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Here is Prob. 19, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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.

Here is the link to a Math SE post of mine where I've included all the relevant definitions.

Prob. 18, Chap. 6, in Baby Rudin: Analysis of Some Curves in the Plane

My Attempt:

As $\phi$ is a continuous, bijective mapping of the compact set $[c,d]$ onto $[a, b]$, so $\phi$ has an inverse $\phi^{-1}$, which is a bijective, continuous mapping of $[a, b]$ onto $[c, d]$, by Theorem 4.17 in Rudin.

And, as $\gamma_2 = \gamma_1 \circ \phi$, so we have $\gamma_1 = \gamma_2 \circ \phi^{-1}$.

Now suppose that $\gamma_1$ is an arc; this means that $\gamma_1$ is 1-1, and since $\phi$ too is 1-1, so is the composite $\gamma_1 \circ \phi$; that is, $\gamma_2$ is an arc.

Conversely, suppose that $\gamma_2$ is an arc; this means that $\gamma_2$ is 1-1, and as $\phi^{-1}$ too is 1-1, so is the composite $\gamma_2 \circ \phi^{-1}$; that is, $\gamma_1$ is an arc.

Thus we have shown that $\gamma_1$ is an arc if and only if $\gamma_2$ is an arc.

Am I right?

Now suppose that $\gamma_1$ is a closed curve. Then $\gamma_1(a) = \gamma_1(b)$. And as $\gamma_2(c) = \gamma_1(\phi(c)) = \gamma_1(a) = \gamma_1(b)$.

What next?

Conversely, suppose that $\gamma_2$ is a closed curve. Then $\gamma_2(c) = \gamma_2(d)$; that is, $\gamma_1(\phi(c)) = \gamma_1(\phi(d))$.

What next?

How to show that $\gamma_1$ is rectifiable if and only if $\gamma_2$ is rectifiable?

And, how to show that $\gamma_1$ and $\gamma_2$ have the same length?

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  • $\begingroup$ You have to prove that since $\phi$ is a continuous bijection, then you have that $\phi(d)=b$. That would finish your proof on the case the curves are closed. $\endgroup$ – Sergio Enrique Yarza Acuña Aug 10 '17 at 13:49
  • $\begingroup$ @SergioEnriqueYarzaAcuña but how to show this? I've thought about it but have not been able to. $\endgroup$ – Saaqib Mahmood Aug 10 '17 at 13:53
  • $\begingroup$ @SaaqibMahmuud: the map $\phi$ is better than a continuous bijection, it's a homeomorphism, which together with the assumption that $\phi(c)=a$ gives you $\phi(d)=b$. $\endgroup$ – Jack Aug 10 '17 at 14:20
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    $\begingroup$ I appreciate your effort for lots of your lengthy proof-verification type questions. However, there are quite a few of them answered but with no feedback at all: no comments, no votes. For instance, this, this, and this. With all due respect, do you not care about your old questions any more? $\endgroup$ – Jack Aug 10 '17 at 14:44
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    $\begingroup$ @SaaqibMahmuud Suppose that $\phi(d)\neq b$. Then, $a<\phi(d)<b$. But since $\phi$ is a bijection, $\exists x\in(c,d)$ such that $\phi(x)=b$. However, since $\psi$ is continuous, by the intermediate value problem, there exists $y\in(a,x)$ such that $\phi(y)=\phi(d)$. This contradicts that $\phi$ is a bijection. $\endgroup$ – Sergio Enrique Yarza Acuña Aug 10 '17 at 15:44
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The argument for the one-to-one (arc) part could be written in a shorter way: the composition of one-to-one functions is one-to-one. Note that in this part, the continuity of $\phi^{-1}$ is not needed (but yes, it would be needed later).

For the "closed curve" part, you have already observed that $\phi$ is a homeomorphism (continuous bijection with continuous inverse); note that by assumption $\phi(c)=a$ one must have $\phi(d)=b$. (This can be done using properties of homeomorphisms.) Thus we see that $\gamma_1(a)=\gamma_1(b)$ if and only if $\gamma_2(c)=\gamma_2(d)$.

Finally for the "same length part", observe that $\phi$ and its inverse $\phi^{-1}$ gives a one-to-one correspondence between partitions $\{s_i\}$ of $[a,b]$ and $\{t_i\}$ such that $$ \sum|\gamma_1(s_i)-\gamma_1(s_{i-1})|=\sum|\gamma_2(t_i)-\gamma_2(t_{i-1})|. $$ Hence the two curves must have the same length by definition.

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  • $\begingroup$ can you please rigorously show how $\phi(d) = b$? $\endgroup$ – Saaqib Mahmood Aug 10 '17 at 15:02
  • $\begingroup$ That deserves to be posted as another question if you are looking for a fully detailed proof. It would be a good question to ask! To be short: a homeomorphism maps boundary to boundary. $\endgroup$ – Jack Aug 10 '17 at 15:19

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