I'm working on a question in Munkres:
If $f$ is a contraction and $X$ is compact, show $f$ has a unique fixed point.
Here's my attempt at a solution so far.
$f$ is continuous, choose $\epsilon = \delta$.
Thus $f^n$ (the composition of $f$ with itself $n$ times) is continuous. So each $A_n = f^n(X)$ is compact since $X$ is compact and $f^n$ is continuous.
Also $A_{n+1} \subset A_n \subset A_{n-1} \subset \ldots$
So $A = \cap A_n$ is nonempty.
I'm not sure how to show that A contains fixed points.
Then define $diam = sup \{d(x,y) : x,y \in X\}$.
So $diam(A) \le diam(A_n) \le \alpha^n diam(X)$ since $\alpha < 1,$ $\alpha^n$ approaches 0. Thus $diam(A) = 0$. So $A$ must contain one point. Obviously I want this point to be the fixed point so I need to show that $A$ must contain its fixed points.
Thanks in advance.
It is an easy exercise to show that $A=\bigcap_n A_n$ satisfies $f(A)=A$. This holds in any compact Hausdorff space. Now $A$ is compact, hence there are $a,b,c,d\in A$ such that $$d(a,b)=\text{diam}(A)=\text{diam}(f(A))=d(f(c),f(d))\le d(c,d)\le d(a,b)$$ This implies $c=d$, thus $A$ is a single point. Clearly, any fixed point of $X$ lies in $A$. Hence there's a unique fixed point.