# Convergent but not Cauchy under two equivalent metrics

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.

• Will you please define "equivalent metrics"? Oct 16, 2017 at 1:15
• If, by equivalent, you mean that they induce the same topology, then this would certainly not be true. Oct 16, 2017 at 1:17
• By equivalent I mean that sequences w.r.t. both metrics converge to the same limit in $X$. But I think it is true that ~ implies $\approx$. I will edit my post immediately. Oct 16, 2017 at 1:19
• How is it not the case that, in metric spaces, every convergent sequence is Cauchy? Oct 16, 2017 at 1:26

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 $B_{r_n}(n) = \{ n \}$.

• But the sequence $(n)_n$ will not converge under the discrete metric. I think what we need, as per the definition of "~" is that the same sequence converges to the same limit under both metrics. Also, how do we know that $d$ induces the same topology as $\rho$? Oct 16, 2017 at 2:09
• For your second question, see my edit. For your first question, note that the definition of equivalence that I give (and I believe you are also using) is silent about sequences that do not converge. When sequences converge, they must converge to the same limit under either metric. What you really need is a sequence that is Cauchy but not convergent, as shown above. Oct 16, 2017 at 2:29
• Sorry, how is it true that $B_{r_n}(n)=\{n\}$ (open ball centred about $n$ of radius $\frac{1}{n(n+1)}$ under $\rho$? Consider, for example, $\rho(r_n, n)=\left|n(n-1)-\frac1n\right|=\left|n^2-n-\frac1n \right|$. Also, if $B_{r_n}(n)=\{n\}$ for this particular sequence, then why does this imply that $\rho$ induces the discrete topology? Moreover, it appears that $(n)_n$ actually converges to $0$ under $\rho$. Let $\epsilon > 0$, then $\exists N\in \mathbb N$ such that $n\ge N$ implies $\rho(n,0)<\epsilon$. Which appears to be true: $\rho(n, 0) = |\frac1n - 0|< \epsilon$ for $n$ large enough. Oct 16, 2017 at 4:30
• 1. $B_{r_n}(n) = \left\{ m : \left\vert \frac 1n - \frac 1m \right\vert < r_n \right\}$, so the value of $\rho (r_n , n)$ is not relevant. 2. If every singleton is an open set, then the topology must be discrete. 3. I exclude $0$ from $\mathbb N$ in the above construction. Note that $\rho (n,0)$ is not defined since it entails division by $0$. Oct 16, 2017 at 21:46

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 $f(x)=\frac {2}{\pi} \arctan (x).$

• Can you please clarify - does $f(n)_n$ actually converge to $1\in\mathbb{R}$? If so then the sequence under $d$ is convergent in $\mathbb{R}$. But if that is the topological space we're considering, then how can it be that $f(n)_n$ does not converge w.r.t. to $d$ if all Cauchy sequences in $\mathbb{R}$ do converge (since $\mathbb{R}$ is complete)? Also, it is strange, but I've seen proofs, as well as exercises asking to prove, that if two metrics have sequences converging to the same limits then there must exist $m,M$ such that $md\le e \le Md$. Oct 16, 2017 at 5:59
• $(f(n))_n$ does indeed converge to $1$ with respect to the metric $e.$ But the sequence that I gave that was neither $d$-nor $e$-convergent that I gave was $not$ $(f(n))_n$ but $(n)_n.$...... It is NOT true that any two metrics with the same convergent sequences are always uniformly equivalent. Oct 16, 2017 at 6:13
• I see. But how do we know that $d$ and $e$ induce the same topology? Oct 16, 2017 at 6:20
• Regarding non-uniform equivalence: Consider $S=\Bbb N$ with $d(x,y)=|x-y|.$ Let $e(x,y)=1$ if $x\ne y$. An open ball of radius $1/2$ centered at any $x\in$, in either metric, is equal to $\{x\}$, so every subset of $X$ is $d$-open and $e$-open. A sequence $(x_n)_n$ is $d$-Cauchy iff it is $e$-Cauchy iff it is $d$-convergent iff it is $e$-convergent iff it is eventually constant (which means that for some $x\in S$ the set $\{n:x_n\ne x\}$ is finite.) But $d, e$ are not uniformly equivalent because $e$ is bounded and $d$ is not. Oct 16, 2017 at 6:27
• Re your most recent comment: $f$ and its inverse $f^{-1}$ are continuous bijections. Since $f$ is continuous, $\lim_{n\to \infty}d(x_n,x)=0\implies$ $\lim_{n\to\infty}|x_n-x|=0 \implies$ $\lim_{n\to \infty}|f(x_n)-f(x)|=0\implies$ $\lim_{n\to \infty}e(x_n,x)=0.$.... And since $f^{-1}$ is continuous, $\lim_{n\to \infty}e(x'_n,x')=0\implies$ $\lim_{n\to \infty}|f(x'_n)-f(x')|=0\implies$ $\lim_{n\to \infty}|f^{-1}(f(x'_n))-f^{-1}(f(x))|=0\implies$ $\lim_{n\to \infty}|x'_n-x'|=0\implies$ $\lim_{n\to \infty}d(x'_n,x')=0.$ Oct 16, 2017 at 6:44