For a continuous injective $f:[0,1)\to\mathbb{R}^2$ with no limit at $t=1$, is the complement of the image of $f$ path connected?

Question

Suppose that $f:[0,1)\to\mathbb{R}^2$ is a continuous injection, and that $\displaystyle\lim_{t\to1^-}f(t)$ does not exist (and is not $\infty$). Is it true that $\mathbb{R}^2\setminus f([0,1))$ is path-connected?

Note that $\displaystyle\lim_{t\to1^-}f(t)$ does not exist means that the two following possiblities are eliminated:

$\bullet$ $\displaystyle\lim_{t\to1^-}f(t)=f(0)$, which means that $f$ is actually a closed curve. In this case $\mathbb{R}^2\setminus f([0,1))$ is obviously not path-connected.

$\bullet$ $\displaystyle\lim_{t\to1^-}f(t)\neq f(0)$ but exists (perhaps being equal to $\infty$). In this case we have a non-closed simple curve $f:[0,1]\to S^2$, and the complement is path-connected according to this question.

Of course, we have $\mathbb{R}^2\setminus f([0,1))=\displaystyle\bigcap_{t\in[0,1)}(\mathbb{R}^2\setminus f([0,t]))$ is the intersection of a descending familly of path-connected sets, but I think the $\displaystyle\lim_{t\to1^-}f(t)=f(0)$ case already tells me that this can't helps much. So is there any clever way to study the complement of the image of $f$? Thank you in advance.

Answer 1

Consider $f : [0, 1) \to \mathbb{R}^2$ defined by

$$ f(t) = r(t) (\cos(2\pi t), \sin(2\pi t)), \qquad \text{where}\quad r(t) = 3 + \sin\left(\frac{\pi}{1-t}\right). $$

Below is the trajectory of $f(t)$, visually demonstrating that $U = \mathbb{R}^2 \setminus f([0,1))$ is disconnected.

Argument. To show that $U$ is indeed not path-connected, assume that

$$ \alpha : [0, 1] \to \mathbb{R}^2$$

is any path from $(1, 0)$ to $(5, 0)$. We want to show that $\alpha$ always passes through $f([0, 1))$. In doing so, we can assume that $\alpha$ avoids the origin. (Whenever $\alpha$ enters and exits a small circle around the origin, we can continuously deform that part by a circular arc.) This then allows to write $\alpha$ as

$$ \alpha(s) = \|\alpha(s)\| (\cos (2\pi \theta(s)), \sin (2\pi \theta(s)) )$$

for some continuous function $\theta : [0, 1] \to \mathbb{R}$. Then $g : [0, 1) \to \mathbb{R}$ defined by

$$ g(s) = \|\alpha(s)\| - r(\theta(s)) $$

has the property that $g([0, 1))$ is an interval. (In fact, $g$ has intermediate value property, meaning that the image of each subinterval of $[0, 1)$ under $g$ is again an interval.) Since

$$ g(0) \leq 1 - 2 < 0 \qquad\text{and}\qquad \liminf_{s \to 1^-} g(s) \geq 5 - 4 > 0, $$

it follows that $g([0, 1))$ contains $0$, or equivalently, $g(s^*) = 0$ for some $s^* \in [0, 1)$. Then

\begin{align*} \alpha(s^*) &= \|\alpha(s^*)\|(\cos(2\pi \theta(s^*)), \sin(2\pi\theta(s^*))) \\ &= r(\theta(s^*)) (\cos(2\pi \theta(s^*)), \sin(2\pi\theta(s^*))) \\ &= f(\theta(s^*)), \end{align*}

hence $\alpha$ passes through $f([0, 1))$ as desired.

Answer 2

The complement need not be path-connected. Start at a point $f(0)$ and then as you increase $t$, go around infinitely many circles while simultaneously increasing the radius (up to at most some finite radius).

An example of such a function is given by $f: [0,1) \to \mathbb{R^2}$ defined by $$f(t) : = \frac12(t+1)(\cos(\exp((1-t)^{-1})),\sin(\exp((1-t)^{-1}))).$$