How do prove that $d_{1}$ and $d_{2}$ are equivalent iff the identity map $f:X\to X$ is continuous

Question

Let $X$ be a non-empty set and $d_{1}$ and $d_{2}$ be metrics on $X$. The following statements are equivalent:

(a) $d_{1}$ and $d_{2}$ are equivalent.

(b) $f:X\to X$ given by $f(x) = x$ is continuous if considered from $(X,d_{1})$ to $(X,d_{2})$, and considered from $(X,d_{2})$ to $(X,d_{1})$.

My attempt

I have been able to prove the implication $(a)\Rightarrow(b)$ as follows.

Once $d_{1}\sim d_{2}$, it results that $\alpha d_{2}(x,y) \leq d_{1}(x,y) \leq \beta d_{2}(x,y)$ for each pair $(x,y)\in X^{2}$, where $\alpha > 0$ and $\beta > 0$. Indeed, for every $\varepsilon > 0$, there corresponds a $\delta_{\varepsilon} = \alpha\varepsilon$ such that for every $x\in X$ one has that
\begin{align*}
d_{1}(x,x_{0}) < \delta_{\varepsilon} \Rightarrow \alpha d_{2}(x,x_{0}) \leq d_{1}(x,x_{0}) < \alpha\varepsilon \Rightarrow d_{2}(x,x_{0}) < \varepsilon.
\end{align*}
which proves the function $f:(X,d_{1})\to(X,d_{2})$ is continuous.

Conversely, for each $\varepsilon > 0$, there corresponds a $\delta_{\varepsilon} = \varepsilon/\beta$ such that for every $x\in X$ one has that
\begin{align*}
d_{2}(x,x_{0}) < \delta_{\varepsilon} \Rightarrow \frac{1}{\beta}d_{1}(x,x_{0}) \leq d_{2}(x,x_{0}) < \frac{\varepsilon}{\beta} \Rightarrow d_{1}(x,x_{0}) < \varepsilon.
\end{align*}
which proves the function $f:(X,d_{2})\to (X,d_{1})$ is continuous.

The problem arises when I try to prove the implication $(b)\Rightarrow(a)$. I have tried to apply the definition of continuity, but it did not lead me anywhere.

Could anyone help me telling how to approach it?

Answer 1

It seems that you misunderstood the meaning of $d_1 \sim d_2$. You define it as follows:

There exist $\alpha, \beta > 0$ such that for $x,y \in X$ one has $\alpha d_{2}(x,y) \leq d_{1}(x,y) \leq \beta d_{2}(x,y)$.

This relation is known as strong equivalence of metrics. With this relation $(b) \Rightarrow (a)$ is not true. The correct definition of $d_1 \sim d_2$ is

For each $x \in X$ and each $\epsilon> 0$ there exists $\delta > 0$ such that for all $y \in X$ one has the implications $d_1(x,y) < \delta \Rightarrow d_2(x,y) < \epsilon$ and $d_2(x,y) < \delta \Rightarrow d_1(x,y) < \epsilon$.

This is nothing else than the explicit formulation of the continuity of $id : (X,d_1) \to (X,d_2)$ and $id : (X,d_2) \to (X,d_1)$.

Clearly strong equivalence implies "ordinary" (also called "topological") equivalence. As an example that the converse fails take any metric space $(X,d)$ with an unbounded metric (this means that there exist $x_n, y_n \in X$ such that $d(x_n,y_n) \ge n$). Define $d'(x,y) = \min(d(x,y),1)$. It is easy to check that $d'$ is a metric on $X$ which is toplogically equivalent to $d$ (simply take $\delta = \min(\epsilon, 1)$).

$d$ and $d'$ are not strongly equivalent: We have $d(x_n,y_n) \ge n$ and $d'(x_n,y_n) = 1$, thus $d(x_n,y_n) \ge n d'(x_n,y_n)$ which shows that no $\beta > 0$ can satisfy $d(x,y) < \beta d'(x,y)$ for all $x,y \in X$.