 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2013/e/1/b/ii/solution

Choose

v_{0} \in V

, using its nonemptiness. For each

u \in U

, the openness of

U \times V

in the product topology supplies open

O_{u} \subseteq X

W_{u} \subseteq Y

such that

(u, v_{0}) \in O_{u} \times W_{u} \subseteq U \times V .

Since

v_{0} \in W_{u}

, every point of

O_{u}

belongs to

U

. Thus

U = ⋃_{u \in U} O_{u}

is open. Interchanging the factors, using a fixed

u_{0} \in U

, proves

V

open. Therefore

U \in T_{X}, V \in T_{Y}

. Nonemptiness matters: an empty product is open whatever the other factor is.

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