This is exercise 5 in Paragraph 1.1.3 from "Complex variables (harmonic and analytic functions)" by Francis J.Flanigan.
Let $\Omega$ be open (in $\Bbb R ^2$). Prove that $\Omega$ is connected if and only if any two points in $\Omega$ may be linked by a path consisting of a finite number of straight-line segments lying entirely in $\Omega$
- Hint. Given $\Omega$ connected and $z_0\in\Omega$, prove that the subset $S$ consisting of all points in $\Omega$ which may be linked to $z_0$ by the specified type of path is not empty (clear!), open and also closed. By exercise 4, $S=\Omega$ , whence any two points of $\Omega$ may be linked
My attempt:
Since $\Omega$ is open, for every $z\in\Omega$ there exists $r\gt0$ such that $B(z;r)\subset\Omega$. Now let $$r_z=sup\{\;r\;; B(z;r)\subset\Omega\;\}$$ Every point $x\in B(z;r_z)$ can be linked to $z$ so if for every two points $z_0,z_n\in\Omega$, a (finite) sequence $z_1,z_2,...,z_{n-1}\in\Omega$ can be found such that $z_{i+1}\in B(z_i;r_{z_i})$ for every $0\le i \le n-1$ then $\Omega$ must be connected. If it weren't and you could find $\Omega_1$ and $\Omega_2$ both open, nonepmty and disjoint and setting $z_0\in\Omega_1$ and $z_n\in\Omega_2$, for every $z_i$, $B(z_i)$ is either in $\Omega_1$ or $\Omega_2$ so all the $z_i$ are in $\Omega_1$ and there is no way to link any of them to $z_n$
My questions are:
- Is the proof correct?
- If it is how can I simplify it?
- How can I use the hint?
- How can it be proven that if $\Omega$ is connected, then there must exist a finite sequence?
Following the hint, Let $\Omega$ be a non-empty open set, $z\in \Omega$. the set of points in $\Omega$ connected to $z$ by a finite number of line segements denoted $\mathcal{O}_z$ is open, since $\Omega$ is open and basic open balls in $\mathbb{R}^2$ are convex (said another way any two points are connected by a single line segments contained in the basic open ball). It is closed since it's complement is open, by a the same reason $\Omega$ is open; the basic open balls in $\mathbb{R}^2$ are convex. Notice $\Omega$ is connected if and only if $\mathcal{O}_z=\Omega$.