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**Sheld** In my book they go through this proof of the following:

An open set $\displaystyle G \subset \mathbb{C}$ is connected iff for any two points $\displaystyle a,b$ in $\displaystyle G, \; \exists$ a polygon from $\displaystyle a$ to $\displaystyle b$ lying entirely in $\displaystyle G$.

We suppose $\displaystyle G$ satisfies this condition and $\displaystyle G$ is not connected to obtain a contradiction.

Pf:

Know $\displaystyle G = A \cup B$ where $\displaystyle A,B$ are both open and closed, $\displaystyle A \cap B = \emptyset$ and $\displaystyle A,B$ are nonempty.

If we let $\displaystyle a \in A$ and $\displaystyle b \in B $

$\displaystyle \exists$ a polygon from $\displaystyle a$ to $\displaystyle b$ such that $\displaystyle P \subset G$

We can assume $\displaystyle P = [a,b]$

Define:

$\displaystyle S = {s \in [0,1] : sb + (1-s)a \in A }$

$\displaystyle T = {t \in [0,1] : tb +(1-t)a \in B}$

$\displaystyle S \cap T = \emptyset, S \cup T = [0,1] , 0 \in S, 1 \in T$

It can be shown that $\displaystyle S$ and $\displaystyle T $are both open, contradicting the connectedness of $\displaystyle [0,1]$.

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Now, I do not see how S and T are both open. Can anyone give me a hint please? Thank you.

-Sheld