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Let $\gamma: [a,b]\to\mathbb{R}^d$ defined by $$\gamma(t)=(\gamma_1(t),\dots,\gamma_d(t)) $$ be a smooth (i.e., $\gamma\in C^\infty (\mathbb{R}))$ and regular ($\gamma^\prime(t)\neq \vec 0$) curve with finite arc-length.

Define a possibly transfinite partition $P=\{a=x_0<x_1<\dots<x_\Omega=b\}$ on $[a,b]$ by setting $x_0=a$, and then $$ x_m= \sup \{t\in[x_{m-1},b]\mid \forall s\in [a,x_{m-1}],\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m>0$ is a successor ordinal, and $$ x_m= \sup \{t\in[x_m',b]\mid \forall s\in [a,x_m'),\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m$ is a limit ordinal, where $x_m'=\sup\{x_l\mid l<m\}$.

Roughly speaking, this partition divides $\gamma(t)$ into injective graphs in respect to last coordinate. The ordinal $\Omega$ indicates how long the partition goes, that is, we define $x_m$ for as long as possible, and the construction eventually terminates when we reach $b$, and $\Omega$ is the subindex ordinal that $b$ was assigned.

My question is "how large can $\Omega$ be?". Clearly, $\Omega<\omega_1$ since there is no order preserving embedding of $\omega_1$ into $\mathbb R$. Can we find a curve $\gamma$ with $\Omega \ge \omega$? Can we bound this ordinal?

Let $\gamma: [a,b]\to\mathbb{R}^d$ defined by $$\gamma(t)=(\gamma_1(t),\dots,\gamma_d(t)) $$ be a smooth (i.e., $\gamma\in C^\infty (\mathbb{R}))$ and regular ($\gamma^\prime(t)\neq \vec 0$) curve with finite arc-length.

Define a possibly transfinite partition $P=\{a=x_0<x_1<\dots<x_\Omega\le b\}$ of an initial segment of $[a,b]$ by setting $x_0=a$, and then $$ x_m= \sup \{t\in(x_{m-1},b]\mid \forall s\in [a,x_{m-1}],\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m>0$ is a successor ordinal, and $$ x_m= \sup \{t\in[x_m',b]\mid \forall s\in [a,x_m'),\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m$ is a limit ordinal, where $x_m'=\sup\{x_l\mid l<m\}$. The ordinal $\Omega $ is the largest ordinal for which these definitions make sense, that is, the sets over which suprema are computed are non-empty, and we have not reached $ b $ yet. Note that we can have $x_\Omega <b $.

Roughly speaking, this partition divides $\gamma(t)$ into injective graphs in respect to last coordinate. The ordinal $\Omega$ indicates how long the partition goes, that is, we define $x_m$ for as long as possible, and the construction eventually terminates when we reach $b$, and $\Omega$ is the subindex ordinal that $b$ was assigned.

My question is "how large can $\Omega$ be?". Clearly, $\Omega<\omega_1$ since there is no order preserving embedding of $\omega_1$ into $\mathbb R$. Can we find a curve $\gamma$ with $\Omega \ge \omega$? Can we bound this ordinal?

Let $\gamma: [a,b]\to\mathbb{R}^d$ defined by $$\gamma(t)=(\gamma_1(t),\dots,\gamma_d(t)) $$ be a smooth (i.e., $\gamma\in C^\infty (\mathbb{R}))$ and regular ($\gamma^\prime(t)\neq \vec 0$) curve with finite arc-length.

Assume that $\gamma_j(t),\gamma_j^\prime(t)\neq 0$ for $1\le j\le d$.

Define a possibly transfinite partition $P=\{a=x_0<x_1<\dots<x_\Omega=b\}$ on $[a,b]$ by setting $x_0=a$, and then $$ x_m= \sup \{t\in[x_{m-1},b]\mid \forall s\in [a,x_{m-1}],\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m>0$ is a successor ordinal, and $$ x_m= \sup \{t\in[x_m',b]\mid \forall s\in [a,x_m'),\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m$ is a limit ordinal, where $x_m'=\sup\{x_l\mid l<m\}$.

Roughly speaking, this partition divides $\gamma(t)$ into injective graphs in respect to last coordinate. The ordinal $\Omega$ indicates how long the partition goes, that is, we define $x_m$ for as long as possible, and the construction eventually terminates when we reach $b$, and $\Omega$ is the subindex ordinal that $b$ was assigned.

My question is "how large can $\Omega$ be?". Clearly, $\Omega<\omega_1$ since there is no order preserving embedding of $\omega_1$ into $\mathbb R$. Can we find a curve $\gamma$ with $\Omega \ge \omega$? Can we bound this ordinal?

