The complement of a simple non closed curve is path connected?

Question

Let $f: [0,1]\rightarrow \mathbb{R}^2$ be a continuous and injective function and $C=f([0,1])$. Is it true that $\mathbb{R}^2\setminus C$ is path connected?

It's pretty obvious that the answer should be positive, but it would be nice if there existed a not quite involved proof of it.

My attempt so far: For every $x\in C$ there exists an $\varepsilon_x>0$ such that $B(x,\varepsilon_x)\cap C$ is connected. (Here by $B(x, \varepsilon)$ we denote the open ball centered at $x$ with radius $\varepsilon$). This is pretty straightforward because $[0,1]$ and $f([0,1])$ are homeomorphic.

The collection $ \{ B(x, \varepsilon_x): x\in C\}$ is an open cover of the compact set $C$, so there exist $x_1, \ldots, x_n$ such that $C\subseteq \bigcup_{i=1}^n B(x_i, \varepsilon_{x_i})=A$. Lets also assume that the $x_i$'s are "consecutive" with $x_1$ and $x_n$ be the starting and ending points of the curve respectively. We notice in particular that both of the sets $B(x_1, \varepsilon_{x_1})\setminus C$ and $B(x_n, \varepsilon_{x_n})\setminus C$ are path connected. This is a nice property that we will need.

An easy (?) inductive argument shows that $A\setminus C$ is path connected. This is all due to $B(x_1, \varepsilon_1)\setminus C$ being path connected combined with the fact that the set $ B(x_2, \varepsilon_2) \cap C^c $ consists of two connected components, both of which intersect with $B(x_1, \varepsilon_1)$ and so on. From this it follows that $\mathbb{R}^2 \setminus C$ is path connected as well.

I think the proof is ok, but it uses a lot of stuff. Am I missing something obvious here? I also tried to go by contradiction. I supposed that $C$ cuts the plane into some path connected components and then I tried to close the curve avoiding self intersections. The plan was to create more than 2 path connected components and reach a contradiction by the Jordan Curve Theorem. This seemed quicker, but it didn't go through.

Answer 1

Consider the following theorem:

Theorem: Let $n$ be fixed. Suppose that $Y$ is a compact space with the property that $\widetilde{H}_*(S^n-f(Y))=0$ for every embedding $f:Y \to S^n$. Then $I \times Y$ also has this property. (*)

The proof of the above theorem is not difficult to follow (given Mayer-Vietoris and knowledge of the basic properties of homology). From it, it follows by induction that

Corollary: If $f:D^r \to S^n$ is an embedding then $\widetilde{H}_*(S^n-f(Y))=0.$ In particular, $S^n-f(D^r)$ is connected.

If you have $f:I \to \mathbb{R}^2$, consider $\widetilde{f}=f \circ i$, where $i: \mathbb{R}^2 \hookrightarrow S^2$ is the "inclusion by steoreographic projection". By the corollary, $S^2-\widetilde{f}(I)$ is connected. Since it is open, when you take out a point it will remain connected (**), and it follows that $\mathbb{R}^2-f(I)$ is connected.

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(*) - Reference: Bredon

(**) - One can see this in several ways. For instance, it is an open connected subset of a manifold, hence path-connected. Taking out a point of a path-connected manifold of dimension greater than $1$ can easily seen to result on a path-connected connected manifold. In particular, connected.

Another way is computing $H_0(X-\{p\})=H_0(X-\{p\}, D-\{p\})=H_0(X,D)=H_0(X),$ where the first equality is by the long exact sequence of homology, the second is excision and the last also by the long exact sequence of homology, and $D$ is a small disk around $p$ ($X$ is a $2$-dimensional manifold).

Answer 2

Hint for an easier proof: Choose $M > 0$ such that $-M <f(x) <M$ for $x\in [0,1].$ If $z\in U = \mathbb R^2\setminus Gr(f),$ then there is a vertical line segment in $U$ connecting $z$ to one of the lines $y=-M, y=M.$