Can a half line separate the plane?

Question

$\mathbb R^2$ carries the Euclidean topology.

Question:
Does there exist a subset $A$ of $\mathbb R^2$, which is homeomorphic to the half open interval $[0, 1)$, and which separates the plane, i.e. $\mathbb R^2 \setminus A$ is not connected?

Notes

1. The most obvious $A$, which are homeomorphic to $[0, 1)$ do not separate the plane, for instance $[0, 1) \times \{0\}$, $[0, \infty) \times \{0\}$, subsets of the circle etc.
2. (removed, see below)
3. If $A$ is homeomorphic to the open interval $(0,1)$, instead, there are trivial examples, which separate the plane (e.g., the $x$-axis), or not (e.g., $(0,1) \times \{0\}$).
4. If $A$ is homeomorphic to the closed interval $[0,1]$, instead, it never separates the plane. This is the Jordan arc theorem, for instance see here for a short and elegant proof based on the Brouwer fixed-point-theorem. In fact, all proofs, which I am aware of, are non-trivial!
5. If the answer to the question is "no", I would therefore expect, that its proof is also non-trivial. Or, perhaps, it can be deduced from the Jordan arc theorem?
6. There are a lot of results in the literature dealing with compact sets separating or not separating the plane (for instance, the famous Jordan curve theorem). But I haven't found anything relevant to non-compact sets. Therefore, I would also appreciate references in this direction.
7. Is it of any relevance to assume that $A$ is bounded or unbounded?
8. Note that subsets separating the plane need not contain closed curves, for instance the Warsaw circle. But, of course, the Warsaw circle is not homeomorphic to $[0,1)$.

Update

1. I was wrong with my Note 2., where I stated that a dense $A$ homeomorphic to $[0, 1)$ exists. Therefore I delete 2. Of course, this does not influence the question itself.
2. the comments so far show that there are A such that $\mathbb R^2 \setminus A$ is not path connected. But all the examples so far don't seem to answer the question.

Answer 1

I think the following works. More generally, I think it shows (with a little extra work) that an embedding of $(0,1)$ can only split the plane into two components if the end points are at infinity (i.e. it extends to an embedding of $S^1$ in $S^2$ containing $\infty$).

Theorem. There is no embedding of $[0,1)$ in $\mathbb{R}^2$ that separates the plane into two (or more) connected components.

Let $\gamma:[0,1)\rightarrow \mathbb{R}^2$ be an embedding, and let $C$ be its image. We say that $C$ locally splits the plane into $n$ components at $c\in C$ if every neighbordhood of $c$ contains neighborhood $N$ of $c$ homeomorphic to $\mathbb{R}^2$ such that $N-C$ has $n$ components. The idea is that for each $t$, $C$ locally splits the plane into either $1$ or $2$ components near $\gamma(t)$, depending on whether $t=0$ or $t>0$. Connectivity of $C$ allows us to stitch up these neighborhoods, which shows that all local components of $\mathbb{R}^2-C$ near $C$ are part of the same global component $\Gamma$.

But the global components of $\mathbb{R}^2-C$ containing $\gamma(t)$ as a limit point are precisely those that show up (as a subset) as a local component near $\gamma(t)$. Thus, $C\subset\overline{\Gamma}$ and and $C$ is disjoint from the closures of all other components of $\mathbb{R}^2-C$. In fact, $C\subset\overline{\Gamma}$ implies $\overline{C}\subset\overline{\Gamma}$, so $\overline{C}-C\subset\Gamma$. However, each component of $\mathbb{R}^2-C$ must have boundary contained in $\overline{C}$, so it is not possible for there to be another component.

Moise Ch10 Theorem 13. Every triangulable set in $\mathbb{R}^2$ is tame.

Triangulable means homeomorphic to a Euclidean simplicial complex, and tame means ambiently so. For us, this means there is an ambient homoemorphism of $\mathbb{R}^2$ sending $C$ to an infinite polygonal chain $P$ with exactly one endpoint. Since $C$ is an embedding, so is $P$, so WLOG, we can assume $\gamma$ is piece-wise linear.

Lemma 1. $C$ locally splits the plane into $1$ or $2$ components near $\gamma(t)$, depending on whether $t=0$ or $t>0$, respectively.

