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I have a question regarding completion of metric spaces on open intervals (I posted similar questions on the community before, but did not receive a good answer, sorry):

Problem: Show that $D_1(x,y)=\left|\tan(\frac{\pi x}{2})-\tan(\frac{\pi y}{2})\right|$ and $D_2(x,y)=|x-y|+\left|\frac{1}{1+x}-\frac{1}{1+y}\right|+\left|\frac{1}{1-x}-\frac{1}{1-y}\right|$ are complete but equivalent metrics on $G=(-1,1)$.

The definition of equivalent metrics is the following in my context: Two metrics are equivalent if convergence of any sequence in one metric implies the convergence of the other, and vise versa. By $\varepsilon-\delta$ approach (can be proven by argue by contradiction), $D_1$ and $D_2$ are equivalent on $G$ if and only if given $x\in G$ and $\varepsilon>0$, there exists $\delta>0$ s.t. for all $y\in G$, $$D_1(x,y)<\delta\text{ implies }D_2(x,y)<\varepsilon\text{ and }D_2(x,y)<\delta\text{ implies }D_1(x,y)<\varepsilon$$

And I also know the fact if two metrics are equivalent, and if one of them is complete, the other one do not have to be complete.

My first attempt is to first prove that the two metrics are complete on $G=(-1,1)$, then prove that they are equivalent. But I'm not sure how to choose Cauchy sequences of $D_1$ and $D_2$ to make them complete. To prove them to be equivalence metrics, I'm not sure how to choose $\delta$ and $\varepsilon$.

My second attempt is to start from an incomplete metric on $(-1,1)$, and extend it to a complete metric by the following method on the same open set: if $G$ is an open set, and $d$ may not be a complete metric, we extend from $(G,d)$ to $(G,D)$, where the latter is a complete metric:

  1. $$D(x,y)=d(x,y)+\left|\frac{1}{d(x,G^C)}-\frac{1}{d(y,G^C)}\right|$$

I use the fact that $d(x,y)=|x-y|$ is not complete on $(-1,1)$, since if I take the Cauchy sequence $x_n=1-\frac{1}{n}$, but $x_n\to 1\notin(-1,1)$ (is this correct?). Then I'd like to apply the method I just mentioned to show that the two metrics are complete whilst equivalent. But I'm stuck here.

Here are my questions:

(a) For my first attempt, can anyone help me finish my argument, or at least give me sketches on how to choose Cauchy sequences and $\varepsilon$ and $\delta$ to prove equivalence?

(b) For my second attempt, I'm trying to seek the relation between the standard metric on $(-1,1)$ and $D_1$ and $D_2$, then apply the completion method I just mentioned. But I'm not sure whether it is feasible. Can anyone follow my completion method?

I'll add more details if I have more progress. If there are mistakes, please let me know. Thank you!

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  • $\begingroup$ Please avoid changing your question in a way that invalidates answers. Since I had already addressed one part of your question, you should not remove it. $\endgroup$
    – user21820
    Oct 4, 2020 at 7:22
  • $\begingroup$ @user21820: Sorry, I'll pay attention to that next time. With your explanation, I'm more comfortable understanding some gaps in previous author's post, and can come up with a proof regarding this. $\endgroup$
    – Mike
    Oct 4, 2020 at 7:27
  • $\begingroup$ I understand; it's just a kind of community thing so nothing to worry about. I'm very glad to hear that you have benefited from my answer. Feel free to inquiry more here or in chat! =) $\endgroup$
    – user21820
    Oct 4, 2020 at 7:33

2 Answers 2

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First I must point out the logical flaw in your reasoning:

But I'm not sure how to choose Cauchy sequences of D1 and D2 to make them complete.

As we discussed in chat, a metric space $(X,d)$ is complete iff ( for every sequence $f$ from $X$ such that $d(f(m),f(n)) → 0$ as $m,n→∞$, there is some $c∈X$ such that $d(f(n),c) → 0$ as $n→∞$ ). So to prove that $(X,d)$ is complete, you do not get to choose Cauchy sequences. A correct proof must instead always look like that:

Given any Cauchy sequence $f$ in $(X,d)$:
$f : ℕ→X$ such that $d(f(m),f(n)) → 0$ as $m,n→∞$.
  ...
  Let $c∈X$ such that ...
  ...
$d(f(n),c) → 0$ as $n→∞$.

The problem is, how to find such a $c$? The key to understanding the two given metric spaces is that, in both of them, moving the points apart in the standard metric (i.e. increasing $|x−y|$) increases their distance in the given metric (i.e. $d(x,y)$). This intuitively suggests that if $d(f(m),f(n)) → 0$, then $f(m)−f(n)$ → 0 as well. But this is not actually true in general, since $f(m),f(n)$ may not be moving strictly towards each other. One way to handle that is to grab a fixed reference point based on the standard metric: Let $c$ be the limit of some subsequence of $f$ that converges in the standard metric (by Bolzano-Weierstrass). Then $c ∈ [-1,1]$. But I claim that $c≠1$. Can you prove it? (Hint: Take any strictly increasing sequence $t : ℕ→ℕ$ such that $f(t(k)) → 1$ as $k→∞$. Then as $k→∞$ and $i,j≥k$, we have $d(f(t(i)),f(t(j))) → 0$, which is impossible since $d(f(t(i)),f(t(k))) → ∞$ as $n→∞$.) Similarly $c≠-1$. Also, I claim that $d(f(n),f(c)) → 0$ as $n→∞$. Try to prove it! (Hint: Let $t$ be some strictly increasing sequence $t : ℕ→ℕ$ such that $t(k) → c$ as $k→∞$. As $n→∞$, there is some $k>n$ such that $t(k)>n$, so $d(f(n),f(t(k))) → 0$ and $f(t(k)) → f(c)$.)

