Convergent but not Cauchy under two equivalent metrics

Question

I want to find an example of a metric space $X$ with two equivalent (in the sense that all sequences in $(X,d)$ converge to the same limit if and only if all sequences in $(X,e)$ converge to the same limit) metrics such that there exists a sequence which is Cauchy with respect to one of the metrics but not with respect to the other.

I had the idea that if a sequence is convergent then it must be Cauchy. However, it turns out not to be the case (?), if such an example as described above exists. I've been thinking about the "French rail road metric", but no, it didn't seem to do the trick.

Would appreciate a hint.

Update: It was pointed out in the comments below that the statement above wouldn't be true if under equivalence it was meant that two metrics induce the same topology. As far as I know, however, two metrics have sequences convergent to the same limit under two metrics if and only if such metrics are also equivalent in the sense that there exist $m,M>0$ such that $m e(x,y)\le d(x,y)\le M e(x,y), \forall x,y\in X$. Which means that equivalent metrics in the sequence sense are also equivalent in the $m,M$-inequality sense, which means that if topology is preserved under both metrics then such metrics must be equivalent in both senses.

I'm somewhat confused, would appreciate a clarification.

Answer 1

A topology $T$ on a set $S$ has a closure operator $Cl_T(A)$ for $A\subset S$, which can be regarded as a function from the set of all subsets of $S$ to the set of all $T$-closed sets. If $T_1,T_2$ are topologies on $S$ and $Cl_{T_1}=Cl_{T_2}$ then the set of $T_1$-closed sets equals the set of $T_2$-closed sets, so $T_1=T_2.$

If $T_d$ is the topology generated by a metric $d$ then $Cl_{T_d}(A)$ is the set of points that are limits (with respect to $d$) of sequences of member(s) of $A.$ So metrics $d,e$ on $S$ generate the same topology iff $Cl_{T_d}=Cl_{T_e}$ iff the set of $d$-convergent sequences equals the set of $e$-convergent sequences.

When $d,e$ are equivalent metrics (i.e. $T_d=T_e$) there may be a sequence which is $d$-Cauchy but not $e$-Cauchy. Such a sequence could not converge with respect to either $d$ or $e$.

For example let $S=\Bbb R$ with the usual topology. Let $f:\Bbb R\to (-1,1)$ be a continuous and strictly monotonic surjection. Let $e(x,y)=|x-y|$ and $d(x,y)=|f(x)-f(y)|.$ Then the sequence $(n)_{n\in\Bbb N}$ is $d$-Cauchy but not $e$-Cauchy. (Caution: Although the sequence $(f(n))_{n\in \Bbb N}$ is converging in $\Bbb R$ to $1$, there is no $x\in S$ for which $f(x)=1$ and there is no $x\in S$ for which $\lim_{n\to \infty} d(n,x)=0$.)

When $md\leq e\leq Md$ for some positive $m,M$ the metrics $d,e$ are called uniformly equivalent. Uniformly equivalent metrics are equivalent. In the example above, $d,e$ are equivalent but not uniformly equivalent.

A common textbook example for the function $f$ in the example above is $f(x)=\frac {2}{\pi} \arctan (x).$

Answer 2

Consider the set of natural numbers $\mathbb N$. Let $d$ be the discrete metric, so that $d(m,n)=1$ if $n \neq m$ and $d(n,n)=0$. Define an alternative metric

$$\rho(m,n)=\left\vert \frac{1}{m} - \frac{1}{n} \right\vert.$$

Both $d$ and $\rho$ induce the discrete topology on $\mathbb N$, and hence are equivalent in the sense that they have the same convergent sequences. More explicitly, say that the metrics $d$ and $\rho$ are equivalent if $x_n \overset{d}{\to}x \iff x_n \overset{\rho}{\to} x$. This is the same as the metrics inducing the same topology.

However, the sequence $\left\{1,2,3,... \right\}$ is $\rho$-Cauchy but not $d$-Cauchy.

To see that $\rho$ induces the discrete topology, let $r_n = \frac{1}{n(n+1)}$, and observe that $B_{r_n}(n) = \{ n \}$.