Connected components of $\mathbb{R}^2 \setminus \gamma$, where $\gamma \subset \mathbb{R}^2$ is a closed path

Question

Let $\gamma$ be a closed path in $\mathbb{R}^2$ ($\gamma : [0,1] \to \mathbb{R}^2$ is continuous and satisfys $\gamma(0)=\gamma(1)$). Are the following statements true?

1. $\gamma$ divides the plane into (at most countably infinite, since every component would contain a rational point) connected components $Q_1, Q_2, Q_3,...$:

$$\mathbb{R}^2 \setminus \gamma = \bigcup_{i=1}^{\infty} Q_i$$

2. There is exactly one unbounded component (which we call) $Q_1$. $Q_i$ is bounded for every $i\geq 2$.

3. The boundary of the union of all bounded components is equal to $\gamma$: $$\partial \bigcup_{i=2}^{\infty} Q_i = \gamma$$ If 3 doesn't hold, do we at least know that the boundary is a subset of $\gamma$?

Essentially I am asking if a generalization of the Jordan curve theorem for not necessarily simply connected paths is true.

Answer 1

1. This is true, for the reason you specified (and since the components partition the space $\mathbb{R}^2 \setminus \gamma$)
2. This is true. The function $|\gamma| : [0,1] \to \mathbb{R}$ is a continuous function with a compact domain, so it has a maximum $R$. Draw a circle centered at $(0,0)$ with radius $R+1$. All the points outside this circle must lie in the same component.
3. This is false. For instance, if $\gamma([0,1]) = [0,1]^2$ is a spacefilling curve, then you'd have $\mathbb{R}^2 \setminus \gamma = \mathbb{R}^2 \setminus [0,1]^2$ is connected, so $\partial(\bigcup_{i=2}^\infty Q_i) = \partial(\emptyset) = \emptyset \neq [0,1]^2$. It will, however, be the case that $\partial\bigcup_{i=2}^\infty Q_i \subseteq \gamma$, which follows from the fact each $Q_i \subset \mathbb{R}^2$ is open.

Answer 2

For (2), since $\gamma$ is continuous and $[0,1]$ is a compact set, its image $\gamma([0,1])$ is a compact set hence entire image of $\gamma$ is contained in a ball of radius $r$ for some $r$. Hence only one unbounded component.