Exercise 9.9

Fix $\check{x}\in E$. Since $E$ is open, there is a open ball $S\subset E$ containing $\check{x}$. By the Corollary to Theorem 9.19, $\check{f}$ is constant on $S$. Hence the set $E^{\prime }$ of all points $ž\in E$ such that $\check{f}(ž)=\check{f}(\check{x})$ is an open subset of $E$. Similarly, the set $E-E^{\prime }$ is also open in $E$ (being the union of open sets on which $\check{f}$ has a constant value not equal to $\check{f}(\check{x})$). By Exercise 2.19(b), $E^{\prime }$ and $E-E^{\prime }$ are two separated sets whose union is $E$. Since $E$ is connected, we must have $E^{\prime }=E$, so $\check{f}$ is constant on $E$.