# Is the complement of an injective continuous map R→R2\mathbb{R}\to \mathbb{R}^2 with closed image necessarily disconnected?

I am interested in the following Jordan curve theorem-esque question:

Suppose that you are given a continuous, injective map $\gamma: \mathbb{R}\to \mathbb{R}^2$ such that the image is a closed subset of
$\mathbb{R}^2$. The complement of the image is an open subset of
$\mathbb{R}^2$ and therefore consists of some number of path connected
components. Must this number always be greater than 1?

There are several cases in which this question reduces to the Jordan curve theorem, and one in which it does not (this last case is where the subtlety seems to lie):

Case 1: If the limits $\lim_{t\to \pm \infty} |\gamma(t)| = \infty$ then we can extend $\gamma$ continuously to a continuous, injective map $S^1\to S^2$ via stereographic coordinates. The result then follows by the Jordan curve theorem.

Case 2: If either of the limits $\lim_{t\to \pm \infty} \gamma(t)$ exist, then since the image of $\gamma$ is closed, the limit must equal $\gamma(t_0)$ for some $t_0 \in \mathbb{R}$. Then $\gamma$ induces a continuous, injective mapping $S^1 \to \mathbb{R}^2$, and the result follows by the Jordan curve theorem.

Case 3: If neither of the limits $\lim_{t\to \pm \infty} \gamma(t)$ exist, and $\lim_{t\to \infty} |\gamma(t)| \neq \infty$ or $\lim_{t\to - \infty} |\gamma(t)| \neq \infty$, then it’s unclear what may happen, but I can’t imagine an example where the complement is connected.

For example, consider a figure-8 type curve such that the ends approach the middle of the figure 8 from both sides like a topologists sine curve pictured below (this picture is edited from a picture in Munkres Topology, 2ed, p381, or the examples discussed in this answer: Is there a continuous injective map from $\mathbb{R}$ that has compact image?). The complement of the image of this curve has three components.

Here is a proof, which assumes familiarity with Chech cohomology and Alexander duality. For the latter, see A.Dold, “Lectures on Algebraic Topology”, formula (8.18): If $X\subset {\mathbb R}^n$ is a closed subset, then

Suppose that $f: {\mathbb R}\to {\mathbb R}^n$ is an injective continuous map whose image is closed in ${\mathbb R}^n$. Then the restriction of the Euclidean metric from ${\mathbb R}^n$ to $Z:=f({\mathbb R})$ is complete; in particular, $Z$ (with the subspace topology) is a Baire space.

Define the subsets $A_\pm\subset Z$ consisting of all points $a\in Z$ for which there exists a sequence $t_n\to \pm\infty$ such that $\lim_{i\to\infty}f(t_i)=a$. Set $A= A_- \cup A_+$. Clearly, both $A_+, A_-$ are closed.
The subset $A^c:= Z- A$ consists of all points $x=f(t)$ such that for some (equivalently, every) $r>0$, $x$ is an interior point of $f([t-r, t+r])$ in $Z$.

Remark. The earlier version of my answer contained an error: I was assuming that $f([0,\infty))$ is closed which need not be the case. See also an edit at the end of this answer.

Lemma 1. $A\ne Z$.

Proof. If not then for every $i\in {\mathbb N}$, $f([-i,i])$ has empty interior in $X$. Then $Z$ is a union of countably many subsets with empty interior. But $Z$ is a Baire space. A contradiction. qed

I will assume from now on that $A\ne \emptyset$, otherwise $f$ is a proper map and $f({\mathbb R})$ is homeomorphic to
the real line. Without loss of generality (by precomposing $f$ with the mat $t\mapsto -t$), we can assume that $A_+\ne \emptyset$.

Lemma 2. If $A_+\ne \emptyset$ then there exists $x=f(t)\in A^c$ and $a=f(t_0)\in A_+$ such that $t > t_0$.

Proof. If $f^{-1}(A_+)$ is unbounded from below, then any $x\in A^c$ will work. If (the closed subset)
$f^{-1}(A_+)$ is bounded from below by some $t_0\in {\mathbb R}$ then the subset $f([t_0,\infty))$ is closed in ${\mathbb R}^n$. I will take $t_0$ to be the infimum of $f^{-1}(A_+)$. In particular, $f(t_0)\in A_+$.
Now, the same argument as in the proof of Lemma 1 shows that $f([t_0,\infty))$ is not contained in $A_+$. Hence,
there exists $t\in (t_0,\infty)$ such that $f(t_0)\notin A_+$. qed

After precomposing $f$ with a translation of ${\mathbb R}$, we may assume that $t_0=0$.

From now on, I will assume that $n=2$, that $A_+\ne \emptyset$, and, moreover, there exists a sequence $t_i\to \infty$ such that $\lim_{i\to\infty}f(t_i)=a=f(0)$. (See above.) Fix a point $b=f(t)\in A^c\subset X=f({\mathbb R}_+)$.

