Part CCSIR NET June 2025uniform-convergence-carries-uniform-continuity-and-none-of-the-others

Uniform convergence carries uniform continuity and none of the others

Let $(f_{n})_{n} \geq_{1}$ be a sequence of real-valued functions on $R$ . Which of the following statements are true?

A.If each $f_{n}$ is uniformly continuous and $(f_{n})_{n} \geq_{1}$ converges to f uniformly, then f is uniformly continuous.✓
B.If each $f_{n}$ is bounded and $(f_{n})_{n} \geq_{1}$ converges to f pointwise, then f is bounded.
C.If each $f_{n}$ is bounded and continuous, $(f_{n})_{n} \geq_{1}$ converges pointwise to a bounded and continuous function f, then the convergence is uniform.
D.If each $f_{n}$ is differentiable and $(f_{n})_{n} \geq_{1}$ converges to f uniformly, then f is differentiable.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests pointwise vs uniform.

See pricing

50 are analysed free — try those first.

The trap it tests

Pointwise vs uniform

The mode of convergence — or of continuity — was the wrong one.

Drill statements like this

Related counterexample: fₙ → f uniformly ⇒ fₙ′ → f′

Uniform convergence carries uniform continuity and none of the others

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