By Moise, we can assume $C$ is a linear complex. Let $t\in [0,1)$, and let $K$ be the union of the edges containing $\gamma(t)$. Define $\varepsilon_t$ to be the minimum distance from $\gamma(t)$ to a vertex in $K-\{\gamma(t)\}$. Then, for $\varepsilon\leq\varepsilon_t$, $N_\varepsilon\cap K$ is connected and the image of an open interval $I$ (w.r.t the topology of $[0,1)$) under $\gamma$. If there are arbitrarily small $\varepsilon\leq\varepsilon_t$ for which $N_\varepsilon\cap C$ is disconnected, then there is a sequence $(t_n)$ with $t_n\not\in \gamma^{-1}(K)\supset I$ such that $\lim_{n\rightarrow\infty}\gamma(t_n)=\gamma(t)$, which contradicts the fact that $\gamma$ is an embedding. Thus, $N_\varepsilon\cap C$ is connected for $\varepsilon$ sufficiently small.

Now within $N_\varepsilon$, $K$ is simply the union of two radii when $t>0$, and one radius when $t=0$. Thus, $N_\varepsilon-C=N_\varepsilon-K$ has one or two components, depending on whether $t=0$ or $t>0$, respectively. $\square$

Lemma 2. There is a single connected component $\Gamma\subseteq\mathbb{R}^2-C$ such that every sufficiently small neighborhood of $c\in C$ has intersection with $\mathbb{R}^2-C$ contained in $\Gamma$.

A choice function $f$ taking $t$ to a $N_\varepsilon$ as in lemma 1 for each $t$ results in a covering of $C$ by $\{f(t)\}_{t\in[0,1)}$ and a covering for $[0,1)$ by $\{\gamma^{-1}(f(t))\}_{t\in[0,1)}$. Each point of $f(t)\cap C$ is a limit point of each of the components of $f(t)-C$. Thus, each component of $f(t)-C$ must intersect a component of $f(t')-C$ for $t'\in \gamma^{-1}(f(t))$. So on any interval $I$ (i.e. a connected set), the number of connected components of $\bigcup_{t\in I}f(t)-C$ cannot exceed the number of connected components of $f(t)-C$ for any $t\in I$. In particular, $\bigcup_{t\in[0,1)}f(t)-C$ is connected. $\square$

Edit. I realized, as pointed out in the comments, that I had assumed every $c\in C$ necessarily had an $\varepsilon$-neighborhood with connected intersection with $C$. This does not seem to be trivial, or necessarily true. I have remedied this with Moise's theorem on the tameness of triangulable manifold embeddings in $\mathbb{R}^2$, which made using Hatcher 2B.1 unnecessary.

Answer 2

$\mathbb C \setminus A$ is connected. The proof is non-trivial and uses Alexander duality plus two theorems of complex analysis.

By $E$ we denote the extended complex plane $\mathbb C \cup \{\infty\}$ which is homeomorphic to $S^2$.

Let us first prove that $E \setminus A$ is connected.

The closure $\overline A^E$ in $E$ is a compactification of $A$ and the remainder $X = \overline A^E \setminus A$ is a nonempty compact connected subset of $E$ (see e.g. my answer to Non-equivalent compactifications of $[0, \infty)$). The complement $E \setminus X$ is an open subset of $E$, thus locally connected. Therefore all components $U_\alpha$ of $E \setminus X$ are open in $E$. Clearly they are not closed because they are contained in $E \setminus X \subsetneqq E$.

We claim that the boundaries $\operatorname{bd}^E U_\alpha$ in $E$ (which are nonempty!) are contained in $X$. Since the $U_\alpha$ are disjoint, we see that $\overline U_\alpha^E \cap U_\beta = \emptyset$ for $\beta \ne \alpha$. Thus $(\operatorname{bd}^E U_\alpha) \cap U_\beta = \emptyset$ for all $\beta$ (for $\beta = \alpha$ this is trivial). This shows $(\operatorname{bd}^E U_\alpha)\cap (E \setminus X ) = \emptyset$ which proves the claim.

Since the $\overline U_\alpha^E$ are connected, we conclude that the $X_\alpha = X \cup \overline U_\alpha^E = X \cup U_\alpha$ are connected.

One of the $U_\alpha$ must contain $A$, say $U_{\alpha_0}$. The set $Y = E \setminus U_{\alpha_0}$ is closed and has the form $Y = X \cup \bigcup_{\beta \ne \alpha}U_\alpha = \bigcup_{\beta \ne \alpha} X_\alpha$. It is connected because all $X_\alpha$ contain $X$.

