Book: Functions of Complex Variables Chapter: 2 Topology in $\mathbb{C}$ Section 2.2: Exercise 2: Show that the sets $S$ and $T$ from Theorem 2.3 are open.
Theorem 2.3: Let $G$ be an open subset of $\mathbb{C}$, then the set $G$ is connected if and only if for any two points $a,b$ in $G$ the polygon that connects a and b lies entirely inside G. (path connected).
What I have so far:
$\\ \\ \textbf{Proof:} \\ \\$
By definition,
$S = \{s \in [0,1]: sb + (1-s)a \in A\}$, $T = \{t \in [0,1]: tb + (1-t)a \in B\}$
We also have that A,B are open/closed subsets of G such that:
$S\cup T = [0,1], S\cap T = \emptyset, G = A\cup B, A\cap B = \emptyset$
Hence, one part of the segment $[a,b]$ lies in $A$, and the other part lies in $B$. Also we know that $0 \in S$, and $1 \in T$. Hence, we can write the sets $S$ and $T$ as
$S = [0,c), T = (c,1], \mbox{ and } S\cup T = [0,1], S \cap T = \emptyset.$
Now, $\forall s \in S$ and $\forall t \in T$, we have that $B^{[0,1]}(s,c - s) \subseteq S \mbox{ and } B^{[0,1]}(t,t - c) \subseteq T$ Hence, $S$ and $T$ are open.
Am I right?
Take $c:[0,1] \to \mathbb{C}$ given by $c(t)=tb+(1-t)a$. This is clearly continuous, and $S=c^{-1}(A)$, $T=c^{-1}(B)$. Since $A,B$ are open, $S$ and $T$ are also open.