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This is probably very simple, but for some reason I don't seem to see why if $\forall x \in \mathbb R^n, \|x-\phi(x)\|>c$ for some $c>0$ then $\|x-y\|>\|\phi(x)-\phi(y)\|$, where $\phi$ is any function, cannot be true. The only thing I would think of to do (that unfortunately doesn't work) is writing $$\|x-y\| \leq \|x-\phi(x)\| + \|\phi(x)-\phi(y)\| + \|y-\phi(y)\| .$$ Any suggestions?

Added: I was thinking about exercise 17.15 from the 2nd edition of W.A. Sutherland's Introduction to Metric & Topological spaces. Where I am supposed to

show that $f \colon X \to X$ is a map of a compact metric space $X$ such that $d(f(x),f(y))

The hint is 

Show that the map $x \mapsto d(x, f(x))$ is continuous on compact $X$ so attains its inf, say $l$ , at some point $x_0 \in X$ . Now show that $l = 0$ by getting a contradiction from the given condition $d(f(x), f(y)) < d(x, y)$ if $l > 0$. Thus $x_0$ is a fixed point of $f $. Uniqueness follows as in the contraction map theorem.

I am thinking of $d$ as the Euclidean metric $\|\cdot\|$ and $X$ a compact subset of $R^n$.

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    @mixedmath: I have added the context of the question as requested.2011-12-10
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    @mixedmath: I think you misunderstood the question, which is very confusingly phrased. If I understand it correctly, a less confusing version might be: Let $\phi$ be any function. Why is it that if $\|x-\phi(x)\|>c$ for all $x$, then $\|x-y\|>\|\phi(x)-\phi(y)\|$ can't hold for all $x$ and $y$?2011-12-10
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    @wilson: You forgot the essential condition "for any distinct $x$ and $y$" in your reformulation of the problem. Obviously the strict inequality can't be true for $x=y$, but we need the strictness of the inequality to derive the desired contradiction.2011-12-10

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If $l\gt0$, then $x_0$ and $f(x_0)$ are distinct. Then

$$l=d(x_0,f(x_0))\gt d(f(x_0),f(f(x_0))\ge l\;,$$

a contradiction.