It is well-known theorem that a connected open subset of $\mathbb C$ is simply connected if and only if its complement in $E$ is connected. This was the purpose of working with $E$. Therefore $U_{\alpha_0}$ is simply connected (actually all $U_\alpha$ are simply connected). Another well-known theorem shows then that $U_{\alpha_0}$ is homeomorphic to an open disk. We conclude that the quotient $E/Y$ is a one-point compactification $U_{\alpha_0} \cup\{*\}$ of $U_{\alpha_0}$, thus homeomorphic to $S^2$. The set $A' = A \cup \{*\}$ is an arc in $E/Y$, thus the Alexander duality theorem shows that $(E/Y) \setminus A' = U_{\alpha_0} \setminus A$ is (pathwise) connected.

We claim that $X \subset \overline{U_{\alpha_0} \setminus A}^{E \setminus A}$. Let $V'$ be an open neighborhood of $x \in X$ in $E \setminus A$. Then $V' = V \cap (E \setminus A)$ with an open neighborhood $V$ of $x$ in $E$. Since $x$ is a cluster point of $A$, $V$ contains some $a \in A$ so that $V \cap U_{\alpha_0}$ is an open neighborhood of $a$ in $U_{\alpha_0}$. Since $A$ does not have interior points, $V \cap U_{\alpha_0}$ must contain a point of $U_{\alpha_0} \setminus A$. Hence $V' \cap (U_{\alpha_0} \setminus A) = V \cap (U_{\alpha_0} \setminus A) \ne \emptyset$.

Since $\overline{U_{\alpha_0} \setminus A}^{E \setminus A}$ is connected, we see that $\overline{U_{\alpha_0} \setminus A}^{E \setminus A} \cup Y = E \setminus A$ is connected. $\phantom{xxxx} \square$

Let us now come to $\mathbb C \setminus A$.

We shall use the same sets $U_\alpha$ as above and what we know about them.

Case 1. $\infty \notin X$ (i.e. $A$ is bounded).

Then $\infty \in U_{\alpha_0} \setminus A$ or $\infty \in U_{\alpha}$ for some $\alpha \ne \alpha_0$. Let $U'_\alpha = U_\alpha \cap \mathbb C$ for all $\alpha$.

$U_{\alpha_0} \setminus A$ and $U_{\alpha}$ are connected open subsets of $E$, thus pathwise connected. Therefore $U'_{\alpha_0} \setminus A$ and the $U'_{\alpha}$ are pathwise connected open subsets of $\mathbb C$ because they arise from $U_{\alpha_0} \setminus A$ and $U_{\alpha}$ by removing at most one point which preserves pathwise connectedness.

We have $\operatorname{bd}^{\mathbb C} U'_\alpha = \operatorname{bd}^E U_\alpha$ and $\overline{U'_{\alpha_0} \setminus A}^{\mathbb C \setminus A} = \overline{U_{\alpha_0} \setminus A}^{E \setminus A}$. Repeating the above arguments shows that $Y' = X \cup \bigcup_{\beta \ne \alpha}U'_\alpha$ is connected and hence also $\overline{U'_{\alpha_0} \setminus A}^{\mathbb C \setminus A} \cup Y' = \mathbb C \setminus A$ is connected.

Case 2. $\infty \in X$.

2.1. $X = \{\infty\}$.

Then $\overline A^E = A \cup \{\infty\}$ is an arc in $E$ and as above (by Alexander duality) we see that $E \setminus \overline A^E = \mathbb C \setminus A$ is connected.

2.2. $X \setminus \{\infty\} \ne \emptyset$.

Let $X' = X \setminus \{\infty\}$. Then all $U_\alpha \subset \mathbb C \setminus X' \subsetneqq \mathbb C$. The same arguments as above show that the (non-empty!) boundaries $\operatorname{bd}^{\mathbb C} U_\alpha \subset X'$ and that $X' \subset \overline{U_{\alpha_0} \setminus A}^{\mathbb C \setminus A}$. Hence all $Z_\alpha = \overline{U_{\alpha_0} \setminus A}^{\mathbb C \setminus A} \cup \overline U_\alpha^{\mathbb C}$ with $\alpha \ne \alpha_0$ are connected. Thus the union $\bigcup_{\alpha \ne \alpha_0} Z_\alpha = \mathbb C \setminus A$ is connected.