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Let ##(X,d)## a metric space and ##f: X→X## an injective, continuous and open function. We define ##d' : XxX→ \mathbb R_{≥0}## as ##d'(x,y)=d(f(x),f(y))##. Prove that ##d'## defines a distance and that ##d'## is topologically equivalent to ##d##.

The attempt at a solution.

Here's the first part: using the fact that d is a distance, we have that

1)##d'(x,y)=d(f(x),f(y))≥0##

2)##d'(x,y)=0## iff ##d(f(x),f(y))=0## iff ##f(x)=f(y)##, ##f## is injective, so ##x=y##.

3)##d'(x,y)=d(f(x),f(y))=d(f(y),f(x))=d'(y,x)##

4)##d'(x,y)=d(f(x),f(y))≤d(f(x),f(z)+d(f(z),f(y))=d'(x,z)+d(z,y)##

As you can see, this is the easy part. I got stuck trying to prove that ##d'## and ##d## are topologically equivalent. To prove this, I should take an open set in ##(X,d)## and prove it is also open in ##(X,d')##. Then, I should take an open set in ##(X,d')## and prove it's open with the metric ##d##. So let ##U## be an open set with ##d'##. Let ##x \in U##, then, there exists ##r_x>0## such that ##B_d(x,r_x) \subset U##. How can I get the appropiate radius ##r'_x## to find a ball ##B_d'(x,r'_x)## is included in ##U##?