Proposition. $\check{H}^1_c(X)\ne 0$. (Here and in what follows, I use only (co)homology with integer coefficients.)

Proof. I will consider a sequence of locally finite open covers ${\mathcal U}_n$ of $X$ by subsets $U_{k,n}$ such that

and that for each $n$, ${\mathcal U}_{n+1}$ refines ${\mathcal U}_n$.

Let $N_n$ denote the nerve of ${\mathcal U}_n$ with the vertex $v_{k}$ corresponding to the open set $U_{k,n}$.

Since there exists an interval $I=(t+\delta, t-\delta)\subset {\mathbb R}_+$ such that $f(I)$ is open in $X$, without loss of generality we may assume that for each $n$, $f(t)$ belongs to the intersection $U_{1,n}\cap U_{2,n}$ of some
$U_{1,n}, U_{2,n}\in {\mathcal U}_n$ such that, moreover,

for all $k\notin \{1, 2\}$. I then define a 1-cocycle $c\in Z^1({\mathcal U}_n)$ by

and $c(U_{j,n}\cap U_{k,n})=0$ otherwise. (Emptyness of the triple intersections above implies that this is indeed a cocycle.) In terms of the nerve $N_n$, the cocycle $c$ is defined by $c([v_1, v_2])=1$ and $0$ for all other edges.

I claim that this cocycle is nontrivial (when $n$ is sufficiently large). I will prove its nontriviality by exhibiting a 1-cycle $\sigma\in Z_1(N_n)$ such that $\langle c, \sigma\rangle =1$. Namely, for each $n$ there exists $y_i=t_i$ such that $f(t_i), a$ belong to a common $U_{k,n}\in {\mathcal U}_n$. By concatenating $f([0, t_i])$ with the line segment $y_i a$ we obtain a (likely nonsimple) loop $\lambda_n$ in ${\mathbb R}^2$. This loop defines a 1-cycle $\sigma$ in $N_n$ as follows. Pick points $s_0=0\le s_1\le s_2\le ... \le s_{p}=t_i$ in $[0, t_i]$ such that for every $j$, $\{f(s_j), f(s_{j+1})\}\subset U_{q_j,n}\in {\mathcal U}_n$. We set $U_{q_p,n}=U_{q_0,n}$. We obtain a simplicial
loop $\sigma$ in $N_n$ with the vertices

Moreover, assuming that

is sufficiently small, we achieve that the edge $[v_{1,n}, v_{2,n}]$ appear exactly one in this loop $\sigma$. Hence
$\langle c, \sigma\rangle =1$.

By the construction, the natural map $\kappa_n: N_{n+1}\to N_n$ satisfies

Thus, the sequence of 1-cycles $(c_n)$ defines a nonzero element of

qed

We can now finish the proof:

Lemma 3. $X=f([0,\infty))$ separates ${\mathbb R}^2$.

Proof. By the Alexander duality,

Since $\check{H}^1_c(X)\ne 0$, so is $\tilde{H}_0({\mathbb R}^2 -X)$. qed

We now are ready to prove:

Theorem. Suppose that $f: {\mathbb R}\to {\mathbb R}^2$ is a continuous injective map with closed image. Then $Z:=f({\mathbb R})$ separates ${\mathbb R}^2$.

Proof. Define the subset $A=A_+ \cup A_-\subset Z$ as before. Then $A=\emptyset$ iff $f$ is a proper map. In this case, by the Alexander duality,

hence, $Z$ separates ${\mathbb R}^2$. If $A\ne \emptyset$, then, as noted earlier, without loss of generality, we may assume that $A_+\ne \emptyset$. Then Lemma 3 implies that ${\mathbb R}^2- (X=f([0,\infty)))$ is a disjoint union $U\sqcup V$ of two nonempty open subsets. By applying Baire category argument as in Lemma 1, we see that neither $U$ nor $V$ is contained in $Z$. Hence, ${\mathbb R}^2 - Z= (U - Z)\sqcup (V- Z)$ is a disjoint union of two nonempty open subsets and, hence, is not connected. qed

Edit. Although we do not need this, here is an interesting fact about the set $A_+$. Define the subset $J:= {\mathbb R} - f^{-1}(A_+)$. It is easy to construct examples where $J$ is bounded below (using a version of the topologist’s sine curve). However, we have:

Lemma 4.
$J\subset {\mathbb R}$ is unbounded above.

Proof. The subset $A^c_+:=f(J)\subset Z$ is open, hence, its complement $Z- A^c_+= A_+$ is closed. Assume for a moment that $J$ is bounded above. Then removing $J$ from ${\mathbb R}$ does not affect the accumulation set $A_+$. It follows that each $a\in Z - A^c_+$ equals the limit

Therefore, for every interval $[-n,n]$, the subset $f([-n,n] -J)$ has empty interior in $A_+$. Thus, we obtain a contradiction with the Baire property as in the proof of Lemma 1. qed