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?



Ordinal of injectivity for a smooth regular curve with a finite arc-length
If I understand the question correctly, the answer is that there is no better bound than $\Omega=\omega\_1$, the first uncountable ordinal. I'll sketch the solution for $d=1$, I believe the others are similar. For any countable ordinal $\alpha$, there is an order-preserving embedding of $\alpha$ into $[a, b]$ with closed image $I$. We can then take any continuous function $\gamma$ on $[a, b]$ satisfying $\gamma^{-1}(\{0\})=I$ (it's not hard to also make $\gamma$ smooth and regular); the associated $\Omega$ will then be at least $\alpha$.


