Prove that $G$ is an open interval for two-valued continuous function $f$

Question

Suppose $G\subset \mathbb R$ is a non-empty open set such that the function $f:G \rightarrow \{0,1\}$ is a two-valued function and is continuous. Show that any two-valued function on $G$ is a constant if and only if the set $G$ is an open interval.

I am concerning about the construction of open set for set $G$ and have no idea to start off with.

Answer 1

Theorem: A metric space $X$ is connected if and only if any continuous function $f:X\to \{0,1\}$ is constant.

Proof:Suppose $X$ is connected and $f:X\to \{0,1\}$ is continuous. If $f$ is not a constant function, then $f$ is onto. Let $A=f^{-1}(0)$ and $B=f^{-1}(1)$. Then $A\cup B=X$,and $A,B\neq \emptyset$. Also note that both are proper subsets of $X$ and are open and closed in $X$, a contradiction.

Suppose $X$ is not connected. Let $A$ and $B$ be the disconnection. Then define $f:X\to \{0,1\}$ such that $$f(x)=\begin{cases} 0, &\text{if $x\in A$}\\ 1,&\text{if $x\in B$} \end{cases}$$

$f$ is a non-constant continuous function(verify).

Another useful theorem is

A subset $I$ of $\Bbb{R}$ is connected if and only if $I$ is an interval.

You can find a proof here.

I hope now you can complete your answer on your own.

Answer 2

If $G$ is an open interval then IVP tells you that if $f$ if takes both the values $0$ and $1$ then it must take all values in between. This proves one way.

For the other way use the fact any open set is a disjoint union of open intervals. If $G$ is a union of two or mote intervals define $f$ to be $1$ on one of them and $0$ on all the others. You will get a continuous function taking both the values $0$ and $1$ (and no other value).