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I'm reading Hubbard's Vector Calculus, Linear Algebra, and Differential forms, and it defines a relaxed parametrization in the following way:

Definition 5.2.3 ("Relaxed" parametrization of a manifold).

Let $M\subset \mathbb{R}^n$ be a $k$-dimensional manifold and let $U\subset \mathbb{R}^k$ be a subset with boundary of $k$-dimensional volume $0$. Let $X\subset U$ be such that $U-X$ is open. Then a continuous mapping $\gamma: U \to \mathbb{R}^n$ parametrizes M if:

  1. $\gamma(U) \supset M;$
  2. $\gamma(U-X)\subset M;$
  3. $\gamma: (U-X) \to M$ is one to one, of class $C^1$ with locally Lipschitz derivative;
  4. the derivative $[D\gamma(\mathbf{u})]$ is one to one for all $\mathbf{u}$ in $U-X$;
  5. X has a $k$-dimensional volume 0, as does $\gamma(X)\cap C$ for any compact subset $C\subset M$.

(note: Hubbard uses volume to refer to volume in the sense of Riemann)

The problem:

Choose an integer $k > 2$, make a list $a_1, a_2, ...$ of the rationals in $[0, 1]$ and consider the set $$U = \bigcup_{i=1}^\infty \left(a_i-\frac{1}{2^{k+i}}, a_i + \frac{1}{2^{k+i}}\right)$$ Show that $U$ is a one-dimensional manifold and that it can be parametrized according to definition 5.2.3.

What I've tried

Showing that it's a 1-manifold is straightforward, because it's an open subset of $\mathbb{R}$. However, The second part of the problem has me stumped. The most obvious choice is the identity map, but (as Hubbard also notes), this fails because $\text{Vol}_1(\partial U)$ is undefined and has nonzero measure. My next attempt was to choose $U = [\inf U, \sup U]$, which of course has boundary volume zero. Then, X would contain its boundary, and some other set of volume zero strategically chosen so that the image of $U-X$ is a subset of $M$. However, this loops back to the original problem, because it basically requires me to take a union of subsets of these open sets as $U-X$, which has the same volume problem.

How do I go about doing this? I'm pretty stuck. Perhaps I shouldn't try and do this constructively?

Prox
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1 Answers1

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Have you figured it out? I'm trying to Prove Theorem 5.2.6(aka Exercise 5.2.8, not achieved yet, I posted a question linked this). This example as a special case maybe I can give a proof as follows:

$U$ is open, and Any open subset of $\Bbb R$ is a countable union of disjoint open intervals. Suppose $U = \cup I_i$, where $\{I_i\}$ is disjoint open intervals.

The following part I believe should be the key part. As Hubbard hint, spread out $\{I_i\}$ so they are well separated.

Obviously, $U \subset [-2, 2] \subset [-4, 4]$ (or any other large enough but bounded interval). "Move" $I_i$ to the center of interval $[8i-12, 8i-4]$ with some map $s_i$. (After this step $\{I_i\}$ is "well separated", but Hubbard only defines volume for bounded set. So in the next I "shrink them" but try to keep the "well separated" property too.) Using some homeomorphism $f: \Bbb{R} \rightarrow (0, 1)$ like this: Is there a bijective map from $(0,1)$ to $\mathbb{R}$?, for each $I_i$ there is a parametrization $\varphi_i: f(s_i(I_i)) \rightarrow I_i$. And $\{f(s_i(I_i))\}$ is disjoint. The rest part is for proving the boundary of $\cup f(s_i(I_i))$ has $0$ volume.

The boundary of $\cup f(s_i(I_i)) = \{0, 1\} \cup \partial s_i(I_i)$, is measurable $0$：any two small open interval covering $0$ and $1$, covers all but finitely many points of $\{0, 1\} \cup \partial s_i(I_i)$. Taking finitely many open intervals with total length as small as possible can cover $\{0, 1\} \cup \partial s_i(I_i)$. Hence the Lesbegue measure of $\partial \big(\cup f(s_i(I_i))\big)$ is $0$. And the discountinuity points set of the indicator function on it is itself. So $\partial \big(\cup f(s_i(I_i))\big)$ has volume. Obviously its volume is $0$. (calculate it via lower Riemann sum.)

In this example, it seems that $X = \varnothing$. So all conditions are trivially true.

onRiv
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