"Relax" parameterizing a dim $k$ submanifold in $ℝ^n$

Question

Given the following definitions and propositions in Hubbard&Hubbard's Vector Calculus, Linear Algebra, and Differential forms

Definition 5.2.1 ( $k$-dimensional volume 0 of a subset of $\mathbb{R}^n$ ).

1. A bounded subset $X \subset \mathbb{R}^n$ has $k$-dimensional volume 0 if $$ \lim _{N \rightarrow \infty} \sum_{\begin{array}{c} C \in \mathcal{D}_N\left(\mathbb{R}^n\right) \\ C \cap X \neq \varnothing \end{array}}(\underbrace{\frac{1}{2^N}}_{\begin{array}{c} \text { sidelength } \\ \text { of } C \end{array}})^k=0 . $$
2. An arbitrary subset $X \subset \mathbb{R}^n$ has $k$-dimensional volume 0 if for all $R$, the bounded set $X \cap B_R(\mathbf{0})$ has $k$-dimensional volume 0 .

Proposition 5.2.2 ( $k$-dimensional volume 0 of a manifold). If integers $m, k, n$ satisfy $0 \leq m<k \leq n$, and $M \subset \mathbb{R}^n$ is a manifold of dimension $m$, any closed subset $X \subset M$ has $k$-dimensional volume 0 .

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 $\gamma: U \rightarrow \mathbb{R}^n$ parametrizes $M$ if

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

How to prove all manifolds can be parametrized?

A hint given is taking local parametrization $$ g_i: \operatorname{B}_{r_i}(0) \subseteq \operatorname{B}_{r^\prime_i}(0) \rightarrow M, $$ spreading away the domains and doctoring up the intersected codomains, as said by Tuček in https://mathoverflow.net/questions/177653/does-every-compact-manifold-exhibit-an-almost-global-chart. But still I can not figure it out (or feeling vague on it) And I am confused on why it requires $C$ compact in item 5 of the definition.

This thread continues an old thread I asked two years ago. (The proof in it is wrong and it's very long but the setup is the same) Prove all (sub)manifold in $\Bbb{R}^n$ can be "relaxed" parametrized

Thank you in advance for any help!

Answer 1

Maybe I found the proof. It seems true for some intuitive examples. I am just not sure if it holds for general(or weird) cases:

Let $M$ be a $C^1$ manifold with dim $k$ in $\Bbb{R}^n$. By Hubbard&Hubbard’s definition and the second countability of manifolds, There are countable many

• open coordinate balls $B_{r_i}(0) \subseteq B_{r_i^\prime}(0) \subseteq \Bbb{R}^k$ with $0 < r_i < r_i^\prime$, abbreviated as $B_i$ and $B_i^\prime$.
• and subsets $V_i \subseteq V_i^\prime \subseteq M$ covering $M$
• and $C^1$ continuous maps $g_i: B_i^\prime \to V_i^\prime$ such that $g_i$ is one-to-one with $g_i(B_i) = V_i$, and the Jacobian matrix $Dg_i$ has rank $k$ at any points in $B_{r_i^\prime}(0)$.

Then spreading out the domains of all $B_i$ and $B_i^\prime$, in a way such that $B_i^\prime$ and $B_j^\prime$ are disjoint for all $i, j$, and for any $R > 0$, the ball $B_R(0)$ intersects only finitely many $B_i^\prime$. More precisely, let $R_i = 4\sum r_i$, and translate $B_i^\prime$ between $B_{R_i}(0)\backslash B_{R_{i-1}}(0)$, illuminated as follow:

Then doctoring up these $g_i$ such that their codomains are disjoint in the following way: first let $U_0 = B_0$ and $X_0 = \varnothing$, then inductively define

$$ U_i = B_i\backslash \cup_{j < i} g_i^{-1}(V_i \backslash g_j(\overline{B_j})) $$

and

$$ X_i = B_i\cap\cup_{j < i} g_i^{-1}(V_i \cap g_j(\partial B_j)) $$

illuminated as follow:

Take $X = \cup X_i$ and $U = X \cup U_i$ and define the parametrization $\gamma: U \to M$ taking any point $x$ in $U_i \cup X_i$ to $g_i(x)$, then:

• $U$ has boundary with $k$ dimension volume $0$ because it’s boundary is $X = \cup X_i$: For any $R > 0$, the intersection $B_R(0) \cap X$ is a finite union of dim $k-1$ manifold. (Here the Jordan content is used, and hence only the finite additivity holds)
• $\gamma(U) = M$ for $\{V_i\}$ covers $M$
• $\gamma|_{U-X}$ same as the union of $g_i|_{U_i}$ and hence one-to-one and $C^1$
• the Jacobian $D\gamma$ is one-to-one because $Dg_i$ is one-to-one (rank $k$)
• $X$ is same as $\partial U$ and hence $k$ dimension volume $0$. Given any $C\subseteq M$ compact, $C$ can be covered by finitely many $V_i$. Hence $\gamma(X) \cap C$ is the union of finitely many $\gamma(X_i) \cap C$ and hence with dim $k$ volume $0$ (also used the finite additivity of Jordan content here, and with $\gamma(X)$ a union of dim $k-1$ manifolds).

A counterexample if $C$ not compact I think should be https://math.stackexchange.com/questions/3977752/showing-that-the-manifold-u-bigcup-i-1-infty-lefta-i-frac12i