MHB Is a Metric Space with a Surjective Contraction Mapping Non-Compact?

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I had several delays which prevented me from posting a problem early. My apologies. Here's this week's problem.

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Problem. Let $(M,d)$ be a metric space for which there is a surjective contraction mapping $f : M \to M$. Show that $M$ is non-compact.

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Remark. There are two important pieces of information that will help those who are having difficulties with this problem. A contraction mapping on a metric space $(X,d)$ is a function $f : X \to X$ for which there is a constant $c$, $0 < c < 1$, such that $d(f(x),f(y)) \le c\, d(x,y)$ for all $x,y\in X$. In the above problem, you are to assume that $M$ has more than one point.

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This week's problem was correctly answered by Fallen Angel. Here is his solution.

Spoiler

Assume $M$ is compact and let $\{U_{i}\}_{i\in I}$ be a open cover and $\{U_{n}\}_{n=1}^{k}$ a finite sub cover.

Since $f$ is contractive, for all $\epsilon >0$ there exists $m_{n}\in \Bbb{N}$ such that $d(f^{m_{n}}(x),f^{m_{n}}(y))<\epsilon$ $\forall (x,y)\in U_{n}$. Take $m=max\{m_{1},m_{2},\ldots ,m_{k}\}$.
Since $\{U_{n}\}_{n=1}^{k}$ is a cover, we got that $d(f^{m}(x),f^{m}(y))<k\epsilon$, $\forall \ x,y\in M$

Now consider $a,b\in M$ with $a\neq b$. Then $d(a,b)=\delta >0$ and take $\epsilon=\dfrac{\delta}{k+1}$.
Then there exists $m$ such that $d(f^{m}(x),f^{m}(y))<\dfrac{k}{k+1}\delta, \ \forall \ x,y\in M$ so $a\notin f^{m}(M)$ or $b\notin f^{m}(M)$.
Hence $f$ can't be surjective.

I'm also providing my solution below:

Spoiler

By way of contradiction, suppose $M$ is compact and there exists a $c$-contraction mapping $f$ from $M$ onto itself. The metric $d$ is a continuous mapping from $(M \times M, D)$ to $\Bbb R$, where $D$ is the metric on $M \times M$ given by $D((x,y),(a,b)):= d(x,a) + d(y,b)$. For by the triangle inequality, if $(x,y), (a,b)\in M \times M$, \[d(x,y) - d(a,b) \le d(x,a) + d(a,y) - d(a,b) \le d(x,a) + d(y,b) = D((x,y),(a,b)).\] By symmetry, $d(a,b) - d(x,y) \le D((x,y), (a,b))$. Therefore \[|d(x,y) - d(a,b)| \le D((a,b),(x,y)),\] showing that $d$ is (uniformly) continuous. Since $M \times M$ is compact, there exists a point $(x_0,y_0)\in M \times M$ such that $d(x_0,y_0) = \text{diam}(M)$. Since $f$ is onto, $x_0 = f(x_1)$ and $y_0 = f(y_1)$ for some $x_1,y_1\in M$. Thus \[\text{diam}(M) = d(x_0, y_0) = d(f(x_1), f(y_1)) \le c\, d(x_1, y_1) \le c\, \text{diam}(M).\] Consequently, $(1 - c)\, \text{diam}(M) \le 0$. Since $0 \le c < 1$, $\text{diam}(M) \le 0$, i.e., $\text{diam}(M) = 0$. This contradicts the assumption that $M$ contains more than one point.

Note. As standard terminology, a contractive mapping on a metric space $(X,d)$ is a function $f : X \to X$ such that $d(f(x),f(y)) < d(x,y)$ for all $x,y\in X$. In Fallen Angel's solution however, a contractive function is a contraction.