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Thread: Nice result of compactness.

  1. #1
    MHF Contributor Drexel28's Avatar
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    Nice result of compactness.

    Problem: Let $\displaystyle (K,d)$ be a compact metric space and $\displaystyle f:K\mapsto K$ be expansive in the sense that $\displaystyle d(f(x),f(y))\geqslant d(x,y)$.

    Prove:

    a) $\displaystyle f$ is an isometry.

    b) $\displaystyle f$ is a homeomorphism.
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  2. #2
    Member Black's Avatar
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    Spoiler:
    (a) It's enough to prove the following:

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

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

    $\displaystyle f_1=f$

    $\displaystyle f_2=f\circ f$

    $\displaystyle \vdots$

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

    $\displaystyle \vdots$


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

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

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

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

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

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

    Therefore,

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


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

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

    Define a sequence as follows:

    $\displaystyle x_1=f(x_0)$

    $\displaystyle x_2=f(x_1)$

    $\displaystyle \vdots$

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

    $\displaystyle \vdots$

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

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

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

    Then $\displaystyle X$ is a discrete set. Since $\displaystyle X$ is an infinite set with no limit point, $\displaystyle K$ is not limit point compact, a contradiction.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Black View Post
    Spoiler:
    (a) It's enough to prove the following:

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

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

    $\displaystyle f_1=f$

    $\displaystyle f_2=f\circ f$

    $\displaystyle \vdots$

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

    $\displaystyle \vdots$


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

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

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

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

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

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

    Therefore,

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


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

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

    Define a sequence as follows:

    $\displaystyle x_1=f(x_0)$

    $\displaystyle x_2=f(x_1)$

    $\displaystyle \vdots$

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

    $\displaystyle \vdots$

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

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

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

    Then $\displaystyle X$ is a discrete set. Since $\displaystyle X$ is an infinite set with no limit point, $\displaystyle K$ 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 $\displaystyle k\in K-f(K)$, since in any metric space the distance between a point and a compact subspace ($\displaystyle f$'s continuity implies $\displaystyle f(K)$'s compactness) is positive we have that $\displaystyle d(k,f(K))=\delta>0$.

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

    Note that you need not have that $\displaystyle f$ 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!!
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