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Let $Ksubset mathbbR^n$ be a compact non-empty set and $f: K to K$ such that $|f(x) - f(y)| geq |x-y|$, for all $x,y in K$. Consider the sequnce $f_n$ given by $f_n = f^n = fcirc cdots circ f$. For $a,b in K$, let $a_n = f_n(a)$ and $b_n = f_n(b)$.

(a) Show that for any $epsilon > 0$, there is $k in mathbbN$ such that
$$|a-a_k| < epsilon, |b-b_k|<epsilon.$$

(b) Show that $f(K)$ is dense in $K$.

(c) Show that $f$ is an isometry

My attempt

(a) I dont have any idea

(b) Given $x in K$, take $x_n = f_n(x)$. By (a), for any $epsilon > 0$, there is $k in mathbbN$ such that $|x - x_k| < epsilon$. The sequence $x_k$ satisfying (a) has a convergent subsequence, since $K$ is compact. But this sequence converges to $x$.

(c) My idea was suppose that $|a - b| > |f(a) - f(b)|$ for some $a,b$ using (b), but I cannot conclude anything.

Can someone help me? Also, I'm not sure about (b). This not seems a hard problem, but I'm stuck on it.

$(a) quad$ Let $varepsilon > 0$. For all $x in K$, denote by $B_x$ the open ball centered in $x$, of radius $varepsilon$. The family of open sets $(B_x times B_y)_x,y in K$ covers $K times K$. Because $K times K$ is compact, there exists a finite family $x_1, ..., x_n$ and $y_1, ..., y_n$ of points of $K$ such that
$$(K times K) subset bigcup_1 leq m leq n (B_x_m times B_y_m) $$

Now consider the $n+1$ points $(a,b)$, $(a_1, b_1)$, ..., $(a_n, b_n)$ of $K times K$. You have $n+1$ points, and $Ktimes K$ is covered by $n$ open sets, so two of these points are in the same open. So there exists $1 leq m leq n$, and $0 leq i < j leq n$ such that $a_i$ and $a_j$ are in $B_x_m$, and $b_i$ and $b_j$ are in $B_y_m$. In particular, $|a_i - a_j| < varepsilon$, and $|b_i - b_j| < varepsilon$. Let $k=j-i$. It is easy to see that, because $f$ is expansive, $|f^i(a)-f^j(a)| geq |a-f^j-i(a)|$, i.e $|a_i-a_j| geq |a-a_k|$. Similarly, $|b_i - b_j| geq |b-b_k|$. So you deduce that
$$|a-a_k| < varepsilon quad textand quad|b-b_k| < varepsilon$$

$(b) quad$ The first question shows that for every $varepsilon > 0$ and $a in K$, there exists $k neq 0$ such that $|a-f^k(a)| < varepsilon$. But $f^k(a)=f(f^k-1(a)) in f(K)$, so you get a point of $f(K)$ at distance less than $varepsilon$ of $a$. This shows that $f(K)$ is dense in $K$.

$(c) quad$ Suppose that there exists $a neq b$ such that $|f(a) - f(b)| > |a-b|$. Let $$varepsilon = frac2$$

By the first question, there exists $k$ such that $|a-a_k| <varepsilon$ and $|b-b_k| < varepsilon$. So you get
$$|a_k-b_k| leq |a-a_k| + |a-b| + |b-b_k| < 2varepsilon + |a-b| = |f(a)-f(b)|$$

This is impossible, because $f$ is expansive so you must have $|f^k(a)-f^k(b)| geq |f(a)-f(b)|$.

This shows that $f$ is an isometry.

Let $epsilon>0$. Cover $K$ with finitely many open balls of radius $epsilon/2$, as we may, since $K$ is compact. The sequence $(a_n)$ must visit at least one such ball, say $B$, infinitely many times. Therefore there is a subsequence $(a_n_k)$ of $(a_n)$ that lives wholly in $B$. In particular, $|a_n_k - a_n_k+1| < epsilon$ for all $k$, where we also used the triangle inequality. By the expansive property of $f$, and by induction, it follows that $|a - a_n_k - n_k+1| < epsilon$ for all $k$, that is, $(alpha_k) = (a_n_k - n_k+1)$ is a subsequence of $(a_n)$ that lives wholly in the $epsilon$-ball centred on $a$. Taking the corresponding subsequence of $(b_n)$, and applying the same argument using compactness of $K$ and expansiveness of $f$, we may pass to further corresponding subsequences of $(a_n)$ and $(b_n)$ that live wholly in the $epsilon$-balls centred on $a$ and on $b$, respectively. In particular, there is an index $k$ such that $a_k$ and $b_k$ have the property required in part a) of the question.

For part b), let $c in K$. We already showed in part a) that points in the sequence $f^n(c)$ may be found that are arbitrarily close to $c$. It follows that the range of $f$ is dense in $K$, as required.

For part c), let $a, b in K$ and suppose, for a contradiction, that $|f(a) - f(b)| = |a - b| + eta$, where $eta > 0$. Then for all $k$, $|a_k - b_k| geq |a - b| + eta$. By a suitable choice of $epsilon$, and by applying the triangle inequality, we can contradict the assertion in part a). Conclude that $|f(a) - f(b)| = |a - b|$.