Consider fₙ(x) = xⁿ on [0,1]. The sequence converges pointwise to a limit function f. Which of the following correctly describes f?
Af(x) = 0 for all x ∈ [0,1], since xⁿ → 0 for large n
Bf(x) = 0 for x ∈ [0,1) and f(1) = 1, making f discontinuous
Cf(x) = x for all x ∈ [0,1], since the identity is the natural limit
Df does not exist because the sequence fails to converge at every point
For any fixed x ∈ [0,1), xⁿ → 0 as n → ∞. But at x = 1, 1ⁿ = 1 for every n, so fₙ(1) = 1. The pointwise limit is thus 0 on [0,1) and 1 at x=1 — a discontinuous step function. Each fₙ is continuous (it's a polynomial), yet the limit is not. This is exactly the counterexample showing that pointwise convergence does not preserve continuity.
Question 2 Multiple Choice
In the ε/3 proof that uniform convergence preserves continuity, what is the critical role of uniform convergence that pointwise convergence cannot fill?
AUniform convergence guarantees each fₙ is bounded, which is needed to apply the triangle inequality
BUniform convergence allows choosing N large enough that |fₙ(y) − f(y)| < ε/3 for all y simultaneously, without N depending on y
CUniform convergence implies the functions are integrable, so the limit integral equals the integral of the limit
DPointwise convergence fails only for unbounded functions, which are automatically excluded here
The ε/3 argument splits |f(x) − f(y)| ≤ |f(x) − fₙ(x)| + |fₙ(x) − fₙ(y)| + |fₙ(y) − f(y)|. The first and third terms require bounding |fₙ(z) − f(z)| for the specific points x and y. With uniform convergence, one N controls all points at once: choose N so that |fₙ(z) − f(z)| < ε/3 for all z. With only pointwise convergence, the N for x might differ from the N for y — and since y may vary, no single N suffices.
Question 3 True / False
If each function in a sequence (fₙ) is continuous and the sequence converges pointwise to f on a closed interval, then f is necessarily continuous.
TTrue
FFalse
Answer: False
False. The counterexample is fₙ(x) = xⁿ on [0,1]: each fₙ is continuous, the convergence is pointwise, and yet the limit is discontinuous at x=1. The theorem requires uniform convergence, not merely pointwise. The failure occurs because pointwise convergence allows the convergence speed to vary with x — near x=1, xⁿ converges arbitrarily slowly — so the limit function can inherit discontinuities.
Question 4 True / False
Uniform convergence of (fₙ) to f on a set E means: for every ε > 0, there exists N such that for all x ∈ E and all n > N, |fₙ(x) − f(x)| < ε.
TTrue
FFalse
Answer: True
This is precisely the definition of uniform convergence. The key feature is that N depends only on ε, not on x — a single N works simultaneously for every point in E. Geometrically, the graphs of fₙ eventually lie inside an ε-band around the graph of f. Pointwise convergence allows N to depend on x: N(x,ε) might grow without bound as x approaches some point, and that is exactly what goes wrong in the xⁿ example near x=1.
Question 5 Short Answer
Explain in your own words why pointwise convergence fails to guarantee continuity of the limit, and what uniform convergence adds that fixes this.
Think about your answer, then reveal below.
Model answer: With pointwise convergence, the sequence can converge arbitrarily slowly near certain points, allowing the limit to 'jump' discontinuously. Uniform convergence forces a single convergence rate across all points simultaneously, which is precisely what the ε/3 argument needs: it lets you choose n large enough to control |fₙ(y) − f(y)| for all y at once, regardless of where y is.
The failure mode is illustrated by xⁿ on [0,1]: near x=1, convergence is arbitrarily slow, so the limit can 'suddenly' equal 1 at x=1 while being 0 nearby. Uniform convergence closes this gap by banning slow-near-some-point convergence. In the proof, once you fix n large enough that fₙ is uniformly within ε/3 of f everywhere, you can then use fₙ's continuity (choose δ for the middle term) knowing the outer terms are already controlled everywhere — including at the nearby point y you're comparing x to.