Possible typo in Brown and Churchill’s “Complex Variables and Applications” The following theorem is presented on pages 56 and 57 in the 6th edition book.
Theorem: If $f’(z)=0$ everywhere in a domain $D$, then $f(z)$ must be constant throughout $D.$
The proof sensibly derived that $u_x=u_y=v_x=v_y=0$ using the fact that $f’(z)=u_x+iv_x=0$ and the Cauchy-Riemann equations. Then, it was reasoned that $\vec{\nabla}u=\vec{0}\Rightarrow D_{\hat{u}}u(x,y)=0$ for any unit vector $\hat{u}$. I understood all of that.
The problem is, they then said exactly: “Consequently, $u$ is a constant along any line segment lying entirely in $D$; and, since there is always a finite number of such line segments, joined end to end, connecting any two points in $D$, the values of $u$ at those points must be the same.”
Question 1: Is the idea that you can keep connecting line segments inside of $D$ forever until you fill up all of $D$ and conclude that every point has the same value?
Question 2: If so, shouldn’t it say infinite instead of finite?
Any input is appreciated.
 A: The text is correct as written. You can indeed connect any two points in a domain (in fact, in any connected open subset of Euclidean space) using finitely many straight lines. The rough idea as to why this is true is that it is "clearly" true on open balls. I won't bother showing that rigorously, but I hope it's intuitively clear from geometric reasoning. In any case, the proof isn't too bad.
Anyway, let $U \subseteq \mathbb R^n$ be open, connected, and nonempty. Pick some $u \in U$ and let $W \subseteq U$ be the set of all points in $U$ that can be connected to $u$ via finitely many straight lines. From now on, I will call such a path a polygonal path. We will show that $W$ is both open and closed in $U$, so as $W$ is nonempty and $U$ is connected, $W = U$ proving that any two points in $U$ can be connected via a polygonal path.
Let's start with showing that $W$ is open. Let $w \in W$. Then there is some open ball $w \in B \subseteq U$, as $U$ is an open subset of Euclidean space. Now, as I remarked above, any two points in an open ball can be connected via a polygonal path. Hence, every single point in $B$ can be connected to $w$ via a polygonal path. As $w$ is connected to $u$ via a polygonal path by assumption (it is in $W$), then by concatenation, every point in $B$ can be connected to $u$ via a polygonal path. Now by definition of $W$, $B \subseteq W$ so $W$ is an open subset of $U$.
Now we show that $W$ is closed, i.e. that $U - W$ is open. The argument is essentially the same. If there was an $x \in U - W$ then take a ball with $x \in B \subseteq U$. If there was a $y \in B \cap W$ then $x$ would be connected to $y$ and $y$ would be connected to $u$ via polygonal paths. Hence, $x$ would be connected to $u$, but we assumed $x \notin W$. Contradiction. Hence, $U - W$ is open so $W$ is open, closed, and nonempty. Thus, as $U$ is connected, $W = U$ and we are done.
Note that the above argument could be slightly refined using equivalence classes, but I felt that this was good enough and a bit more elementary, in case you are unfamiliar with these. Regardless, here's the idea. Let $a \sim b$ in $U$ if $a$ and $b$ can be connected via a polygonal path. Then the above argument shows that the equivalence classes of $U$ under $\sim$ are open. Since equivalence classes are disjoint, they form a partition of $U$ into open subsets. But as $U$ is connected, there can be only one equivalence class, so every pair of points in polygonally path connected. This is essentially the exact same proof I gave above, but with the language of equivalence classes.
