Connected Domain is Connected by Staircase Contours

Theorem
Let $D \subseteq \C$ be an open set.

Then $D$ is a connected domain iff for all $z, w \in \C$, there exists a staircase contour in $D$ with start point $z$ and endpoint $w$.

Necessary Condition
Suppose $D$ is a connected domain.

If $z, w \in D$, there exists a path $\gamma: \left[{0 \,.\,.\, 1}\right] \to D$ with $\gamma \left({0}\right) = z$ and $\gamma \left({0}\right) = w$.

From the Paving Lemma, it follows that there exist $\epsilon \in \R_{>0}$ and a subdivision $\left\{ {x_0, x_1, \ldots, x_n }\right\}$ of $\left[{0 \,.\,.\, 1}\right]$ such that:


 * $\displaystyle \bigcup_{k \mathop = 0}^n B_\epsilon \left({\gamma \left({x_k}\right) }\right) \subseteq S$

and for all $k \in \left\{ {0, 1, \ldots, n - 1}\right\}$:


 * $\gamma \left({\left[{x_k \,.\,.\, x_{k + 1} }\right] }\right) \subseteq B_\epsilon \left({\gamma \left({x_k}\right) }\right)$

Let $\operatorname{Re} \left({x_{k + 1} - x_k}\right)$ denote the real part of $x_{k + 1} - x_k$, and $\operatorname{Im} \left({x_{k + 1} - x_k}\right) $ denote the imaginary part of $x_{k + 1} - x_k$.

For all $k \in \left\{ {0, 1, \ldots, n - 1}\right\}$, define the smooth paths $\gamma_{2k+1}, \gamma_{2k+2}: \gamma: \left[{0 \,.\,.\, 1}\right] \to D$ by:


 * $\gamma_{2k+1} \left({t}\right) = x_k + t \left({\operatorname{Re} \left({x_{k + 1} - x_k}\right) }\right)$
 * $\gamma_{2k+2} \left({t}\right) = x_k + \operatorname{Re} \left({x_{k + 1} - x_k}\right) + it \left({\operatorname{Im} \left({x_{k + 1} - x_k}\right) }\right)$

These are the line segments connecting $x_k$, $x_k + \operatorname{Re} \left({x_{k + 1} - x_k}\right)$, and $x_k+1$.

It follows that $x_k + \operatorname{Re} \left({x_{k + 1} - x_k}\right) \in B_\epsilon \left({\gamma \left({x_k}\right) }\right)$, as:

Then Open Ball is Convex Set shows that $\gamma_{2k+1} \left({t}\right), \gamma_{2k+2} \left({t}\right) \in B_\epsilon \left({\gamma \left({x_k}\right) }\right) \subseteq D$ for all $t \in \left[{0 \,.\,.\, 1}\right]$.

It follows that the images of $\gamma_{2k+1}$ and $\gamma_{2k+2}$ are subsets of $D$.

Define $C_k$ as the directed smooth curve that is parameterized by $\gamma_k$, so $C_{2k-1}$ has start point $x_{k - 1}$, and $C_{2k}$ has endpoint $x_k$.

Define $C$ as the concatenation of $C_1, \ldots, C_{2n}$.

Then $C$ is a staircase contour in $D$ with start point $z$ and endpoint $w$.


 * [[File:ConnectedDomainStaircase.png]]

Illustration of the open balls inside the connected domain $D$.

The path $\gamma$ between $w$ and $z$ is grey, and the constructed staircase contour $C$ is blue.

Sufficient Condition
Suppose that for all $z, w \in \C$, there exists a staircase contour in $D$ with start point $z$ and endpoint $w$.

Let $\gamma: \left[{a\,.\,.\,b}\right] \to D$ be a parameterization of $C$, where $\left[{a\,.\,.\,b}\right]$ is a closed interval.

Then $\gamma$ is a path in $D$ with $\gamma \left({a}\right) = z$ and $\gamma \left({b}\right) = w$.

Define $\gamma_0: \left[{0\,.\,.\,1}\right] \to D$ by $\gamma_0 \left({t}\right) = \gamma \left({a + t\left({b-a}\right) }\right)$.

Then $\gamma_0$ is also a path in $D$ with $\gamma_0 \left({0}\right) = z$ and $\gamma_0 \left({1}\right) = w$.

By definition of path-connected, it follows that $D$ is path-connected.

Hence, $D$ is a connected domain.