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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.

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Hint: The thing to notice here is that all elements $x\in \partial U$ satisfy $d(x,x_0) = 1$. So by assumption $f(x)\notin U$ for all $x\in \partial U$. This then implies that $f(\overline U) \cap U = f(U) \cap U$.

In general, if you read the condition "$X$ compact, Hausdorff", a thing on your mind should be that all continuous maps are automatically closed.

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    @KWO $X$ compact and Hausdorff. A compact subspace of a non-Hausdorff space need not be closed.2012-08-12