Consider fₙ(x) = xⁿ on [0,1], which converges pointwise to f(x)=0 for x∈[0,1) and f(1)=1. A student says this sequence also converges uniformly because 'every fₙ is continuous and they converge at every point.' What is wrong?
ANothing is wrong — pointwise convergence at every point implies uniform convergence
BThe argument ignores that the pointwise limit is discontinuous — uniform convergence of continuous functions always produces a continuous limit, so this sequence cannot converge uniformly
CThe argument is wrong because uniform convergence only applies to differentiable functions
DThe student is correct for closed intervals like [0,1] but wrong for open intervals
Uniform convergence preserves continuity: if each fₙ is continuous and fₙ → f uniformly, then f is continuous. Since the pointwise limit here is discontinuous (jumps at x=1), the convergence cannot be uniform. The failure of uniform convergence can also be confirmed via the supremum criterion: sup_{x∈[0,1]} |xⁿ − f(x)| does not go to zero.
Question 2 Multiple Choice
Which condition correctly captures uniform convergence of fₙ to f on S?
AFor each x∈S, there exists N(x,ε) such that n>N implies |fₙ(x)−f(x)| < ε
BFor every ε>0, there exists N(ε) independent of x such that for all x∈S, n>N implies |fₙ(x)−f(x)| < ε
Csup_{x∈S}|fₙ(x)−f(x)| is bounded for all n
DThere exists a fixed N such that |fₙ(x)−f(x)| < ε for all n and all x
Option A is pointwise convergence — each x has its own N that can grow without bound. Option B is uniform convergence — one N serves all x simultaneously. The critical distinction is whether N depends on x. Option C only says the differences are bounded, not that they converge to zero.
Question 3 True / False
Uniform convergence of a sequence of continuous functions guarantees that the limit function is also continuous.
TTrue
FFalse
Answer: True
This is the key theorem separating uniform from pointwise convergence. If fₙ are all continuous on S and fₙ → f uniformly, then f is continuous on S. The proof uses the uniform N to control all three parts of the ε/3 argument simultaneously — something pointwise convergence cannot do because different points need different N values.
Question 4 True / False
If fₙ → f pointwise on S, then the limit function f inherits most of the properties of the individual fₙ (continuity, integrability, differentiability).
TTrue
FFalse
Answer: False
Pointwise convergence does not preserve these properties. The canonical counterexample: fₙ(x) = xⁿ on [0,1] is a sequence of continuous, differentiable functions, but the pointwise limit is discontinuous. Uniform convergence is required to guarantee that such properties pass to the limit.
Question 5 Short Answer
Explain the key difference between pointwise and uniform convergence using the idea of a 'deadline N.' Why does this difference matter for preserving continuity?
Think about your answer, then reveal below.
Model answer: In pointwise convergence, each point x gets its own deadline N(x,ε) — different points can take arbitrarily long to converge. In uniform convergence, a single deadline N(ε) applies to all points simultaneously: once n exceeds N, every point in the domain is within ε of the limit at the same time. This shared deadline is what allows continuity to be preserved: near a point x₀, we can choose n large enough that all nearby points are simultaneously close to their respective limit values, making the limit function continuous at x₀.
The key insight is that continuity requires controlling behavior at an entire neighborhood of points simultaneously — exactly what uniform convergence provides and pointwise convergence does not.