If $d_1$ and $d_2$ are two equivalent metric on $X$ then is $X$ complete with respect $d_1$ if and only if it is complete with respect $d_2$? I know that completeness is not a topological property so that if $d_1$ and $d_2$ are two equivalent metric of $X$ then I can state that $X$ is complete with respect $d_1$ if and only if it is complete with respect $d_2$ by the homomorphy of $(X,d_1)$ and $(X,d_2)$ via the identity map. However if $\cal T_1$ and $\cal T_2$ are two equal topology on $X$ then a sequence converges with respect one of theese if and only if it converges with respect the each other (right?) so that I can conclude that a Cauchy sequence $(x_n)_{n\in\omega}$ converges with respect $d_1$ if and only if it converges with respect $d_2$ so that finally $X$ is complete with respect $d_1$ if and only if it is complete with respect $d_2$.
So first of all I ask if the statement is true and so if it is true then I ask if the argumentation I gave are correct; whereas if the result is not true then I ask a counterexample: could someone help me, please?
 A: $\mathbb{R}$ is complete with respect to the metric $d(x,y)=|x-y|$ but not with respect to the metric $\rho(x,y)=|\arctan x-\arctan y|$. These metrics are equivalent in the sense that they yield the same topology on $\mathbb{R}$.
The sequence $x_n=n$ is not Cauchy with respect to $d$ but it is with respect to $\rho$. The completion of $\mathbb{R}$ w.r.t. $\rho$ is the extended real line $\mathbb{R}=\{-\infty,\infty\}\cup\mathbb{R}$.
A: It is not true.
If we endow all subspaces of $\mathbb R$ with the Euclidean metric $d_e(x,y) = \lvert x - y \rvert$ (which means that they receive their standard topology), then $(\mathbb R,d_e) $ is complete, but $((0,\infty),d_e)$ is not.
The exponential map $\exp : \mathbb R \to (0,\infty)$ is a homeomorphism. Define a metric $d$ on $(0,\infty)$ by $d(x,y)= \lvert \exp(x)- \exp(y) \rvert$. Since $\exp$ is a homeomorphism, the metric $d$ induces the standard topology on $\mathbb R$ which means that $d$ is equivalent to $d_e$. However, the metric space $(\mathbb R, d)$ is not complete. In fact, the sequence $x_n = -n$ is a Cauchy sequence because with $r = \min(n,m)$ we get
$$d(x_n,x_m) = \lvert \exp(-n)- \exp(-m) \rvert = \lvert e^{-n}- e^{-m} \rvert = e^{-r}\lvert e^{-n+r}- e^{-m+r} \rvert \\ \le e^{-r}(e^{-n+r}+ e^{-m+r}) \le 2 e^{-r} .$$
But obviously $(x_n)$ does not converge to any $\xi \in \mathbb R$.
Update:
The equivalence of $d$ and $d_e$ follows from the following general observation:

*

*Let $(X_1,d_1)$ and $(X_2,d_2)$ be metric spaces with metric topologies $\tau_1, \tau_2$ and let $h : (X_1,\tau_1) \to (X_2,\tau_2)$ be a homeomorphism. Then $d'_1(x,y) = d_2(h(x),h(y))$ defines a metric on $d'_1$ on $X_1$ which is equivalent to $d_1$.

That $d'_2$ is a metric is easily verified (the proof works for any injective, not necessarily continuous, function $h : X_1 \to X_2$). The function $h$ is by definition an isometry $h :(X_1,d'_1) \to (X_2,d_2)$, in particular a homeomorphism $h : (X_1, \tau'_1) \to (X_2,\tau_2)$. This shows that
$$id = h^{-1} \circ h : (X_1, \tau_1) \stackrel{h}{\to} (X_2,\tau_2) \stackrel{h^{-1}}{\to} (X_1, \tau'_1)$$
is a homeomorphism which means that $\tau_1 = \tau'_1$.
Let us next observe

*

*If $(X_1,d'_1)$ is complete iff $(X_2,d_2)$ is complete.

This is obvious because $h$ is an isometry $h :(X_1,d'_1) \to (X_2,d_2)$.
In my example the metric space $((0,\infty),d_e)$ is not complete since each sequence $(\xi_n)$ in $(0,\infty)$ converging to $0$ when regarded as a sequence in $\mathbb R$ is a non-convergent Cauchy-sequence in $((0,\infty),d_e)$ since $0 \notin (0,\infty)$. Thus also $(x_n) = (\exp^{-1}(\xi_n)) = (\ln \xi_n)$ is a non-convergent Cauchy-sequence in $(\mathbb R,d)$. In my example I took $\xi_n = e^{-n}$ which gives $x_n = -n$.
