Let $X\subset \mathbb{R^2}$ be bounded and convex.Show that $\mathbb{R^2} \setminus X$ is path connected

Question

Let $X\subset \mathbb{R^2}$ be bounded and convex.Show that $\mathbb{R^2} \setminus X$ is path connected.

I understand the proposition, I was trying proof by contradiction,but I don´t get inspired, My other idea is let $x,y\in \mathbb{R^2} \setminus X$ and notice that since $X$ is bounded exits $r \in \mathbb{R}$ such that $ X \subset B_{r}(0)$ and from here begin to construct a path from $x$ to $y$ without pass for $X$. but I don´t know how use that $X$ is convex.

Any sugestion, idea, intuition was very helpful.And in case of be possible, may possibly give advice for be more crative in the problems that requiere a construction. Thanks in advice

Answer 1

Let $P$ and $Q$ be two points in the complement of $X$ and let $O\in X$. Join $O$ to $P$ and $Q$. Observe that on the line $OP$, on the side of $P$ opposite to that where $O$ belongs, there are no other points of $X$ (otherwise convexity would be violated at $P$). For the same reason, on the side of $OQ$ opposite to that where $O$ belongs there are no other points of $X$. By boundedness you can now pick a circle centered at $O$ with radius much larger than $|OP|$ and $|OQ|$ such that $X$ is entirely contained in a disk of half the radius centered at $O$. Let this circle intersect $OP$ at $P_1$ and $OQ$ at $Q_1$. Then $PP_1Q_1Q$ is a path connecting $P$ and $Q$, entirely contained in the complement of $X$.

Answer 2

Since $X$ is bounded, there exists $r > 0$ such that $\lVert x \rVert < r$ for all $x \in X$. Choose a point $\xi \in X$. Then $\lVert x - \xi \rVert < r + \lVert \xi \rVert = R$ for all $x \in X$.

The circle $C_R(\xi) = \{x \in \mathbb R^2 \mid \lVert x - \xi \rVert = R \}$ with center $\xi$ and radius $R$ is connected since it is the image of $\phi : [0,2\pi] \to \mathbb R^2, \phi(t) = \xi + (R \cos(t), R \sin(t))$. Obviously $C_R(\xi) \cap X = \emptyset$.

For $y \in \mathbb R^2 \setminus X$ and $t \in \mathbb R$ define $p(t) = \xi + t(y - \xi)$. Let $R_y = \{ p(t) \mid t \ge 0 \}$ denote the ray beginning at $\xi$ and going through $y$. It intersects $C_R(\xi)$ in the point $p(R/\lVert y - \xi \lVert)$. The set $R_y \cap X$ is a bounded convex subset of $R_y$, hence an interval containing $\xi$ as a boundary point. Therefore also the complement $A_y = R_y \setminus X \subset T_y$ is an (unbounded) interval, in particular it is connected. By construction it contains $y$. Hence $$\mathbb R^2 \setminus X = C_R(\xi) \cup \bigcup_{y \in \mathbb R^2 \setminus X} A_y $$ is connected.