edit: Another question about formatting. My questions on math.stackexchange keep cutting off the last few lines of my post. How do I fix this? The remaining lines show up in my edit box but not in the saved, published question.
I have two metric spaces, $(X,d)$ and $(X,d')$.
$$d(x,y)=\frac{d'(x,y)}{1-d'(x,y)}$$
$$d'(x,y)=\frac{d(x,y)}{1+d(x,y)}$$
This was an exercise: "If $(X,d')$ is complete, is $(X,d)$ necessarily complete?"
I had no idea how to approach this problem, but my TA gave me these comments when I received the problem set back:
It [$(X,d)$] is [complete]. You want to show that any Cauchy sequence in $(X,d)$ is Cauchy under $d'$. This follows from $d'\le d$. Thus any Cauchy sequence in $(X,d)$ has a limit under $d'$ by completeness of $(X,d')$. Now you need to show that this is also a limit under $d$.
I'm confused about the steps after $d'\le d$. First of all, what do "limit under $d'$" and "limit under $d$" mean? This terminology is not used in my book. As far as my understanding of metric spaces goes, since $d\ge d'$, then a sequence could have $d'(x,y)<\epsilon$ but not necessarily have $d(x,y)<\epsilon$. On the other hand, I think it would be possible to show that if $d'(x,y)<\epsilon_1$, then $d(x,y)<\epsilon_2$.
Thanks, Jeff
If $\rho$ is a metric on a set $X$, then any notion 'under $\rho$' means the notion considered in the metric space $(X,\rho)$.
$d'\le d$ because it is $d$ divided by a number $\ge 1$. Now, if $(a_n)$ is Cauchy under $d$, then it is also Cauchy under $d'$ [because ...: $\ d'(a_n,a_m)\le d(a_n,a_m)<\varepsilon$]. As $X$ is assumed to be complete under $d'$, this $d'$-Cauchy sequence will have a limit point $a$, according to $d'$.
It is not ab ovo guaranteed that the notion of limit would coincide for different metrics on the same set.
However, now we have $a_n \overset{d'}\longrightarrow a$, i.e. $\ d'(a_n,a)\,\to 0$. But then, by the formula of $d(x,y)$ we also get $$d(a_n,a)\,\longrightarrow\,\frac01=0\,.$$ So, $a_n$ converges to the same limit under $d$.