Is the open interval (0,1) compact as a subset of ℝ with the standard topology?
AYes, because it is bounded and any open cover can be reduced to finitely many sets
BNo, because the cover {(1/n, 1) : n = 2, 3, 4, ...} has no finite subcover
CNo, because (0,1) is not closed, and only closed sets can be compact
DYes, because every continuous function on (0,1) attains its maximum and minimum
The collection {(1/n, 1)} for n ≥ 2 covers (0,1) — every x ∈ (0,1) lies in (1/n, 1) for all large enough n. But no finite sub-collection covers (0,1): any finite collection omits (1/N, 1) for some large N, leaving points near 0 uncovered. So (0,1) fails the compactness definition. Option C conflates closed and compact — closedness is relevant only when the set sits inside a compact ambient space. Option D is false: f(x) = 1/x is continuous on (0,1) but attains no maximum.
Question 2 Multiple Choice
A continuous function f: K → ℝ attains its maximum value on K for every continuous f. What does this tell us about K?
AK must be closed, since the maximum must be achieved at a boundary point
BK must be compact, since continuous images of compact sets are compact, and compact subsets of ℝ are closed and bounded
CK must be connected, since attaining a maximum requires no gaps in the domain
DK must be both compact and path-connected, so that the intermediate value theorem also applies
The extreme value theorem states that continuous functions on compact spaces attain their maximum and minimum. The key theorem here: continuous images of compact spaces are compact. If f(K) is compact in ℝ, it is closed and bounded, hence contains its supremum — so f attains its maximum. Conversely, if K is not compact, one can find a continuous function that doesn't attain its maximum (as (0,1) shows). Connectedness (options C, D) is needed for the intermediate value theorem, not the extreme value theorem.
Question 3 True / False
If K is compact and C ⊆ K is closed, then C is compact.
TTrue
FFalse
Answer: True
Given any open cover {Uᵢ} of C, add the open set K\C (open because C is closed in K) to obtain an open cover of K. Since K is compact, this extended cover has a finite subcover. Remove K\C from that finite subcover to get a finite sub-collection of the original {Uᵢ} that covers C. So C is compact. This is one of the two main preservation theorems: compact sets pass compactness to their closed subsets.
Question 4 True / False
Compactness is a metric-space concept, so the same set can be compact under one metric and non-compact under another on the same underlying set.
TTrue
FFalse
Answer: False
Compactness is a topological property defined purely in terms of open sets, making no reference to any metric. The open-cover definition applies in any topological space. It is true that different metrics on the same set can induce different topologies — and a change in topology can change compactness — but the concept itself is topological, not metric. This is precisely why the abstract open-cover definition matters: it works in spaces where no metric exists.
Question 5 Short Answer
Explain in your own words why (0,1) fails to be compact while [0,1] is compact. What does this contrast reveal about what compactness captures?
Think about your answer, then reveal below.
Model answer: (0,1) fails compactness because points can 'escape toward the boundary': the cover {(1/n,1)} has no finite subcover because any finite collection misses points arbitrarily close to 0. [0,1] is compact because its boundary points 0 and 1 are in the set, blocking that escape route; by Heine-Borel, any open cover of [0,1] reduces to a finite one. Compactness captures the absence of any way to 'escape' — to infinity or to a missing boundary point.
Non-compactness corresponds to a kind of leak: you can build covers where finitely many sets are always insufficient because points crowd toward some inaccessible limit. Compactness blocks all such leaks. This is why compactness is so powerful in analysis — it converts infinite coverings into finite ones, enabling extremal arguments (maximum and minimum) that fail on non-compact spaces like (0,1) or all of ℝ.