Proving that two metrics are topologically equivalent

[mahler1]

Homework Statement .

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 $U$?

 

[pasmith]

Homework Statement .

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 $U$?

Two metrics d and d' are topologically equivalent if and only if the identity function is both (d,d')-continuous and (d',d)-continuous.

The identity is (d,d')-continuous at x \in X if and only if for every \epsilon > 0 there exists \delta > 0 such that for all y \in X, if d(x,y) < \delta then d'(x,y) < \epsilon.

The identity is (d',d)-continuous at x \in X if and only if for every \epsilon > 0 there exists \delta > 0 such that for all y \in X, if d'(x,y) < \delta then d(x,y) < \epsilon.

At some point you should expect to use the facts that f is continuous and open (both with respect to d), neither of which you have yet used.

 

[mahler1]

Two metrics d and d' are topologically equivalent if and only if the identity function is both (d,d')-continuous and (d',d)-continuous.

I think I could do it with the property you've suggested:

First, I'll show that $Id:(X,d)→(X,d')$ is continuous.
Let $x \in X$ and let $ε>0$. We know that $f$ is continuous, so, for that given $ε$, there exists $δ_{x,ε}$ such that $d(x,y)< δ_{x,ε}$→$d(f(x),f(y))<ε$. But then, $d'(x,y)=d(f(x),f(y))<ε$. This shows the identity from $(X,d)$ to $(X,d')$ is continuous.

Now it remains to prove that $Id:(X,d')→(X,d)$ is continuous. So let $x \in X$ and let $ε>0$. We know f is open, so $f(B(x,ε))$ is an open set in $(X,d)$. This means that there exists $δ'$ such that $B_{d}(f(x),δ') \subset f(B_{d}(x,ε))$. If $d'(x,y)=d(f(x),f(y))<δ'$$→ f(y) \in f(B_{d}(x,ε)) → y \in B_{d}(x,ε) → d(x,y)<ε$