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Let $\gamma_1$ be a curve in $\mathbb{R}^k$, defined on $[a, b]$; let $\phi$ be a continuous 1-1 mapping of $[c, d]$ onto $[a, b]$, such that $\phi(c) = a$; and define $\gamma_2(s) = \gamma_1(\phi(s))$. Prove that $\gamma_2$ is an arc, a closed curve, or a rectifiable curve if and only if the same is true of $\gamma_1$. Prove that $\gamma_2$ and $\gamma_1$ have the same length.

My attempt:

(1) Claim: $\gamma_2$ is an arc $\iff$ $\gamma_1$ is an arc. Proof: ($\Rightarrow$) first suppose $\gamma_2$ is one-to-one. We need to show that, $\gamma_1$ is also 1-1, i.e. $\forall x,x^\prime \in [a,b],$ If $\gamma_1(x)=\gamma_1(x^\prime)$, then $x=x^\prime$. Since $\phi$ is bijective and $x,x^\prime \in [a,b]$, $\exists ! u,v \in [c,d]$ such that $x=\phi(u)$ and $x^\prime=\phi(v)$, respectively. So $\gamma_1(\phi(u))=\gamma_1(\phi(v))$. $\gamma_2$ is one-to-one, i.e. $\forall s,s^\prime \in [c,d],$ If $\gamma_2(s)=\gamma_1(\phi(s))= \gamma_1(\phi(s^\prime)) =\gamma_1(s^\prime)$, then $s=s^\prime$. Since $u,v \in [c,d]$, $\gamma_1(\phi(u))=\gamma_1(\phi(v)) \Rightarrow u=v \Rightarrow x=x^\prime$. Thus, $\phi(u)=\phi(v)=x=x^\prime$. ($\Leftarrow$) conversely suppose $\gamma_1$ is 1-1. We need to show that, $\gamma_2$ is also 1-1, i.e. $\forall x,x^\prime \in [c,d],$ If $\gamma_2(x)=\gamma_2(x^\prime)$, then $x=x^\prime$. $\gamma_2(x)=\gamma_2(x^\prime)=\gamma_1(\phi(x))=\gamma_1(\phi(x^\prime))$. Since $\phi(x), \phi(x^\prime) \in [a,b]$ and $\gamma_1$ is one-to-one, we have $\phi(x)=\phi(x^\prime)$. Since $\phi$ Is bijective, $x=x^\prime$. Thus, $\forall x,x^\prime \in [c,d], \gamma_2(x)=\gamma_2(x^\prime) \Rightarrow x=x^\prime$. Is this proof correct?

(2) Claim: $\gamma_2$ is closed curve $\iff$ $\gamma_1$ is closed curve. Proof: we’ll first show, If $\phi$ is continuous, bijective and $\phi(c)=a$ , then $\phi(d)=b$. Proof: Assume towards contradiction, i.e. $\phi(d) \neq b$, $a \lt \phi(d) \lt b$. Since $\phi$ is bijective, $\exists ! x\in(c,d)$ such that $\phi(x)=b$. Since $\phi$ is continuous on $[c,d]$, it’s also continuous on $[c,x]$. By theorem 4.23, $\phi(c)=a \lt \phi (d) \lt \phi(x)=b, \exists y \in (c,x)$ such that $\phi(y)=\phi(d)$. By injectivity property of $\phi$, $\phi(y)=\phi(d) \Rightarrow y=d$. Which is a contradiction, since $y \lt d$. Thus our initial assumption must be wrong. This completes the proof.

($\Rightarrow$) suppose $\gamma_2$ is closed. So, $\gamma_2(c)=\gamma_2(d)$. $\gamma_2(c)= \gamma_1(\phi(c)) = \gamma_1(a)$, by hypothesis of the problem. $\gamma_2(d)= \gamma_1(\phi(d))= \gamma_1(b)$, by above claim. Thus, $\gamma_1(a)=\gamma_1(b)$, $\gamma_1$ is closed. ($\Leftarrow$) conversely suppose $\gamma_1$ is closed, i.e. $\gamma_1(a)= \gamma_1(b)$. So, $\gamma_2(c)= \gamma_1(\phi(c))= \gamma_1(a)$ and $\gamma_2(d)= \gamma_1(\phi(d))= \gamma_1(b)$. Thus, $\gamma_2(c)= \gamma_2(d)$. Is this proof correct?

