Solution

The inequalities imply

B_{d_{1}} (x, r / β) \subseteq B_{d_{2}} (x, r) and B_{d_{2}} (x, α r) \subseteq B_{d_{1}} (x, r) .

Thus the two metrics induce the same open sets and are equivalent.

The converse fails. On

R

, let

d_{1} (x, y) = ∣ x - y ∣, d_{2} (x, y) = ∣ arctan x - arctan y ∣.

The metrics are equivalent because

arctan : R \to (- π /2, π /2)

is a homeomorphism, but

\frac{d _{2} ( 0 , n )}{d _{1} ( 0 , n )} ⟶ 0,

so no positive lower comparison constant exists. This is an equivalent metrics need not be bi-Lipschitz equivalent example.

Equivalent metrics on the codomain also need not give the same uniform convergence. Take

X = N

Y = R

, and use the two metrics above. Define

f (k) = k, f_{n} (k) = {2 n, k, k = n, k \neq = n .

Then

sup_{k} d_{2} (f_{n} (k), f (k)) = arctan (2 n) - arctan n ⟶ 0,

f_{n} \to f

uniformly for

d_{2}

, whereas

sup_{k} d_{1} (f_{n} (k), f (k)) = n,

so convergence is not uniform for

d_{1}

Solved by gpt-5.6-sol high.

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