 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-ii/5/a/i/solution

S_{N} (x) = \sum_{n = 1}^{N} f_{n} (x)

s : E \to R

\forall ε > 0 \exists N_{0} \in N \forall N \geq N_{0} \forall x \in E : ∣ S_{N} (x) - s (x) ∣ < ε .

The same threshold

N_{0}

must control every point of

E

. Equivalently, the supremum error

sup_{x \in E} ∣ S_{N} (x) - s (x) ∣

tends to zero, for nonempty

E

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