(3) claim: $\wedge(\gamma_1) \lt \infty \iff \wedge(\gamma_2) \lt \infty$ and $\gamma_2$ & $\gamma_1$ have the same length. While solving this problem, I first thought that, I should show $\wedge(\gamma_2) \leq \wedge(\gamma_1) \lt \infty$ and $\wedge(\gamma_1) \leq \wedge(\gamma_2) \lt \infty$ so that $\wedge(\gamma_1)=\wedge(\gamma_2)$. But I end up doing the following. Proof: ($\Rightarrow$) first suppose $\wedge(\gamma_1) \lt \infty$. $\wedge(\gamma_1)= \sup \wedge(P,\gamma_1)$; $\wedge(P,\gamma_1)= \sum_{i=1}^n | \gamma_1 (x_i) - \gamma_1 (x_{i-1})|$. $P=\{x_0, ..... , x_n\}$. Since $\phi$ is bijective, $\exists ! y_i \in [c,d]$ such that $\phi(y_i)=x_i, \forall i \in J_n \cup \{0\}$. It’s clear that $\phi(y_0)=x_0=a$ and $\phi(y_n)=x_n=b$. To each partition $P=\{x_0, ..... , x_n\}$ of $[a,b]$ corresponds a partition $Q=\{y_0, ... , y_n\}$ of $[c,d]$, so that $x_i=\phi(y_i)$. All partition of $[c,d]$ are obtained in this way. So $\wedge(Q,\gamma_2)= \sum_{i = 1}^n | \gamma_2(y_i)-\gamma_2(y_{i-1})| = \sum_{i = 1}^n | \gamma_1(\phi(y_i))-\gamma_2(\phi(y_{i-1}))| = \sum_{i = 1}^n | \gamma_1(x_i)-\gamma_1(x_{i-1})| = \wedge(P, \gamma_1)$. Thus $\wedge(\gamma_1)=\sup \wedge(P,\gamma_1)= \sup \wedge(Q,\gamma_2) = \wedge(\gamma_2) \lt \infty$. Hence, $\gamma_1$ and $\gamma_2$ have the same length. [Que: My last step, to be more specific, second equality from left side, is not satisfactory to me. It’s not rigorous. Can you make it rigorous?]. ($\Leftarrow$) conversely suppose $\wedge (\gamma_2) \lt \infty$. $\wedge (\gamma_2) = \sup \wedge (Q,\gamma_2)$; $\wedge (Q,\gamma_2) = \sum_{i=1}^n | \gamma_2 (y_i) - \gamma_2 (y_{i-1})|$. $Q=\{y_0, ..... , y_n\}$. To each partition $Q=\{y_0, ..... , y_n\}$ of $[c,d]$ corresponds a partition $P=\{x_0, ..... , x_n\}$ of $[a,b]$, so that $x_i = \phi(y_i)$. All partition of $[a,b]$ are obtained in this way. $\wedge(P,\gamma_1)= \sum_{i = 1}^n | \gamma_1(x_i)-\gamma_1(x_{i-1})| = \sum_{i = 1}^n | \gamma_1(\phi(y_i))-\gamma_2(\phi(y_{i-1}))| = \sum_{i = 1}^n | \gamma_2(y_i)-\gamma_2(y_{i-1})| = \wedge(Q, \gamma_2)$. Thus, $\wedge(\gamma_2)=\sup \wedge(Q,\gamma_2)= \sup \wedge(P,\gamma_1) = \wedge(\gamma_1) \lt \infty$. Hence, $\gamma_1$ and $\gamma_2$ have the same length.[Que: same question as above]. Is this proof correct?