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  • $\begingroup$ By the way, this approach of utilizing sequences instead of ε-δ definitions can be useful in matching intuitive reasoning in real analysis, as it is not 'backwards' (i.e. you don't have to go backwards from some error margin ε to some window δ). Of course, it is also good to translate the sequence-based proof to an ε-δ one just to make sure you fully understand how exactly the sequence usage succinctly captures a lot of the ε-δ reasoning. $\endgroup$
    – user21820
    Oct 4, 2020 at 7:40
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Let $d$ be the usual metric on $G$. By the convergence definition of metric equivalence, it suffices to show that for each $i=1,2$ any sequence $(x_n)$ of points of $G$ converges in $(G,D_i)$ to a point $x$ of $G$ iff $(x_n)$ converges to $x$ in $(G,d)$.

$(\Leftarrow)$ It follows from an observation that $D_i(y,z)$ has a form $\sum |f_j(y)-f_j(z)|$ for some finite family of continuous functions on $(G,d)$ for each $y,z\in D$.

$(\Rightarrow)$ it follows from an inequality $D_i(y,z)\ge C_i d(y,z)$ for and some constant $C_i>0$ and each $y,z\in G$. Clearly, we can pick $C_2=1$. We can pick $C_1=\tfrac{\pi}{2}$, because by Lagrange’s theorem for each $y,z\in G$ there exists $u$ between $y$ and $z$ such that $$\left|\tan \frac{\pi y}{2}-\tan \frac{\pi z}{2} \right|=\left|\left(\tan\frac{\pi u}{2}\right)’\right||y-z|=\left|\frac{\pi}{2\cos^2\tfrac{\pi u}{2} }\right||y-z|\ge \frac{\pi}{2}|y-z|.$$

It remains to show completeness of $(G, D_i)$ for each $i$. Let $(x_n)$ be any Cauchy sequence in $(G, D_i)$. It is easy to see that for each $x\in G$, when $y\in G$ tends to $1$ or $-1$ in the usual metric on $[-1,1]$, then $D_i(x,y)$ tends to infinity. Therefore there eixists $H<1$ such that the sequence $(x_n)$ belongs to $[-H,H]$. Since $D_i(y,z)\ge C_i d(y,z)$ for each $y,z\in G$, it follows that $(x_n)$ is a Cauchy sequence on a compact metric subspace $[-H,H]$ of $(G,d)$. So $(x_n)$ converges in $(G,d)$ to some point $x\in H$. By the equivalence of metrics $D_i$ and $d$, $(x_n)$ converges to $x$ in $(G,D_i)$ too.

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    $\begingroup$ First, thank you for the answer. I have some questions regarding your answer: For showing equivalence, how do you show convergence from $D_1$ to $D_2$,and vise versa, since I do not see $D_2$ in your answer? When you are proving the sufficient side of equivalence, why can you use the inequality $D_i(y,z)\ge C_id(y,z)$? When showing completeness, why does such inequality implies each $(G,D_i)$ is complete? $\endgroup$
    – Mike
    Oct 3, 2020 at 19:21
  • $\begingroup$ Another question: Is there an alternative way to prove this, which I listed as the second attempt in my original post? The method is laballed "1" by me. It basically turns an incomplete metric into a complete one under the same open set. I'm not sure whether this will work, but can you say such method is feasible? $\endgroup$
    – Mike
    Oct 3, 2020 at 19:23
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    $\begingroup$ @Mike For showing the equivalence, we show that both $D_1$ and $D_2$ are equivalent to $d$. The inequality $D_i(y,z)\ge C_id(y,z)$ assures that when $D_i(y,z)$ is small then $d(y,z)$ is small too, so the convergence of a sequence to $x$ with respect to $D_i$ implies its convergence to $x$ with respect to $d$. But to show the completeness of $(G,D_i)$, we need to show somehow that sequence which is Cauchy in $(G,D_i)$ is convergent in $(G,d)$, and this is done at the end of the answer. $\endgroup$ Oct 3, 2020 at 19:48
  • $\begingroup$ Another question is that why convergence in $(G,D_1)$ and $(G,D_2)$ is equivalent to convergence in $(G,d)$, where $d$ is the standard metric? Are there any proofs regarding this? $\endgroup$
    – Mike
    Oct 4, 2020 at 2:48
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    $\begingroup$ I guess the derivative should be $|\frac{\pi}{2\cos^{2}\frac{\pi u}{2}}|$ $\endgroup$ Oct 5, 2020 at 2:50

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