# Nice result of compactness.

• Feb 24th 2010, 07:27 PM
Drexel28
Nice result of compactness.
Problem: Let $(K,d)$ be a compact metric space and $f:K\mapsto K$ be expansive in the sense that $d(f(x),f(y))\geqslant d(x,y)$.

Prove:

a) $f$ is an isometry.

b) $f$ is a homeomorphism.
• Feb 25th 2010, 08:46 PM
Black
Spoiler:
(a) It's enough to prove the following:

If there exists $r,s \in K$ such that $d(f(r),f(s))>d(r,s)$, then there exists $t,u \in K$ such that $d(f(t),f(u)).

Proof: Suppose not and $\forall x,y \in K
$
, $d(f(x),f(y)) \ge d(x,y).$ Define a sequence $(f_n)$ as follows:

$f_1=f$

$f_2=f\circ f$

$\vdots$

$f_n=f \circ f_{n-1}$

$\vdots$

Since we have for $x,y \in K$, $d(f(x),f(y)) \ge d(x,y)$, then $d(f_n(x),f_n(y)) \ge d(x,y).$ Since $K$ is a compact metric space, it is sequentially compact, so the sequence $(f_n(x))$ has a Cauchy subsequence $(f_{n_j}(x))$. So for $\varepsilon > 0$ pick $m$ such that for $k \ge 1$

$d(f_{n_{m+k}}(x),f_{n_m}(x)) < \frac{\varepsilon}{2}$

$d(f_{n_{m+k}}(y),f_{n_m}(y)) < \frac{\varepsilon}{2}$.

Then we 'go back' $n_m$ times to get

$d(f_{n_{m+k}-n_m}(x),x) \le d(f_{n_{m+k}}(x),f_{n_m}(x)) <\frac{\varepsilon}{2}$

$d(f_{n_{m+k}-n_m}(y),y) \le d(f_{n_{m+k}}(y),f_{n_m}(y)) <\frac{\varepsilon}{2}$.

Therefore,

$d(f(x),f(y)) \le d(f_{n_{m+k}-n_m}(x),f_{n_{m+k}-n_m}(y))$
$\le d(x,y)+d(f_{n_{m+k}-n_m}(x),x)+d(f_{n_{m+k}-n_m}(y),y)$
$,
and since $\varepsilon$ is arbitrary, $d(f(x),f(y))< d(x,y)$, a contradiction.

(b) Since $f$ is an isometry, it is clearly injective and continuous. If $C \subseteq K$ is closed, then it is compact. Therefore, $f(C)$ is compact, which implies that $f(C)$ is closed ( $K$ is Hausdorff).

Suppose $f$ is not surjective. Then let $x_0 \in K-f(K)$. Since $f(K)$ is closed, $K-f(K)$ is open. Then for $\varepsilon >0, \, \, B_{\varepsilon}(x_0) \subseteq K-f(K)$, so for $x \in f(K)$, $d(x_0,x) \ge \varepsilon$.

Define a sequence as follows:

$x_1=f(x_0)$

$x_2=f(x_1)$

$\vdots$

$x_n=f(x_{n-1})$

$\vdots$

and let $X_n=\{x_1,x_2,\dots , x_n\}.$ I claim that $d(x_i,x_j) \ge \varepsilon$ for $i \not= j$. Prove by induction.

Since $x_1 \in f(K)$, $d(x_0,x_1) \ge \varepsilon$. Suppose it's true for the elements of $X_n$ for $n \ge 1$. We must show that $d(x_i,x_{n+1}) \ge \varepsilon$ where $x_i \in X_n$. It's true for $i=0$ since $x_{n+1} \in f(K)$. If $i>0$, then $d(x_i,x_{n+1})=d(f(x_{i-1}),f(x_n))=d(x_{i-1},x_n) \ge \varepsilon$ since $x_{i-1},x_n \in X_n$. Let

$X=\bigcup_{n \in \mathbb{N}}X_n$.

Then $X$ is a discrete set. Since $X$ is an infinite set with no limit point, $K$ is not limit point compact, a contradiction.
• Feb 25th 2010, 08:55 PM
Drexel28
Quote:

Originally Posted by Black
Spoiler:
(a) It's enough to prove the following:

If there exists $r,s \in K$ such that $d(f(r),f(s))>d(r,s)$, then there exists $t,u \in K$ such that $d(f(t),f(u)).

