Exercise 16.6

Show that the countable collection

{(a,b) × (c,d)a < b and c < d and a,b,c,d are rational}

is a basis for 2.

Answers

Proof. It was proven in Exercise 13.8 part (a) that the set

B = {(a,b)a < ba and b rational}

is a basis for the standard topology on . It then follows that

D = {B × CB,C B}

is a basis for the standard topology on 2 by Theorem 15.1. Clearly, we have

D = {(a,b) × (c,d)a < b and c < d and a,b,c,d are rational} ,

which shows the desired result. □

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2019-12-01 00:00
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