Is there always an equivalent metric which is not complete?

I have seen that completeness is not a topological property like compactness or connectedness. I have seen some examples also showing that there are two equivalent metrics one of which is complete and the other one is incomplete. I want to know some general result. Consider any metric space $(X,d)$. Let $d$ be complete. Does there exist an equivalent metric $d'$ in $X$ which is incomplete? When $d$ is compact, such $d'$ does not exist (Since compactness implies completeness). So answer me when $d$ is noncompact.

• What is your definition of equivalence of metrics? – Lee Mosher Sep 24 '14 at 14:57
• If the two metrics generate the same uniform structure, i.e. the identity is uniformly continuous in both directions, then one metric is complete iff the other metric is complete. So I think by equivalent santanu means that they generate the same topology. – Stefan Hamcke Sep 24 '14 at 15:14

Let $X$ be a non-compact metrizable space with compatible metric $d$ bounded by $1$. Note that a metrizable space is compact iff it is countably compact, and so there is a decreasing sequence $\langle F_n \rangle_{n \in \mathbb{N}}$ of nonempty closed sets with empty intersection. Consider the mapping $\rho : X \times X \to \mathbb{R}$ defined by $$\rho ( x , y ) = \sum_n 2^{-n} \color{blue}{\left( | d ( x, F_n ) - d(y,F_n) | + \left[ \min \{ d(x,F_n) , d(y,F_N) \} \right] d(x,y) \right)}. \tag{1}$$ It can be shown that $\rho$ is a compatible metric for $X$ (this is largely because the part of (1) in blue defines a pseudometric on $X$ which is continuous as a function $X \times X \to \mathbb{R}$ for each $n$), and that $\operatorname{diam}_\rho (F_n) \leq 2^{-n}$ for each $n$. So if $\langle x_n \rangle_n$ is a sequence such that $x_n \in F_n$ for each $n$, then it is Cauchy with respect to $\rho$, but does not converge (since a limit would have to belong to $\bigcap_n F_n = \varnothing$).