Consider covering ℝ with the open intervals (−n, n) for n = 1, 2, 3, … This collection covers every real number, yet ℝ is not compact. What does this example demonstrate?
AThat ℝ has too many points to be covered by open sets at all
BThat some open covers of ℝ have no finite subcollection that still covers all of ℝ — so ℝ fails the definition of compactness
CThat compactness requires the intervals to be nested
DThat ℝ is not a topological space because it is unbounded
Compactness requires that EVERY open cover has a finite subcover — not just some nice ones. The intervals {(−n, n)} cover ℝ, but no finite subcollection does: any finite selection has a largest N, and real numbers beyond N are uncovered. This single open cover with no finite subcover is enough to prove ℝ is not compact. Compactness fails as soon as you can find even one 'escaping cover' with no finite subcover.
Question 2 Multiple Choice
If f: X → Y is a continuous surjection and X is compact, what can you conclude about Y?
AY is compact — the continuous image of a compact space is compact
BY is compact only if f is also injective (a homeomorphism)
CNothing — compactness is not preserved under continuous maps
DY is compact only if Y is a subset of ℝⁿ with the standard topology
The continuous image theorem states that the continuous image of a compact space is compact. The proof uses only the definitions: pull back any open cover of Y to an open cover of X (possible because f is continuous and surjective), extract a finite subcover of X, and push it forward to cover Y. No metric, no coordinates — just open sets. This is one of the most important consequences of the open-cover definition of compactness.
Question 3 True / False
A subset of ℝⁿ is compact if and mainly if it is closed and bounded. This is the definition of compactness.
TTrue
FFalse
Answer: False
False. 'Closed and bounded in ℝⁿ' is the Heine–Borel characterization of compactness, specific to ℝⁿ with the standard topology. The definition of compactness — valid in every topological space — is that every open cover has a finite subcover. In a general topological space, 'bounded' has no meaning (there is no metric), so the open-cover definition is primary. Heine–Borel is a theorem that the two conditions coincide in ℝⁿ, not the definition.
Question 4 True / False
Every finite topological space is compact.
TTrue
FFalse
Answer: True
True. If X = {p₁, p₂, …, pₙ} is finite, then any open cover assigns each pᵢ to at least one open set. Selecting one open set per point gives a finite subcover — with at most n sets. The argument works for any open cover, so X trivially satisfies the definition. Compact spaces generalize this finiteness property to infinite spaces: even infinitely many points, some 'finite sampling' of open sets already covers the whole space.
Question 5 Short Answer
Why is the open cover definition of compactness more fundamental than 'closed and bounded' for general topological spaces?
Think about your answer, then reveal below.
Model answer: Because 'bounded' requires a notion of distance (a metric), which a general topological space does not have. The open cover definition uses only the topology — the collection of open sets — with no additional structure. It therefore applies in any topological space, including abstract spaces with no metric, no coordinates, and no notion of size or distance. 'Closed and bounded' only makes sense once you have a metric.
The open cover definition is primary because topology is the study of properties that depend only on open sets, not on distances. Compactness, as defined by open covers, is a topological invariant: homeomorphic spaces are either both compact or both non-compact. 'Closed and bounded' is not a topological invariant — a set can be bounded in one metric and unbounded in another. The open cover definition captures the essence of 'finiteness' in purely topological terms.