 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-4/3h/solution

f

(0, 1)

precisely when

\forall ε > 0 \exists N \forall n \geq N \forall x \in (0, 1) : ∣ f_{n} (x) - f (x) ∣ < ε .

The same

N

must work throughout the interval; equivalently

sup_{0 < x < 1} ∣ f_{n} (x) - f (x) ∣ \to 0

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