# Exercise 7.32#

See also Subspace topology. The author refers to an arbitrary open set of $$X$$ as $$B$$, and of $$Y$$ as $$A$$. Keep this flip in mind; we think of $$X$$$$Y$$ (alphabetically) but this doesn’t match $$A$$$$B$$. The Wikipedia article has slightly better variable names.

For part 1., take $$B = X$$. Taking $$B = Y$$ (as the author suggests) is incorrect because there’s no guarantee that $$Y ∈ \bf{Op}$$.

For part 2. we know that $$Y$$ is a member of $$\bf{Op}_{?∩Y}$$ by part 1. and that ∅ is a member by taking $$B = ∅$$.

We know that we have binary/finite intersections between any $$A_1, A_2$$ because there must be some $$B_1, B_2$$ such that $$A_1 = B_1 ∩ Y$$ and $$A_1 = B_2 ∩ Y$$. Since $$B_1 ∩ B_2 ∈ \bf{Op}$$ because we have arbitrary intersections, we must have:

$A_3 = A_1 ∩ A_2 = (B_1 ∩ Y) ∩ (B_2 ∩ Y) = (B_1 ∩ B_2) ∩ Y ∈ \bf{Op}_{?∩Y}$

To show that we have arbitrary unions, we must show that given $$I$$ as a set where we are given an open set $$A_i ∈ \bf{Op}_{?∩Y}$$ for each $$i$$ then their union $$⋃_{i∈I}A_i ∈ \bf{Op}_{?∩Y}$$. We know that for every $$A_i$$ there must be some corresponding $$B_i$$ such that $$A_i = B_i ∩ Y$$, so we can also write arbitrary unions of $$A_i$$ as $$⋃_{i∈I}A_i = ⋃_{i∈I}(B_i ∩ Y) = (⋃_{i∈I}B_i) ∩ Y$$. We know that $$⋃_{i∈I}B_i ∈ \bf{Op}$$ because it has arbitrary unions, so $$⋃_{i∈I}A_i ∈ \bf{Op}_{?∩Y}$$.

For part 3. we must show that for every $$B ∈ \bf{Op}$$, the preimage $$f^{-1}(B) ∈ \bf{Op}_{?∩Y}$$. Take any $$B$$ that has no elements in common with $$Y$$; the preimage will be ∅ ∈ $$\bf{Op}_{?∩Y}$$. Otherwise the preimage $$f^{-1}(B)$$ will be some open set $$A = B ∩ Y ∈ \bf{Op}_{?∩Y}$$ that has all or a subset of the elements in $$B$$.