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Suppose that $(X, d)$ is a compact metric space and that $f: X \rightarrow X$ is a continuous function satisfying $d(x,y) \leq d(f(x), f(y))$ for all $x, y \in X$. Show that $f(X) = X$.

Here is a cheap argument (for which I am not entirely confidence with): Since $X$ is a compact metric space and $f$ is continuous, $X \backslash f(X)$ is an open subset of $X$, which is either empty or non-empty. If empty then we are done. Otherwise it has a point $x_{0}$. Then there exists a $\delta >0$ such that a ball centered at $x_{0}$ of radius $\delta$ is contained in $X \backslash f(X)$. Pick a point $t$ in this ball. Then by uniform continuity of $f$, we have for any $\epsilon >0$, $\epsilon > d(f(x_{0}), f(t)) \geq d(x_{0}, t)$ with the last inequality from the hypothesis. But then since $\epsilon$ is arbitrary, we have $x_{0} = t$ a contradiction since a ball must be open.

Daniel Akech Thiong
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  • I don't understand this part:

    Then by uniform continuity of $f$, we have for any $\epsilon >0$, $\epsilon > d(f(x_{0}), f(t)) \geq d(x_{0}, t)$ with the last inequality from the hypothesis.

    Why would that $\delta$ work for any $\epsilon$?

     – Ilham Apr 21 '15 at 16:01
  • You could easily modify this argument I think (sorry I don't have time to do it as an answer):

    http://math.stackexchange.com/questions/170989/a-isometric-map-in-metric-space-is-surjective

     – Ilham Apr 21 '15 at 16:04
  • Your argument is not correct. You are assuming that $t$ is the same for each $\varepsilon >0$ but $t$ depends on $\epsilon$ too.

    Let's continue your proof. Let $x_n=f(x_{n-1})$. The following holds:

    $d(x_{n

     – Ergos Apr 21 '15 at 16:04
  • @Ergos I think if you continue his proof you can just post it as an answer, to "resolve" the question. – Ilham Apr 21 '15 at 16:11

1 Answers1

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Let's continue your proof. Let $x_0$ as in your proof. Define the sequence $x_n=f(x_{n-1})$. The following holds:

$$ d(x_{n+m},x_n) \ge d(x_{n+m-1},x_{n-1}) \ge .... \ge d(x_m,x_0) > \delta $$

In particular, the sequence $x_n$ has no convergent subsequence. Contradiction because $X$ is compact.

Ergos
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