Metric spaces question (finding an equivalent incomplete metric) Suppose $(X, \tau)$ is a metrisable topology and let $d$ be any metric that metrises it. Suppose further that $(x_n)$ in $X$ is a sequence such that $\alpha =\text{inf}_{m \neq n} \: d(x_n,x_m) >0$.
I want to show that for a suitable sequence $(\beta_n) $ in $\mathbb{R}_{>0}$,
$$\bar{d}(x,y) = \text{min } \{ d(x,y), \text{inf }_{m,n} d(x,x_m) + |\beta_m - \beta_n| \alpha + d(x_n,y)  \}$$
is an incomplete equivalent metric.
Any ideas/hints??
 A: I think any sequence of distinct positive numbers $(\beta_n)$ satisfying $\beta_n \to 0$ will do. The assumption shows that $(x_n)$ is not a Cauchy sequence in the metric $d$, so it does not converge in $(X,\tau)$. The definition of $\bar{d}$ implies that $\bar{d}(x_m,x_n) \le |\beta_m-\beta_n| \alpha$ for all $m,n$, so if $\beta_n \to 0$, we know that $(x_n)$ is a Cauchy sequence with respect to $\bar{d}$. This shows that $\bar{d}$ has a non-converging Cauchy sequence, so it is an incomplete metric.
In order to show that $d$ and $\bar{d}$ are equivalent, observe first that the definition implies $\bar{d} \le d$. Now if $x \in X$ is not in the sequence $(x_n)$, then $\delta := \inf_n d(x,x_n)>0$ (since the assumption implies that $(x_n)$ has no accumulation points.) So for all $y$ with $d(x,y)<\delta$ or $\bar{d}(x,y)<\delta$ we actually have that $d(x,y) = \bar{d}(x,y)$. If $x=x_m$ for some $m$, then by the assumption that $\beta_m \ne \beta_n$ for $m\ne n$, and that $\beta_n \to 0$, we get that $\delta := \inf_{n \ne m} |\beta_m - \beta_n|\alpha > 0$, and so again for any $y$ with $d(x,y)<\delta$ or $\bar{d}(x,y)<\delta$ we get $d(x,y) = \bar{d}(x,y)$. These arguments show that neighborhood with respect to $d$ and $\bar{d}$ are the same, so the topologies induced by $d$ and $\bar{d}$ coincide.
