You want to prove that a continuous function f: X → ℝ achieves its maximum on X. You take a maximizing sequence (xₙ) with f(xₙ) → sup f. What property of X allows you to extract a convergent subsequence and complete the proof?
AX must be connected, so f cannot skip values on the way to its supremum
BX must be path-connected, allowing continuous curves between any two points
CX must be sequentially compact, guaranteeing a subsequence xₙₖ → x* in X so continuity gives f(x*) = sup f
DX must be bounded so the sequence cannot escape to infinity
Boundedness alone is insufficient — a bounded sequence can still converge outside the set (e.g., in an open interval). Sequential compactness guarantees two things: first, a convergent subsequence exists; second, the limit is in X. Then continuity of f gives f(x*) = lim f(xₙₖ) = sup f. This argument pattern — take a sequence, extract a convergent subsequence, identify the limit as a solution — is the core technique that makes sequential compactness practically valuable in analysis.
Question 2 Multiple Choice
In which setting are sequential compactness and compactness (open-cover definition) guaranteed to be equivalent?
AIn all topological spaces — the two definitions always describe the same property
BIn metric spaces — the metric structure allows sequences to capture the full topology
CIn Hausdorff spaces — the separation axiom is sufficient for the equivalence
DOnly in finite topological spaces — infinite spaces can always be constructed to separate the two
The equivalence holds in metric spaces, where the metric allows sequences to detect all open-set behavior through total boundedness and completeness arguments. In general topological spaces, the equivalence fails: there exist compact spaces that are not sequentially compact, and sequentially compact spaces that are not compact. Without a metric, sequences are too coarse an instrument to fully detect the topology — you need nets or filters to capture everything open covers can express.
Question 3 True / False
A space can be sequentially compact without being compact (in the open-cover sense) when considered as a general topological space.
TTrue
FFalse
Answer: True
The equivalence between sequential compactness and compactness is a theorem specific to metric spaces, not a universal truth. In general topological spaces, there are sequentially compact spaces that are not compact (uncountable products can exhibit this), and compact spaces that are not sequentially compact. This is why topology distinguishes the two concepts: each captures a different aspect of 'finiteness' in a space, and they only coincide when the metric provides enough structure for sequences to do the full job.
Question 4 True / False
The open interval (0, 1) fails to be sequentially compact because it contains unbounded sequences.
TTrue
FFalse
Answer: False
Every sequence in (0, 1) is bounded — the issue is not boundedness but whether limit points stay in the space. The sequence 1/n is bounded (all terms lie in (0, 1)) but converges to 0, which is outside (0, 1). So no subsequence of 1/n can converge within (0, 1). Sequential compactness requires that the limit of the convergent subsequence belongs to the space itself. Closedness is the missing property here — (0, 1) lacks its boundary points, which are exactly where sequences can 'escape.'
Question 5 Short Answer
Why is it insufficient to know a sequence in (0, 1) is bounded in order to conclude that (0, 1) is sequentially compact?
Think about your answer, then reveal below.
Model answer: Sequential compactness requires that every sequence has a convergent subsequence whose limit is in the space. A bounded sequence in (0, 1) may have a convergent subsequence, but the limit might be 0 or 1 — points outside (0, 1). The space must be closed to guarantee limit points stay inside. Boundedness ensures subsequences exist (by Bolzano-Weierstrass in ℝ), but closedness ensures the limit is in the space. Sequential compactness requires both.
The Bolzano-Weierstrass theorem guarantees convergent subsequences for bounded sequences in ℝ, but sequential compactness is a property of the space, not of ℝ. The closed interval [0, 1] is sequentially compact precisely because it is both bounded (so subsequences converge in ℝ) and closed (so the limits stay inside). Remove closedness by taking (0, 1) and the boundary points 0 and 1 become escape routes. This is why 'closed and bounded' (Heine-Borel) characterizes compact sets in ℝⁿ.