Show that, for any $0<r<R$, there exists a function $\phi\colon\mathbb R^n\to[0,1]$ such that $$\phi(x):= \begin{cases} 1 & \text{when }x\in B(x,r), \\ 0 & \text{when }x\in B(x,R)^c. \end{cases}$$ where recall that $B(x,r)=\{y\in\mathbb R^n; |y-x|<r\}$.
My Attempt: I found this problem so trivial. We can always find a function $\phi(x) = \begin{cases} 1 , x \in B(x,r) \\ 0 , x \in B(x,R)^c \\ \frac{1}{2} \text{otherwise }\end{cases}$
Is there any twist ? Can anyone please tell me ?
The question is at stands makes no sense at all, there is a weird double use of $x$ here (it must be the centre of some fixed balls $B(x,r), B(x,R)$ but it's also used as a variable. Moreover, of course this function must be at least continuous (maybe even smooth, if we're in differentiable topology) or there is indeed no work involved at all. So find $\phi: \Bbb R^n \to [0,1]$ continuous so that
$$\phi(y)=\begin{cases} 0 & y \in B(x,r)\\ 1 & y \in B(x,R)^\complement \\ \end{cases}$$
It's geometrically quite obvious what has to be done here: we only have to define $\phi(y)$ for $y$ with $r \le \|x-y\| < R$, as the rest is prescribed..
