 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2010/e/1/b/i/solution

We prove (i)

\Rightarrow

(ii). If

B = U \cap F

with

U

open and

F

closed, then

\overline{B} \subseteq F

. Hence

B \subseteq U \cap \overline{B} \subseteq U \cap F = B,

B = U \cap \overline{B}

. By the subspace topology definition this makes $B$ open in its closure.

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