Connected Domain is Connected by Staircase Contours

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Theorem

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


Then $D$ is a connected domain if and only if:

for all $z, w \in \C$, there exists a staircase contour in $D$ with start point $z$ and end point $w$.


Proof

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:

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

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

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

where $B_\epsilon \left({\gamma \left({x_k}\right) }\right)$ is the open $\epsilon$-ball about $\gamma \left({x_k}\right)$.


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:

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

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:

\(\displaystyle \left\vert{\operatorname{Re} \left({x_{k + 1} - x_k}\right) }\right\vert\) \(\le\) \(\displaystyle \left\vert{x_{k + 1} - x_k}\right\vert\) Modulus Larger than Real Part
\(\displaystyle \) \(<\) \(\displaystyle \epsilon\)

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 $\gamma_k$.

Thus $C_{2 k - 1}$ has start point $x_{k - 1}$, and $C_{2 k}$ has end point $x_k$.

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

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

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 red.

$\Box$


Sufficient Condition

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

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.

$\blacksquare$


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