Question:
If d,e are equivalent metrics on X, then $\left ( X,d \right )$ and $\left ( X,e \right )$ have the same Cauchy sequences.
I've first assumed to the contrary that the metric spaces have different Cauchy sequences.
Any hint is appreciated.
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asked Nov 2, 2016 at 13:32
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$\begingroup$ $d(u_n,u_p)\leq \alpha e(u_n,u_p)$. $\endgroup$– hamam_AbdallahCommented Nov 2, 2016 at 13:33
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1$\begingroup$ It is probably easier to show this without contradiciton. Let $\langle x_n\rangle$ be a Cauchy sequence in $(X,d)$. Now what does this sequence need to satisfy to be a Cauchy sequence in $(X,e)$? $\endgroup$– Mees de VriesCommented Nov 2, 2016 at 13:34
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$\begingroup$ If $\forall \epsilon >0$, there exists $N\left ( \epsilon \right ) \in \mathbb{N} $such that $d\left ( x_{m},x_{n} \right ) < \epsilon \forall m,n \geq N\left ( \epsilon \right )$@MeesdeVries $\endgroup$– MathematicingCommented Nov 2, 2016 at 13:36
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$\begingroup$ Alright. What would make sense as a choice for $\epsilon$? Of course, you have to use the equivalence of the norms. $\endgroup$– Mees de VriesCommented Nov 2, 2016 at 13:37
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$\begingroup$ I am unsure how to make useful of the equivalence norm to determine a candidate for $\epsilon$...@MeesdeVries $\endgroup$– MathematicingCommented Nov 2, 2016 at 13:45
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1 Answer
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This is false.
On $\mathbb R$ let $d(x,y)=|x-y|$ and $e(x,y)=|\tan^{-1}x - \tan^{-1}y|.$ Then $d$ and $e$ are equivalent, that is, they generate the same topology. But $(n)_{n\in \mathbb N}$ is an $e$-Cauchy sequence and not a $d$-Cauchy sequence.
What is true is that two metrics $d,e$ are equivalent iff they have the same convergent sequences. A convergent sequence is a Cauchy sequence that has a limit point in the space. Note that $(n)_{n\in \mathbb N}$ in the example above, is an $e$-Cauchy sequence but it has no limit point in $\mathbb R.$
Metrics $d,e$ are called uniformly equivalent if and only if there exist positive $k_1,k_2$ with $k_1d(x,y)\leq e(x,y)\leq k_2 d(x,y)$ for all $x,y.$ Uniformly equivalent metrics are equivalent metrics, and do have the same Cauchy sequences.
answered Nov 3, 2016 at 5:43
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$\begingroup$ Does the converse to the original question hold? I.e. if two metric spaces (X,d) and (X,e) have the same Cauchy sequences, does it follow that $d \sim e$? $\endgroup$– SdVCommented Oct 6, 2017 at 20:06
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1$\begingroup$ @SWdeVries.YES. Let $(X,d)$ and $(X,e)$ have the same Cauchy sequences. Let $(x_n)_n$ be a $d$-Cauchy sequence, with $p\in X$ such that with $ \lim_{n\to \infty}d(x_n,p)=0$. Let $y_{2n}=x_n$ and $y_{2n-1}=p.$ Then $(y_n)_n$ is a $d$-Cauchy sequence so $(y_n)_n$ is an $e$-Cauchy sequence so $ \lim_{n\to \infty}e(y_{2n},p)=0.$ Now repeat with $d$ and $e$ interchanged . So $(X,d)$ and $(X,e)$ have the same convergent sequences so $d\sim e.$ $\endgroup$ Commented Oct 7, 2017 at 1:16
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$\begingroup$ @DanielWainfleet Can you please clarify why you had to construct $(y_n)_n$ in order to prove that $(x_n)_n$ is an $e$-Cauchy sequence? How does it follow that if $(y_n)_n$ is an $e$-Cauchy sequence then $(x_n)_n$ converges to $p$? $\endgroup$– sequenceCommented Oct 15, 2017 at 21:07
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1$\begingroup$ @sequence. As to why I had to: Maybe there is another method.... An $e$-Cauchy sequence in which infinitely many terms are equal to $p$ $must$ converge to $p$. This is true in $\Bbb R$. For a metric space it's the same except we write $e(a,b)$ instead of $|a-b|.$ $\endgroup$ Commented Oct 16, 2017 at 2:09
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1$\begingroup$ @Thrash. Uniformly equivalent metrics have the same Cauchy sequences. But two equivalent metrics may have the same Cauchy sequences without being uniformly equivalent. If $e(x,y)=\min (1, d(x,y))$ then $d,e$ are equivalent and have the same Cauchy sequences, but if $d$ is unbounded then $d,e$ are not uniformly equivalent. E.g. if $d(x,y)=|x-y|$ on $\Bbb R$.... $\endgroup$ Commented Sep 20, 2018 at 1:51