 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2012/e/e2/b/solution

T

to be all unions of members of

B

, including the empty union. Then

\emptyset \in T

and the covering assumption gives

X \in T

. Arbitrary unions of such sets are again unions of basis members.

For two members

U = ⋃_{i} B_{i}

and

V = ⋃_{j} B_{j}^{'}

, distribute intersection over union:

U \cap V = ⋃_{i, j} (B_{i} \cap B_{j}^{'}) .

Each intersection is a union of members of

B

by the second assumption, so

U \cap V \in T

. Repeating gives closure under all finite intersections. These are precisely the topology axioms. Every

B \in B

belongs to

T

, and every set in

T

is a union of such members by construction. Thus

T = {⋃ C : C \subseteq B}, B is a basis for T .

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