Proof: Suppose not and $\forall x,y \in K
$
, $d(f(x),f(y)) \ge d(x,y).$ Define a sequence $(f_n)$ as follows:

$f_1=f$

$f_2=f\circ f$

$\vdots$

$f_n=f \circ f_{n-1}$

$\vdots$

Since we have for $x,y \in K$, $d(f(x),f(y)) \ge d(x,y)$, then $d(f_n(x),f_n(y)) \ge d(x,y).$ Since $K$ is a compact metric space, it is sequentially compact, so the sequence $(f_n(x))$ has a Cauchy subsequence $(f_{n_j}(x))$. So for $\varepsilon > 0$ pick $m$ such that for $k \ge 1$

$d(f_{n_{m+k}}(x),f_{n_m}(x)) < \frac{\varepsilon}{2}$

$d(f_{n_{m+k}}(y),f_{n_m}(y)) < \frac{\varepsilon}{2}$.

Then we 'go back' $n_m$ times to get

$d(f_{n_{m+k}-n_m}(x),x) \le d(f_{n_{m+k}}(x),f_{n_m}(x)) <\frac{\varepsilon}{2}$

$d(f_{n_{m+k}-n_m}(y),y) \le d(f_{n_{m+k}}(y),f_{n_m}(y)) <\frac{\varepsilon}{2}$.

Therefore,

$d(f(x),f(y)) \le d(f_{n_{m+k}-n_m}(x),f_{n_{m+k}-n_m}(y))$
$\le d(x,y)+d(f_{n_{m+k}-n_m}(x),x)+d(f_{n_{m+k}-n_m}(y),y)$
$,
and since $\varepsilon$ is arbitrary, $d(f(x),f(y))< d(x,y)$, a contradiction.

(b) Since $f$ is an isometry, it is clearly injective and continuous. If $C \subseteq K$ is closed, then it is compact. Therefore, $f(C)$ is compact, which implies that $f(C)$ is closed ( $K$ is Hausdorff).

Suppose $f$ is not surjective. Then let $x_0 \in K-f(K)$. Since $f(K)$ is closed, $K-f(K)$ is open. Then for $\varepsilon >0, \, \, B_{\varepsilon}(x_0) \subseteq K-f(K)$, so for $x \in f(K)$, $d(x_0,x) \ge \varepsilon$.

Define a sequence as follows:

$x_1=f(x_0)$

$x_2=f(x_1)$

$\vdots$

$x_n=f(x_{n-1})$

$\vdots$

and let $X_n=\{x_1,x_2,\dots , x_n\}.$ I claim that $d(x_i,x_j) \ge \varepsilon$ for $i \not= j$. Prove by induction.

Since $x_1 \in f(K)$, $d(x_0,x_1) \ge \varepsilon$. Suppose it's true for the elements of $X_n$ for $n \ge 1$. We must show that $d(x_i,x_{n+1}) \ge \varepsilon$ where $x_i \in X_n$. It's true for $i=0$ since $x_{n+1} \in f(K)$. If $i>0$, then $d(x_i,x_{n+1})=d(f(x_{i-1}),f(x_n))=d(x_{i-1},x_n) \ge \varepsilon$ since $x_{i-1},x_n \in X_n$. Let

$X=\bigcup_{n \in \mathbb{N}}X_n$.

Then $X$ is a discrete set. Since $X$ is an infinite set with no limit point, $K$ is not limit point compact, a contradiction.

You're first solution is almost identical to mine. The second bit differs slightly.

Assume there exists some $k\in K-f(K)$, since in any metric space the distance between a point and a compact subspace ( $f$'s continuity implies $f(K)$'s compactness) is positive we have that $d(k,f(K))=\delta>0$.

Define, $\left\{f^n(k)\right\}_{n\in\mathbb{N}}$ where $f^n(k)=\underbrace{f(f(f(\cdots(k)))\cdots)}_{n\te xt{ times}}$. This is a sequence in $K$ and since every compact metric is sequentially compact there exists some $\left\{f^{n_m}(k)\right\}_{m\in\mathbb{N}}$ which is convergent, and thus Cauchy. In particular, there exists some $N\in\mathbb{N}$ such that $d\left(f^{n_{N+1}}(k),f^{n_N}(k)\right)<\frac{\del ta}{2}$. But, by assumption $d\left(f^{n_{N+1}}(k),f^{n_N}(k)\right)\geqslant d(f^{n_N}(k),f^{n_{N-1}}(k))\geqslant\cdots\geqslant d(f(k),k)>\frac{\delta}{2}$. This is of course a contradiction.

Note that you need not have that $f$ is an isometry to prove that it's surjective.

There is another way to prove the first one that is a tad more elegant, but a tad more verbose.

Regardless, good job!!