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Let $A \subseteq \mathbb{R}^2$ be a subset of the plane which is homeomorphic to $\mathbb{R}$. How many connected components does $\mathbb{R}^2 \setminus A$ have?

My conjecture is that only one or two components are possible. Is it true? How can I prove this?

(To see these options are in fact possible, consider $A=(0,1) \times \{ 0\}$ and $A=\mathbb{R} \times \{0\}$).

Update: It says here* that the Jordan Brouwer separation theorem implies that a closed subset of $\mathbb{R}^n$ homeomorphic to $\mathbb{R}^{n-1}$ disconnects the space into two connected components.

So, we need only to consider the case where $A$ is not closed in $\mathbb{R}^2$.

*Here is an explanation, following Stefan Hamcke's comment:

Consider the homeomorphism $h:\mathbb{R}^{n-1} \to A$. We can view $h$ as a map $\mathbb{R}^{n-1} \to \mathbb{R}^{n}$. Since $A$ is closed (by assumption) $h$ is a closed map. A closed map having the property that the pre-image of every point is compact is proper (Proof). The requirement on pre-images is satisfied here trivially, since $h$ is injective.

As a proper map, $h$ extends to the one-point compactifications, so we can view it as an embedding $h:\mathbb{S}^{n-1} \to \mathbb{S}^{n}$. Thus, theorem 2B.1 (Hatcher, Algebraic Topology) implies $\mathbb{S}^{n} \setminus h(\mathbb{S}^{n-1})$ has two connected components, which in turn implies $\mathbb{R}^n \setminus A$ have two components.

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Asaf Shachar
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  • This is just an idea, but if you got these two maps $f$ and $f^{-1}$ defining the homeomorphism, one of them ought to extend to the plane. Then you know that continuous functions preserve connected components, which should give you an upper-bound. – T.J. Gaffney Apr 26 '16 at 15:36
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    If $A$ is closed, then the map $\Bbb R^{n-1}\to\Bbb R^n$ is proper and thus extends to a map $h$ between their one-point-compactifications, so $h$ can be seen as an embedding of $S^{n-1}$ into $S^n$. Then proposition 2B.1 in Hatcher's Algebraic Topology implies that there are two path components in $S^n\setminus h(S^{n-1})$, and these induce the two components of $\Bbb R^n\setminus A$. – Stefan Hamcke Apr 26 '16 at 17:09
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    By the way, properness of $h$ follows more easily by considering a compact set $K$, then $K\cap A$ is closed in $K$ and thus compact. Now since $h$ is a homeomorphism to $A$, $h^{-1}(K) = h^{-1}(K\cap A)$ is compact as well. – Stefan Hamcke Apr 26 '16 at 17:51
  • Interestingly it is possible for a one to one continuous function from R to the plane to disconnect the plane into any number of components. The trick is that there are sequences of loops that "pinch off" regions by converging to the curve in the limit. Therefore the function does not have a continuous inverse... – btilly Apr 26 '16 at 18:17

1 Answers1

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It is proven in

S. Eilenberg, An invariance theorem for subsets of $S^n$. Bull. Amer. Math. Soc. 47, (1941). 73–75.

that if $A, B\subset S^n$ are homeomorphic subsets of $S^n$ (which need not be closed) then the number of connected components of $S^n-A$ and $S^n -B$ is the same.

It follows from this theorem, at least, that a bounded subset of $R^2$ homeomorphic to $R$ cannot separate $R^2$. If $R$ is properly embedded in $R^2$ then it separates $R$ in $2$ components (since it is closed). The remaining case, which I do not quite see how to deal with is when $A\subset R^2$ is homeomorphic to $R$, is unbounded but is not closed. I will come back to this when I have more time.

There are relevant papers by Sitnikov from 1950s, where he proves duality theorems for general subsets of $S^n$, but I do not have access to them.

Edit. Everything indeed works, one obtains, in fact a more general result:

Theorem. If $B\subset R^n$ is homeomorphic to $R^{n-1}$, $n\ge 2$, then either $B$ is closed and, hence separates $R^n$ in two components or $B$ does not separate.

To prove this one considers the subset $B'=B\cup \{\infty\}$ in $S^n= R^n \cup \{\infty\}$. The separation properties of $B$ and $B'$ (in $R^n$ and $S^n$ resp.) are, of course, equivalent. There are two cases to consider: $B'$ is compact. Then everything follows from Jordan separation theorem in $S^n$. Assume, therefore, that $B'$ is not compact. One proves the following

Lemma. $H_{n-1}^c(B', {\mathbb Q})=0$, where $H_k^c$ denotes the Chech homology with compact support, i.e. $$ \lim_K H_k(K, {\mathbb Q}) =0, $$ where the direct limit is taken over all compacts $K\subset B'$ and $H_k$ denotes Chech homology (with rational coefficients).

Given this lemma, one then uses Theorem 8 in

S. Kaplan, Homology properties of arbitrary subsets of Euclidean spaces. Trans. Amer. Math. Soc. 62, (1947) 248–271.

which is a more elaborate version of Eilenberg's paper. Here $B'$ as above is $S^n - A$ in Kaplan's notation. When you unravel what this theorem 8 says in our setting, it shows that if $B'$ is not connected, then $H^c_{n-1}(B')\ne 0$, which contradicts the Lemma.

Moishe Kohan
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