# Isometry and compactness.

Let $$K\subset \mathbb{R}^{n}$$ 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 $$f_{n} = f^{n} = f\circ \cdots \circ f$$. 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 $$k \in \mathbb{N}$$ 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 $$k \in \mathbb{N}$$ 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 $$K\times K$$ 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 $$|a-a_k| < \varepsilon \quad \text{and} \quad|b-b_k| < \varepsilon$$

$$(b) \quad$$ 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 $$\varepsilon = \frac{|f(a) - f(b)| - |a,b|}{2}$$

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

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 $$|f(a) - f(b)| = |a - b|$$.

• I didn't see TheSilverDoe's nice answer, until after I'd posted this. – Simon Mar 20 at 13:02