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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 $P=\{a=x_0<x_1<\dots<x_\Omega\le b\}$ 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?

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  • $\begingroup$ @BenMcKay But the cardinality of $P$ won't be infinite. $\endgroup$
    – Alan Watts
    Commented Mar 2, 2016 at 3:58
  • $\begingroup$ @BenMcKay also I can't see why it has a finite arc-length. $\endgroup$
    – Alan Watts
    Commented Mar 2, 2016 at 4:25
  • $\begingroup$ Each point $t \in P$ is a point at which $\gamma(t)=\gamma(a)$. If there are infinitely many, they accumulate at at point $t \in P$ at which $\gamma'(t)=0$. $\endgroup$
    – Ben McKay
    Commented Mar 2, 2016 at 7:25
  • $\begingroup$ If you allow $\gamma'(b)=0$ and only want $C^{\infty}[a,b)$, make $\gamma$ start at the origin, turn around circle at unit speed, then around a smaller circle, tangent to the first at the origin, and so on. This only gives $C^2$ $\gamma$. Replace the circles by a smooth loop that has infinite order tangency to the horizontal axis at the origin, and repeat with every smaller copies of that loop, shrinking fast enough to get finite total length. $\endgroup$
    – Ben McKay
    Commented Mar 2, 2016 at 7:26
  • $\begingroup$ @BenMcKay but the Hawaiian earring you describe is in $C^\infty[a,b)$, isn't it? How replacing it with smooth loops helps? Another question: What is order of tangency ? However, it's only an example for $|P|=\aleph_0$ as far as I understood. Can we construct something similar but now with cardinality of continuum ? $\endgroup$
    – Alan Watts
    Commented Mar 2, 2016 at 8:24

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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 $\alpha$.

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