The question is:
Let $(X,d)$ be a compact metric space. Let $f : X \rightarrow X$ be continuous. Fix a point $x_0 \in X$, and assume that $d(f(x), x_0)\geq1$ whenever $x\in X$ is such that $d(x,x_0)=1$. Prove that $U\setminus f(U)$ is an open set in $X$, where $U=\{ x\in X: d(x,x_0) <1\}$.
This is one of the analysis qual question. (I am going to sit for one in two days time) I am having problem visualize the set in the question, whenever I have this sort of feeling, I will have great difficulties answering the question. Any suggestion? Thanks.