1/N] exposed. In [0,1], the point 0 must be covered by some open set containing it; any such set covers an interval [0, ε) for some ε > 0, blocking the escape route. This is the geometric intuition behind Heine-Borel: being closed plugs all boundary escape routes; being bounded prevents escape to infinity.
Question 5 Short Answer
Explain what it means for a topological space to NOT be compact, using the open-cover definition, and give an example.
Think about your answer, then reveal below.
Model answer: A space is not compact if there exists at least one open cover from which no finite subcollection still covers the space. Example: ℝ covered by (n-1, n+1) for each integer n has no finite subcover, so ℝ is not compact.
Negating the definition: 'every open cover has a finite subcover' becomes 'there exists an open cover with NO finite subcover.' To show non-compactness, exhibit such a cover. For ℝ: the cover {(n-1, n+1) : n ∈ ℤ} is open, but any finite subcollection only covers a bounded portion of ℝ, leaving infinitely many integers uncovered. For (0,1): the cover {(1/n, 1) : n ≥ 1} has no finite subcover because the left endpoints approach 0 from above, and any finite subcollection misses points near 0. The existence of even one such 'adversarial' cover is sufficient to prove non-compactness.