user264745
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    Yes, I am now definitely more interested in editing tags and finding duplicates. As you can imagine, most new posts nowadays are some forms of duplicate anyway. I don't get +2 for the edit though (once you hit 2k+ you no longer have that). – Arctic Char Dec 19 '21 at 12:45
  • @ArcticChar what’s your justification for removing proof explanation tag? – user264745 Dec 19 '21 at 12:47
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    My understanding is that [tag:proof-explanation] is used when you have doubt in a proof in a text, not a proof you write yourself. If you write a proof/solution and you have doubt on that, [tag:solution-verification] is the one you used. – Arctic Char Dec 19 '21 at 12:48
  • @ArcticChar two different interpretation of a single statement. Official proof explanation tag don’t say anything about proof by myself and proof by text. What exactly do you mean by text? book, paper? What if that proof I copied from some text? – user264745 Dec 19 '21 at 12:54
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    Please have a look at this. If the proof is from some text that you are trying to understand, then of course the tag is suitable. – Arctic Char Dec 19 '21 at 12:57
  • But it makes no sense that you write the proof yourself, and you need someone else to explain to you what the proof is writing right? – Arctic Char Dec 19 '21 at 12:58
  • @ArcticChar I am explicitly asking to make a step rigorous. I known that step is correct, because it has been used in one of the theorem(theorem 6.19) of Baby Rudin. – user264745 Dec 19 '21 at 13:05
  • Well, now I am more confused. You start with "My attempt", so I assume everything below is your own work. Anyway, if you have a specific proof of Rudin that you wish to understand, that's no point hiding it in your solution verification question. Ask another question, citing exactly the wording of Rudin, and ask for explanation. – Arctic Char Dec 19 '21 at 13:11
  • @ArcticChar “ I assume everything below is your own work”- I’m making stuff by myself I rely on definitions and some techniques. That step is one of the technique I used. – user264745 Dec 19 '21 at 13:18
  • @ArcticChar practically, in this post, I’m only asking about that step. – user264745 Dec 19 '21 at 13:20
  • No, in this post you ask if "your proof is correct". You are asking other check if your proof is correct, and at some point you used something from Rudin, and you ask how that is justified. But l still don't see what exactly Rudin's theorem that you used, and what you do not understand. Again, ask two separate questions please. – Arctic Char Dec 19 '21 at 13:29
  • @ArcticChar opss... I’m not* making. In theorem 6.19, $U(P,f,\alpha) = U(Q,g, \beta)$, partition $Q$ of [A,B] derived from partition $P$ of [a,b]. We know that $\inf U(P,f,\alpha) = \int_a^b f d\alpha$ and $\inf U(Q,g, \beta) = \int_a^b g d\beta$, we need to show that $\int_a^b f d\alpha = \int_a^b g d\beta$, so $\inf U(Q,g, \beta)$ must be equal to $\inf U(P,f,\alpha)$, given $U(Q,g, \beta)= U(P,f,\alpha)$. The difference is that for $g$ $\inf$ is taken over partition Q of [A,B] and for $f$ $\inf$ is taken over partition P of [a,b]. (All notation is similar to theorem 6.19) – user264745 Dec 19 '21 at 13:42
  • @ArcticChar if you look carefully, above(theorem 6.19) scenario is similar to (3). – user264745 Dec 19 '21 at 13:47

1 Answers1

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We know that $\phi$ has a continuous 1-1 inverse $\varphi$, and that the composition of one-to-one functions is one-to-one. Hence, since $\gamma_1 = \gamma_2 (\varphi(x))$ we see that $\gamma_1$ and $\gamma_2$ are both arcs (one-to-one) if either is. Since necessarily $\phi(d) = b$, we see that $\gamma_1(a) = \gamma_1(b)$ if and only if $\gamma_2 (c) = \gamma_2 (d)$. Hence both are closed curves if either is. Finally, since $\phi$ and $\varphi$ establish a one-to-one correspondence between partitions $\{s_i\}$ of $[a, b]$ and $\{t_i\}$ of $[c, d]$ such that $\sum | \gamma_1(s_i) - \gamma_1 (s_{i-1})| = \sum | \gamma_2(t_i) - \gamma_2 (t_{i-1})|$, it follows that the two curves have the same length.

Mr. Proof
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  • Thank you for the answer. I think your post basically summarize this post https://math.stackexchange.com/q/2389104/861687. I wasn’t familiar with the property composite of 1-1 functions is 1-1. How to rigorously prove $\gamma_1=\gamma_2(\varphi(x))$? Here is my try: we know $\gamma_2= \gamma_1 \circ \phi$. So, $\gamma_1= (\gamma_1 \circ \phi) \circ \varphi = \gamma_1 \circ (\phi \circ \varphi)$. I believe $\phi \circ \varphi$ is identity function. Hence $\gamma_1= \gamma_1$ always true. This completes the proof. – user264745 Dec 19 '21 at 17:27
  • Now to the last part. I think there is huge jump. The expression $\sum | \gamma_1(s_i) - \gamma_1 (s_{i-1})| = \sum | \gamma_2(t_i) - \gamma_2 (t_{i-1})|$, don’t show two curves have the same length. We have to show $\sup \sum | \gamma_1(s_i) - \gamma_1 (s_{i-1})|$ and $\sup \sum | \gamma_2(t_i) - \gamma_2 (t_{i-1})|$ are the same. In former case sup is taken over all partition of [a,b] and in the later case sup is taken over all partition of [c,d]. Now how do I rigorously show sup of former and later sum is equal? – user264745 Dec 19 '21 at 17:27
  • @user264745 It is clear. Since they are just partitions then the sums over partitions are clear. – Mr. Proof Dec 19 '21 at 17:54
  • yeah, but both are different partitions. I’m sorry but I don’t get it. – user264745 Dec 19 '21 at 17:58
  • @user264745 we have a one-to-one correspondence between the two partitions. Just draw some intervals and make the correspondence and I am sure it will be clear for you. – Mr. Proof Dec 19 '21 at 18:11
  • Okay. Is cardinality of set of all partition of [c,d] is equal to set of all partition of [a,b]? – user264745 Dec 19 '21 at 18:15
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    @user264745 Sure. We define the cardinality by a correspondence mapping. – Mr. Proof Dec 19 '21 at 18:16
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    Then it's clear sup over all partition of [a,b] and over partition of [c,d] are equal. – user264745 Dec 19 '21 at 18:22