Does an immersed curve have to be "curve-like" somewhere? Let $C$ be a smooth immersed curve, i.e. an image of a smooth immersion $\varphi\colon \mathbb{R} \to \mathbb{R}^n$. We will say that $C$ is curve-like at a point $p \in C$ if, for every two smooth immersions $\alpha, \beta\colon \mathbb{R} \to C$ with $\alpha(0) = \beta(0) = p$, we have neighborhoods $V_\alpha, V_\beta$ of 0 so that $\alpha(V_a) = \beta(V_b)$.
An embedded curve is curve-like at every point. However, a curve with "self-intersections", like the $\infty$ symbol, fails to be curve-like at those self-intersections.
The existence of space-filling curves shows that the image of $\mathbb{R}$ by a continuous map does not have to be curve-like anywhere. When the parameterization $\varphi$ is guaranteed to be a smooth immersion, does $C$ need to have at least one curve-like point?
 A: No -- it is possible for $C$ to be non-curvelike everywhere.
Let $g$ be a fixed smooth bump funcion $\mathbb R\to\mathbb R$ supported on the open unit interval and define $\hat g(x)=g(x-\lfloor x\rfloor)$, an infinite row of bumps.
To warm up, let us first consider this set:
$$ D = \{(x,0)\mid x\in\mathbb R\} \cup \bigcup_{n=1}^\infty\{(x,e^{-n}\hat g(nx))\mid x\in\mathbb R\} $$
It consists of the $x$ axis overlaid by a series of bumps that get narrower or narrower, with smaller and smaller amplitude. The amplitudes decrease vastly faster than the widths.
Even thought the wiggles only touch the $x$-axis at rational coordinates, $D$ is not curve-like at any point on the $x$-axis. If we have an irrational point we can still let $\alpha$ be the $x$-axis itself, and let $\beta$ include a sequence of non-overlapping bumps that collectively approach $p$. If $\beta$ is parameterized by the $x$-coordinate, it will be smooth, and it doesn't coincide with $\alpha$ in any neighborhood of $p$.

First construction: With this construction in mind, we can construct a smooth curve $\varphi$ whose image is nowhere curvelike:
Start by choosing an surjection $B: \mathbb N\to \mathbb Z\times \mathbb N$ such that the first component of $B(m)$ is always smaller than $m$.


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*For $t\le 0$ we have $\varphi(t)=(t,0)$. Ho hum.

*For each $m\in\mathbb N$, let $(k,n)=B(m)$ and construct $\varphi(t)$ for $6m<t\le 6m+6$ as follows:


*

*On the interval $6m+5\le t\le 6m+6$, let $\varphi(t)$ follow the already chosen path for $\varphi$ on $[k,k+1]$, but offset by the wiggle function $e^{-n}\hat g(nx)$ added either to the $y$ or the $x$ coordinate depending on whether the $[k,k+1]$ interval is considered "horizontal-ish" or "vertical-ish" (and this classification is then taken over for $[6m+5,6m+6]$).

*On the interval $6m < t < 6m+5$ choose a smooth connector curve from where the previous step left off to the location and direction we need to reach at $t=6m+5$. If necessary, we can construct the connector such that each of its five unit intervals is clearly either horizontal-ish or vertical-ish for when we add bumps to it later.



Now each $p\in C$ will either correspond to an integer $t$ (and then $C$ is clearly not curve-like at $p$, since differently bumpy segments of $\varphi$ join up there by construction), or else there's a neighborhood of $p$ that has a subset that looks locally like $D$ (modulo a diffeomorphism of $\mathbb R^2$).

A more explicit construction thought up later. Let $\psi$ be a surjection from $\mathbb Z$ to the set of finite subsets of $\mathbb N$. For example $\psi(n)$ could be $\varnothing$ for negative $n$ and consist of the positions of $1$ bits in the binary representation of $n$ when $n\ge 0$. Then set
$$ \phi(k+x) = \biggl[ 1 + \sum_{n\in \psi(k)} e^{-n}\hat g(nx)\biggr]\cdot(\cos 2\pi t,\sin 2\pi t)$$
for all $k\in\mathbb Z, x\in[0,1)$.
Now $C$ is bounded, and each of the derivatives of $\varphi$ is globally bounded too.
