I have a homework problem from Conway's Functions of One Complex Variable that I am stuck on. The problem statement is as follows:
Let $G$ be an open subset of $\mathbb{C}$ and let $P$ be a polygon in $G$ from $a$ to $b$. Use Theorems 5.15 and 5.17 to show that there is a polygon $Q \subset G$ from $a$ to $b$ which is composed of line segments which are parallel to either the real or imaginary axes.
Def. "Polygon." If $w$ and $z$ are in $\mathbb{C}$ then we denote the straight line segment from $z$ to $w$ by $[z,w] = \{ tw + (1-t)z : 0 \leq t \leq 1 \}$. A polygon from $a$ to $b$ is a set $P = \bigcup_{k=1}^n [z_k,w_k]$, where $z_1 = a$, $w_n=b$, and $w_k=z_{k+1}$ for $1 \leq k \leq n-1$.
Thm. 5.15. Suppose $f : X \to \Omega$ is continuous and $X$ is compact; then $f$ is uniformly continuous.
Thm. 5.17. If $A$ and $B$ are nonempty disjoint sets in $X$ with $B$ closed and $A$ compact, then $0 < d(A,B) = \inf \{ d(a,b) : a \in A, \; b \in B \}$.
To me, the most straightforward (if tedious) way of doing this problem is to use the openness of $G$ to find $\varepsilon$ neighborhoods which completely cover the polygon $P$, and then decompose the line segments inside each neighborhood into parts which are parallel to the real or imaginary axes. However, this is clearly not the strategy the author wants. I'm assuming, referring to thms 5.15 and 5.17, that the polygon would play the role of the compact set, perhaps $\mathbb{C} \backslash G$ would be the closed set... but I'm totally lost as to how uniform continuity could play into this proof at all. Any hints/suggestions are very much appreciated. Thanks!
You can trim $P$ so that it never overlaps itself (call this trimmed set $P'$ and define a simple function like $f: P' \to \mathbb{R}$ so that $f(x) = $ distance along $P'$ from $a$ to $x$. Then $f$ is continuous as long as $P'$ does not contain line segments that cross (what we filtered out at the beginning).
Since $P'$ is compact, $f$ is uniformly continuous.
Note that $G^c$ is closed (by definition). Then by 5.17, $d(G^c,P) > 0$.
Let $\epsilon = d(G^c,P) > 0$ and pick $\delta>0$ as in uniform continuity so that $d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon$.
Now you can use $\delta$ to partition $P'$ and take each segment of the partition and connect the points with vertical and horizontal lines, knowing that these new lines are never longer than $\epsilon$.
I am not sure this is any better than the direct proof you referred to, but I can't think of a different way to use uniform continuity in this context.