 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-1/10g/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

This is always true. Given

ε > 0

, choose

N_{A}, N_{B}

such that

∣ f_{n} (x) - f (x) ∣ < ε

A

for

n \geq N_{A}

and on

B

for

n \geq N_{B}

. For

n \geq max (N_{A}, N_{B})

the same inequality holds at every point of

X = A \cup B

. Hence

f_{n} \to f

X

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