# Fixed point of a shrinking map. Proof.

Could you please verify my proof?

Let $f:X \to X$ be a shrinking map on a compact metric space $(X,d)$.

In other words: $d(f(x),f(y)) < d(x,y)$ for all $x,y \in X$.

Prove: $f$ has a unique fixed point.

## Proof:

Define $f^n=f^{n-1} \circ f.$ Let $A_n = f^n(X)$.

(1) By induction I proved that $\bigcap_{i=1}^n A_i$ is non empty

(2) Each of the $A_i$'s is a continous mapping of a compact set hence it is compact. A compact subset of a Hausdorff space (metric spaces are Hausdorff) is closed and thus $A_i$ is closed for all $i$.

(3) $\{A_n\}_{n \in \mathbb{N}}$ is a collection of closed sets with the finite intersection property by (1) thus their intersection is non-empty. Let $A = \bigcap_{n \in N} A_n$ denote that intersection.

(4) $x \in A \implies x \in f^n(X)$ for all $n$. It follows that for every $n$ there exist $x_n \in X$ such that $f^{n}(x_n)=x$.

(5) By sequential compactness (which is equivalent to compactness for metric spaces) $x_n$ has a convergent subsequence $x_{p_n} \to a$. By continuity $f(x_{p_n}) \to f(a)$. (there is a a gap here!) But $f^{p_n}(x_{p_n}) = x$ by construction and we have $f(a)=x.$

(6) By (4) and (5) $A \subseteq f(A)$.

(7) It simple to prove that if $x \in f(A)$ then $x \in A$ thus $f(A) \subseteq A$.

(8) by (6) and (7) $A=f(A)$. The metric is a continuous function in both variables. $f$ is a continuous map from a compact space $A \times A$ to $\mathbb{R}$ (which is ordered) thus the metric obtains a maximum value. let $(x,y) \in A \times A$ be the point such that $diam(A)=d(x,y)$.

(9) Let $(a,b) \in A$ be such that $f(a) = x, f(b) =y$. Assume $x \neq y$

$$d(x,y) = d(f(a),f(b)) < d(a,b) \le diam(A) = d(x,y)$$

In (5), how do you argue that $a\in A$? Or how do you obtain $A \subset f(A)$ in (6)? – Daniel Fischer Dec 18 '13 at 20:32
In (5) you say 'by continuity'. But of what? Notice that on the LHS you have $f^n$ which depends on $n$, not a single function. – Marek Dec 18 '13 at 20:40