Let $f\colon \mathbb R^1\to \mathbb R^3$ be a continuous and injective map. Is $\mathbb R^3\setminus f(\mathbb R^1)$ a path-connected space?



Space-discriminating injective curve
Yes. In fact, $\mathbb R^3$ could be any 3-manifold and $f(\mathbb R^1)$ any countable union of embedded segments.


**Lemma 1**. Let $U$ be an open ball in $\mathbb R^3$ and $f:I\to\mathbb R^3$ an embedding. Then $U\setminus f(I)$ is path-connected.


*Proof.* See [Hatcher](http://www.math.cornell.edu/~hatcher/AT/ATpage.html), Proposition 2B.1 on page 169. This proposition proves (among other things) that $S^3$ minus any embedded segment is path-connected. Extract the proof of this particular statement and repeat it word-by-word with $U$ in place of $S^3$. (The proof uses only the following facts about $U$: $U$ minus any closed sub-segment of $f(I)$ is open; $U$ minus any point of $\mathbb R^3$ is simply connected.) 


**Lemma 2.** Let $f:I\to\mathbb R^3$ be an embedding and $s:I\to\mathbb R^3$ a path with endpoints outside $f(I)$. Then for any $\varepsilon>0$ there exists a path $s\_\varepsilon:I\to\mathbb R^3$ which is $\varepsilon$-close to $s$ (in $C^0$), connects the same endpoints, and avoids $f(I)$.


*Proof.* Divide the domain of $s$ into intervals so small that the diameters of their images are less than $\varepsilon/10$. Let $p\_0,p\_1,\dots,p\_n\in\mathbb R^3$ be the images of the division points ($p\_0$ and $p\_n$ are the endpoints of $s$). For each $p\_i$, let $q\_i$ be a point outside $f(I)$ such that $|p\_i-q\_i|<\varepsilon/10$ (for $i=0$ and $i=n$, choose $q\_i=p\_i$). Such a point $q\_i$ exists since $f$ is injective and hence cannot cover a set with nonempty interior. Let $U\_i$ be the ball of radius $\varepsilon/3$ centered at $p\_i$. By Lemma 1, $U\_i\setminus f(I)$ is path-connected. Therefore we can connect $q\_i$ to $q\_{i+1}$ by a path contained in $U\_i$ and avoiding $f(I)$. The (suitably parametrized) product of these paths is the desired $s\_\varepsilon$, q.e.d.


Now return to the original problem. The set $f(\mathbb R)$ is a countable union of sets $J\_k$, $k\in\mathbb N$, where each $J\_k$ is an injective image of a segment. We want to prove that any two points $p,q\in\mathbb R^3$ can be connected by a path avoiding $\bigcup J\_k$.


Let $s\_1$ be a path from $p$ to $q$ avoiding $J\_1$ (such a path exists by Lemma 1). Let $\varepsilon\_1$ be the minimum distance from $s\_1$ to $J\_1$, divided by 10. By Lemma 2, there exists a path $s\_2$ which is $\varepsilon\_1$-close to $s\_1$ and avoids $J\_1$. Let $\varepsilon\_2$ be the minimum of $\varepsilon\_1$ and the minimum distance from $s\_1$ to $J\_1$, divided by 10. Then there is a path $s\_3$ which is $\varepsilon\_2$-close to $s\_2$ and avoids $J\_2$, and so on. The resulting sequence $s\_1,s\_2,\dots$ converges (since $C^0$ is complete) to some path $s$ which avoids every set $J\_k$.


*Remark*. The closure of $f(\mathbb R)$ can separate the space, for example, $f(\mathbb R)$ can be a dense subset of the torus.


