Prove that topological equivalence is an equivalence relatio

Prove that topological equivalence is an equivalence relation for metric spaces.

Solution

Suppose that d₁ and d₂ are topologically equivalent, and that f is continuous from (X, d₁) to (Y, d_Y ). Then , for every open set U in (Y, dY ), f ¹ (U) is open in (X, d1). Since d1 and d2 are topologically equivalent, f ¹ (U) is also open in (X, d2), so that f is continuous from (X, d₂) to (Y, d_Y ). Conversely, suppose that f continuous from (X, d1) to (Y, d_Y ) implies f continuous from (X, d2) to (Y, dY ). Since this does not depend on f or Y , we are free to choose the image space as (X, d1) and the function from X to X as the identity function. Then for any open set U in (X, d1), f ¹ (U) = U is open in both (X, d1) and (X, d2), and d1 and d2 are topologically equivalent