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Is there an easy example of one and the same space $X$ with two different metrics $d$ and $e$ such that one and the same sequence $\{x_n\}$ is a Cauchy sequence for both metrics, but converges only for one of them?

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    $\begingroup$ You should define some additional properties you want your metric to have. Otherwise something like "every two different points have distance one" will also work as an example. $\endgroup$ – Dirk Apr 5 at 7:12
  • $\begingroup$ @Dirk In a discrete space a Cauchy sequence is eventually constant, hence if a Cauchy sequence converges in one discrete metric, it converges in all of them to the same limit. Thus, it will not work as an example. $\endgroup$ – uniquesolution Apr 5 at 7:14
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    $\begingroup$ @uniquesolution Yeah, but on the other hand, every converging sequence (in whatever metric you like) that doesn't get constant will not converge in a discrete metric. $\endgroup$ – Dirk Apr 5 at 7:18
  • $\begingroup$ @Dirk, Every convergent sequence is Cauchy in any metric space. Hence in a discrete space every convergent sequence will also get eventually constant. Therefore there is no converging sequence that does not get eventually constant in a discrete space. $\endgroup$ – uniquesolution Apr 5 at 7:19
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    $\begingroup$ Goblin has an excellent point. You say the "same space $X$". If by "space" you are referring to the topology, not just the set, then the answer is no. $\endgroup$ – Paul Sinclair Apr 5 at 16:44
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Take $X=[0,\infty)$, let $d(x,y)=\lvert x-y\rvert$ and let$$e(x,y)=\begin{cases}\lvert x-y\rvert&\text{ if }x,y\neq0\\\lvert x+1\rvert&\text{ if }x\neq0\text{ and }y=0\\\lvert y+1\rvert&\text{ if }x=0\text{ and }y\neq0\\0&\text{ if }x,y=0.\end{cases}$$Then the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ is a Cauchy sequence with respect to both metrics, but it is convergent only in $(X,d)$.

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  • $\begingroup$ Is there a reason this example needs to use $X = [0, \infty)$ rather than $X = \mathbb{R}$? $\endgroup$ – Michael Seifert Apr 5 at 12:13
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    $\begingroup$ If, in my answer, you work with $\mathbb R$ instead of $[0,\infty)$, then $e$ will not be a distance, because then $e(0,-1)=0$. $\endgroup$ – José Carlos Santos Apr 5 at 12:47
  • $\begingroup$ The second metric basically makes it $\{-1\}\cup (0,\infty)$, which can't really be said to be a metric for the same space. $\endgroup$ – Matt Samuel Apr 5 at 23:57
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    $\begingroup$ @MattSamuel That seems like an easier way to define this function. Let $f(0)=-1$ and $f(x)=x$ for $x>0$, and define $e(x,y)=|f(x)-f(y)|$. This is a metric because $f$ is injective. $\endgroup$ – Mario Carneiro Apr 6 at 3:47
  • $\begingroup$ For $X = \mathbb{R}$, one could use $e(0,x)=e(x,0)=|x|+1$ when $x\neq 0$. $\endgroup$ – Litho Apr 6 at 9:13
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You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=\mathbb{R_{\ge 0}}$, $d_1$ be the Euclidean metric and $d_2(x, y)=|\hat{x}-\hat{y}|$, where $ \hat{x}=-1$ if $x=0$ and $ \hat{x}=x$ otherwise. Then the sequence $x_n=\frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.

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  • $\begingroup$ Is this somehow a standard example? Since you other poster also jsed it? Which book is it from $\endgroup$ – lalala Apr 5 at 18:50
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    $\begingroup$ AFAIK, this isn't from some standard reference (at least I didn't refer to one). I guess that's just a coincidence. After all, $x_n=\frac 1n$ is often the first (non-trivial) convergent sequence when you are asked to think of one and moving $0$ to $-1$ just seems like a natural thing to do if you have to move $0$ away $\endgroup$ – Poon Levi Apr 5 at 21:25
  • $\begingroup$ But how did you see immediately that this metric is a metric? (Maybe its obvious if one has a geometric interpretation like you seem to have). Triangular inequality doesnt seem obvious to me. $\endgroup$ – lalala Apr 6 at 8:26
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    $\begingroup$ $d_2$ is essentially the Euclidean metric on a new set $\{-1\}\cup(0,\infty)$. But since we can't change the base set, we instead use a map $x\mapsto\hat{x}$ to rename $0$ to $-1$. $\endgroup$ – Poon Levi Apr 6 at 8:40
  • $\begingroup$ Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm) $\endgroup$ – lalala Apr 6 at 8:45
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Depending on what you mean, the answer is "no". In particular, if by "two metrics on the one space" you mean two metrics on the same set that induce the same topology, then the answer is no because convergence is a topological property.

However the other answers are correct if the induced topologies are allowed to be different.

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Having "the same X" is somewhat meaningless, since there's no shared structure other than cardinality. Given any other set Y with metric $d_Y$ and a injective map $\phi: X \rightarrow Y$, you can define $e(x_1,x_2) = d_Y(\phi(x_1),\phi(x_2))$. So all you have to do is find two spaces with the same cardinality where one has a convergent Cauchy sequence and the other has a non-convergent one. Or find a convergent Cauchy sequence and a non-convergent Cauchy sequence in the same space, then map one sequence to the other.

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