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Math Help - Nice result of compactness.

  1. #1
    MHF Contributor Drexel28's Avatar
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    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.
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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 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))<d(t,u).

    Proof: Suppose not and \forall x,y \in K<br />
, 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)
    <d(x,y)+\varepsilon ,
    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.
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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 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))<d(t,u).

    Proof: Suppose not and \forall x,y \in K<br />
, 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)
    <d(x,y)+\varepsilon ,
    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!!
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