Let $X$ be an open subset of $\mathbb{R}^n$ and for any subset $A\subseteq \mathbb{R}^n$, let $\chi_A$ be the characteristic function of $X$, i.e.: $$ \chi_X(x):= \begin{cases} 1 & : x \in A\\ 0 & : x \not\in A. \end{cases} $$
Can $\chi_X$ be written as $\chi_{(\frac1{2},\infty)}\circ f$ where $f:\mathbb{R}^n\to (0,1)$ continuously?
This puts the following conditions on $f$: if $x \in X$, then $\chi_X(x)=\chi_{(\frac1{2},\infty)}(f(x))=1$ which means that $f(x) > \frac12$, while $x \notin X$ implies $\chi_X(x)=\chi_{(\frac1{2},\infty)}(f(0))=0$, so $f(x) \le \frac12$. So $X=f^{-1}[(\frac12, \infty)]$ and so $X$ is open when $f$ is continuous. If $X$ is also closed, connectedness of $\Bbb R^n$ implies that this can only happen for $X=\emptyset$ or $X=\Bbb R^n$.