Self-maps of compact metric spaces

Let $X$ be a compact metric space and $f : X \to X$ an isometry. Show that $f(X) = X$.

Let $K$ be a nonempty compact set in a metric space with distance function $d$. Suppose that $\varphi : K \to K$ satisfies

for all $x \neq y$ in $K$. Show there exists precisely one point $x \in K$ such that $x = \varphi (x)$.

Prove that if a set $M$ is compact in a metric space $X$, and a mapping $f: M \rightarrow M$ satisfies the inequality $\rho (f(x), f(y)) < \rho (x, y)$ for any $x \neq y$, then there exists a unique fixed point $x \in M$ (compare with problem 1.74 and Theorem 1.4).

Theorem 1.4 (contraction mapping principle): let $f: X \rightarrow X$ be a contraction mapping (i.e., there exists $q \in [0, 1)$ such that $\rho (f(x_{1}), f(x_{2})) \leqslant q \cdot \rho (x_{1}, x_{2})$ for all $x_{1}, x_{2} \in X$) on a complete metric space $X$. Then there exists a unique point $x_{0} \in X$ such that $f(x_{0}) = x_{0}$; this point is the limit of the recursively defined sequence $x_{n} = f(x_{n-1})$ with an arbitrary starting point $x_{1} \in X$.

Prove that in a metric space any isometric embedding of a compact into itself is an isometry of this compact (and show that this is false for an arbitrary closed bounded set).

Let $X$ be a complete metric space, $M$ a compact set in it, $\varphi : M \rightarrow M$ a mapping satisfying the inequality $\rho (\varphi (x), \varphi (y)) \geqslant \rho (x, y)$. Prove that $\varphi$ is an isometry of the set $M$.

Let $X$ be a compact metric space and $T$ a mapping from $X$ into itself such that

Show that $T$ has a unique fixed-point.

Define $f(x) = \pi / 2+x-\arctan x$ for all real numbers $x$. Show that

Show that $f$ does not have a fixed-point. Does this contradict the preceding problem?