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??

  • $\begingroup$ Is $(\beta_n)$ supposed to be independent of $(x_n)$? If not, why don't you absorb the factor $\alpha$ into $(\beta_n)$? $\endgroup$ – joriki Dec 5 '12 at 12:58
  • $\begingroup$ No, $\beta_n$ can depend on $x_n$. In fact, I imagine it would... $\endgroup$ – chango Dec 5 '12 at 13:05
  • $\begingroup$ Then you could do without the factor $\alpha$, right? $\endgroup$ – joriki Dec 5 '12 at 13:06
  • $\begingroup$ Yeah, totally. But I think its there to help thinking about the problem, that's all. $\endgroup$ – chango Dec 5 '12 at 13:08
  • $\begingroup$ I could be good to consider the example when $X$ is the real line with usual metric, and $x_n=n$. Taking $\beta_n:=n^{-1}$, we get what we want. $\endgroup$ – Davide Giraudo Dec 11 '12 at 22:28

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.

  • $\begingroup$ I think the second part of your argument might be wrong. If $x=x_m$ for some $m$ it doesn't follow that $\delta>0$. Take $\beta_n = 1/n$. Your sequence is convergent and therefore Cauchy so $\delta= 0$ always. $\endgroup$ – chango Dec 21 '12 at 19:20
  • $\begingroup$ If $x=x_m$, and $\beta_n = 1/n$ then we still have $\delta = \inf_{n\ne m} |\beta_m - \beta_n| \alpha \ge (1/m - 1/(m+1))\alpha > 0$. Or am I misunderstanding you? $\endgroup$ – Lukas Geyer Dec 22 '12 at 7:52
  • $\begingroup$ Of course $\delta = 0$, just take $m$ and $n$ really big! $\endgroup$ – chango Dec 22 '12 at 17:25
  • $\begingroup$ You can't vary $m$, because $\delta$ depends on $x=x_m$, so it does depend on $m$. I should probably fix the notation. $\endgroup$ – Lukas Geyer Dec 22 '12 at 19:40
  • $\begingroup$ I am not sure about what you mean. Is the infimum you are taking to calculate $\lambda$ only over $n$? $\endgroup$ – chango Dec 24 '12 at 14:53

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