Toronto Math Forum
MAT3342020F => MAT334Lectures & Home Assignments => Chapter 1 => Topic started by: Kuba Wernerowski on October 02, 2020, 01:35:40 PM

My proof follows roughly the same logic as the outline in the textbook, but I'm not sure if it's quite rigorous enough.
Choose $p_a, p_b \in D$. Since $D$ is a domain, $\exists$ a polygonal curve $P_1 P_2 \cup P_2 P_3 \cup \cdots P_{n1}P_n$ connecting $p_a$ and $p_b$.
Since $D$ is open, each point line segment $P_1, \ldots P_n$ have an open disc $A_j$ centered at $A_j$ where $j = 1, 2, \ldots, n$.
Construct the polygonal curve such that for each pair of endpoints, $P_j, P_{j+1}$, their respective open discs $A_j, A_{j+1}$ have the property that $A_j \cap A_{j+1} ≠ \emptyset$ for $j=1, 2,\ldots, n1$.
Then, since $u$ has the property that for each of those open discs, $u(A_j)$ = some constant, $c_j$.
$u(A_1) = c_1$, $u(A_2) = c_2$, and given that $A_1 \cap A_2 \neq \emptyset$, then $c_1 = c_2$.
Repeat this argument for each pair of $A_j$ and $A_{j+1}$ until we reach $A_{n1} \cap A_n \neq \emptyset \implies u(A_{n1}) = u(A_n)$.
To conclude the proof, $p_a \in A_1$ and $p_b \in A_n$ meaning that $u(p_a) = u(p_b)$, and since $p_a, p_b$ were chosen arbitrarily, $u$ is constant on $D.$.
Any feedback / criticisms are much appreciated ;D

You should remember that each segment of the polygonal curve may be served not by two discs, but several discs (think why)

For any polygonal segment $P_i P_j$ where that's the case, could we divide that segment into smaller subsegments $P_{i_1} P_{i_2} \cdots P_{i_{n1}}P_j$ and put open, overlapping discs over each subsegments end/start points?

Indeed