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Suppose $Gsubset mathbb R$ 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.

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

Proof:Suppose $X$ is connected and $f:Xto {0,1}$ is continuous. If $f$ is not a constant function, then $f$ is onto. Let $A=f^{-1}(0)$ and $B=f^{-1}(1)$. Then $Acup B=X$,and $A,Bneq emptyset$. 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 $f:Xto {0,1}$ such that $$f(x)=begin{cases}
0, &text{if $xin A$}\
1,&text{if $xin B$}
end{cases}$$

$f$ 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.

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).