 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2010/e/e2/b/ii/solution

x \in A = g^{- 1} (B)

and

y R x

, then

g (y) = g (x) \in B

, so

y \in A

. Thus

A

contains every equivalence class it meets: it is a saturated subset. Always

A \subseteq p^{- 1} (p (A))

; conversely, a point of the latter set is equivalent to some

x \in A

and hence belongs to

A

. Therefore

p^{- 1} (p (A)) = A .

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