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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1$\begingroup$ 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. $\endgroup$– DirkApr 5, 2019 at 7:12
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$\begingroup$ @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. $\endgroup$– uniquesolutionApr 5, 2019 at 7:14
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1$\begingroup$ @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. $\endgroup$– DirkApr 5, 2019 at 7:18
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$\begingroup$ @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. $\endgroup$– uniquesolutionApr 5, 2019 at 7:19
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3$\begingroup$ 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 SinclairApr 5, 2019 at 16:44
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4 Answers
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Let $X=[0,\infty)$, 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 converges only in $(X,d)$.
answered Apr 5, 2019 at 7:11
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$\begingroup$ Is there a reason this example needs to use $X = [0, \infty)$ rather than $X = \mathbb{R}$? $\endgroup$ Apr 5, 2019 at 12:13
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2$\begingroup$ If, in my answer, you work with $\mathbb R$ instead of $[0,\infty)$, then $e$ will not be a distance, because then $e(0,-1)=0$. $\endgroup$ Apr 5, 2019 at 12:47
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1$\begingroup$ The second metric basically makes it $\{-1\}\cup (0,\infty)$, which can't really be said to be a metric for the same space. $\endgroup$ Apr 5, 2019 at 23:57
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3$\begingroup$ @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. $\endgroup$ Apr 6, 2019 at 3:47
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$\begingroup$ For $X = \mathbb{R}$, one could use $e(0,x)=e(x,0)=|x|+1$ when $x\neq 0$. $\endgroup$– LithoApr 6, 2019 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 $(X, d_2)$.
answered Apr 5, 2019 at 7:32
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$\begingroup$ Is this somehow a standard example? Since you other poster also jsed it? Which book is it from $\endgroup$– lalalaApr 5, 2019 at 18:50
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1$\begingroup$ 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 $\endgroup$ Apr 5, 2019 at 21:25
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1$\begingroup$ 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. $\endgroup$– lalalaApr 6, 2019 at 8:26
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1$\begingroup$ $d_2$ 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 $-1$. $\endgroup$ Apr 6, 2019 at 8:40
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$\begingroup$ Thanks! Yes. It is just a renameing. (I am not so used to metrics which do not derive from a norm) $\endgroup$– lalalaApr 6, 2019 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.
answered Apr 5, 2019 at 13:26
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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.
answered Apr 5, 2019 at 16:50