Why is f(x) = 1/x not uniformly continuous on (0, 1), even though it is continuous at every point in (0, 1)?
ABecause 1/x is unbounded on (0, 1), making it impossible to control globally
BBecause (0, 1) is not compact — sequences approaching 0 have no convergent subsequence inside the domain, allowing f to vary without bound near the missing endpoint
CBecause f is discontinuous at x = 0, which contaminates behavior on (0, 1)
DBecause f has an unbounded derivative everywhere, and any function with an unbounded derivative fails uniform continuity
The key is that (0, 1) is not compact — the endpoint 0 is missing. Sequences approaching 0 leave the domain, so there is no point in (0, 1) where continuity can constrain the function's behavior there. This is precisely what Bolzano-Weierstrass needs: a convergent subsequence whose limit is still in the domain. Without compactness, that fails and the contradiction argument collapses. Option A is a symptom, not the root cause — the theorem's proof hinges on compactness, not boundedness per se.
Question 2 Multiple Choice
A student claims: 'f(x) = sin(1/x) is uniformly continuous on [0.1, 1] because it is periodic.' What is the correct analysis?
AThe student is right — periodic functions are always uniformly continuous on closed intervals
BThe student reaches the right conclusion for the wrong reason: sin(1/x) is uniformly continuous on [0.1, 1] because [0.1, 1] is compact, and every continuous function on a compact set is uniformly continuous
CThe student is wrong: sin(1/x) is not uniformly continuous on [0.1, 1] because it oscillates rapidly
DThe student is wrong: uniform continuity requires monotonicity, which sin(1/x) lacks
sin(1/x) is continuous on [0.1, 1] (no singularity in this range), and [0.1, 1] is closed and bounded, hence compact by Heine-Borel. Therefore by the theorem, it is uniformly continuous. The student's justification (periodicity) is irrelevant and wrong — periodicity alone guarantees nothing about uniform continuity. The correct reason is the Heine-Cantor theorem: compact domain + continuity = uniform continuity.
Question 3 True / False
The function f(x) = x² is uniformly continuous on [0, 100] but not uniformly continuous on [0, ∞).
TTrue
FFalse
Answer: True
[0, 100] is closed and bounded, hence compact by Heine-Borel. Since x² is continuous, the theorem guarantees uniform continuity there. On [0, ∞), which is not compact, x² fails uniform continuity: as x grows large, |f(x + δ) − f(x)| = |2xδ + δ²| ≈ 2xδ can exceed ε for any fixed δ by choosing x large enough. Compactness is what makes the difference.
Question 4 True / False
Uniform continuity on a compact set is 'obvious' from continuity — no special proof is needed, because compact sets are inherently well-behaved.
TTrue
FFalse
Answer: False
This is a common intuitive shortcut that obscures real mathematical content. The proof requires a genuine argument using Bolzano-Weierstrass. Without it, you cannot rule out the possibility that the required δ shrinks to zero across the domain. Compactness is needed precisely to guarantee that every sequence has a convergent subsequence whose limit is in the domain — so that continuity at the limit gives a contradiction. Compactness is doing essential logical work, not just making things 'nicer.'
Question 5 Short Answer
Explain in your own words why compactness is essential to this theorem — what goes wrong when the domain is not compact?
Think about your answer, then reveal below.
Model answer: Without compactness, sequences in the domain can 'escape' toward a missing boundary point or toward infinity. The contradiction proof produces sequences (xₙ), (yₙ) with |xₙ − yₙ| → 0 but |f(xₙ) − f(yₙ)| ≥ ε. Bolzano-Weierstrass (which requires compactness) gives a convergent subsequence xₙₖ → p with p still in the domain. Continuity at p then forces both f(xₙₖ) and f(yₙₖ) to converge to f(p), producing the contradiction. Without compactness, the limit p might not be in the domain, and continuity at p cannot be invoked.
The failure of uniform continuity on (0, 1) for f(x) = 1/x illustrates this exactly: take xₙ = 1/n and yₙ = 1/(n+1). Then |xₙ − yₙ| → 0 but |f(xₙ) − f(yₙ)| = |n − (n+1)| = 1. The sequence xₙ → 0, but 0 is not in the domain — so we cannot use continuity at 0 to close the argument. Compactness plugs precisely this gap.