Let $\gamma: [a,b]\to\mathbb{R}^d$ defined by $$\gamma(t)=(\gamma_1(t),\dots,\gamma_d(t)) $$ be a smooth (i.e., $\gamma\in C^\infty (\mathbb{R}))$ and regular ($\gamma^\prime(t)\neq \vec 0$) curve with finite arc-length.

Define a possibly transfinite partition $P=\{a=x_0<x_1<\dots<x_\Omega=b\}$ on $[a,b]$ by setting $x_0=a$, and then $$ x_m= \sup \{t\in[x_{m-1},b]\mid \forall s\in [a,x_{m-1}],\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m>0$ is a successor ordinal, and $$ x_m= \sup \{t\in[x_m',b]\mid \forall s\in [a,x_m'),\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m$ is a limit ordinal, where $x_m'=\sup\{x_l\mid l<m\}$.

Roughly speaking, this partition divides $\gamma(t)$ into injective graphs in respect to last coordinate. The ordinal $\Omega$ indicates how long the partition goes, that is, we define $x_m$ for as long as possible, and the construction eventually terminates when we reach $b$, and $\Omega$ is the subindex ordinal that $b$ was assigned.

My question is "how large can $\Omega$ be?". Clearly, $\Omega<\omega_1$ since there is no order preserving embedding of $\omega_1$ into $\mathbb R$. Can we find a curve $\gamma$ with $\Omega \ge \omega$? Can we bound this ordinal?

Let $\gamma: [a,b]\to\mathbb{R}^d$ be a smooth (i.e. $\gamma\in C^\infty (\mathbb{R}))$ and regular ($\gamma^\prime(t)\neq \vec 0$) curve with arc-length defined by $$\gamma(t)=(\gamma_1(t),\dots,\gamma_d(t)).$$

Assume that $\gamma_j(t),\gamma_j^\prime(t)\neq 0$ for $1\le j\le d$.

Define a possibly transfinite partition $P=\{a=x_0<x_1<\dots<x_\Omega=b\}$ on $[a,b]$ by setting $x_0=a$, and then $$ x_m= \sup \{t\in[x_{m-1},b]\mid \forall s\in [a,x_{m-1}],\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m>0$ is a successor ordinal, and $$ x_m= \sup \{t\in[x_m',b]\mid \forall s\in [a,x_m'),\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m$ is a limit ordinal, where $x_m'=\sup\{x_l\mid l<m\}$.

Roughly speaking, this partition divides $\gamma(t)$ into injective graphs in respect to last coordinate. The ordinal $\Omega$ indicates how long the partition goes, that is, we define $x_m$ for as long as possible, and the construction eventually terminates when we reach $b$, and $\Omega$ is the subindex ordinal that $b$ was assigned.

My question is "how large can $\Omega$ be?". Clearly, $\Omega<\omega_1$ since there is no order preserving embedding of $\omega_1$ into $\mathbb R$. Can we find a curve $\gamma$ with $\Omega \ge \omega$? Can we bound this ordinal?

Let $\gamma: [a,b]\to\mathbb{R}^d$ defined by $$\gamma(t)=(\gamma_1(t),\dots,\gamma_d(t)) $$ be a smooth (i.e., $\gamma\in C^\infty (\mathbb{R}))$ and regular ($\gamma^\prime(t)\neq \vec 0$) curve with finite arc-length.

Assume that $\gamma_j(t),\gamma_j^\prime(t)\neq 0$ for $1\le j\le d$.

Define a possibly transfinite partition $P=\{a=x_0<x_1<\dots<x_\Omega=b\}$ on $[a,b]$ by setting $x_0=a$, and then $$ x_m= \sup \{t\in[x_{m-1},b]\mid \forall s\in [a,x_{m-1}],\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m>0$ is a successor ordinal, and $$ x_m= \sup \{t\in[x_m',b]\mid \forall s\in [a,x_m'),\gamma_j(t)\neq\gamma_j(s) \quad \forall 1\le j\le d-1\}, $$ whenever $\Omega\ge m$ is a limit ordinal, where $x_m'=\sup\{x_l\mid l<m\}$.

Roughly speaking, this partition divides $\gamma(t)$ into injective graphs in respect to last coordinate. The ordinal $\Omega$ indicates how long the partition goes, that is, we define $x_m$ for as long as possible, and the construction eventually terminates when we reach $b$, and $\Omega$ is the subindex ordinal that $b$ was assigned.

My question is "how large can $\Omega$ be?". Clearly, $\Omega<\omega_1$ since there is no order preserving embedding of $\omega_1$ into $\mathbb R$. Can we find a curve $\gamma$ with $\Omega \ge \omega$? Can we bound this ordinal?