The well known fixed-point theorem by Banach reads as follows:
Let $(X,d)$ be a complete metric space, and $A\subseteq X$ closed. Let $f: A\to A$ be a function, and $\gamma$ a constant with $0\leq\gamma <1$, such that $d(f(x), f(y))\leq\gamma\cdot d(x,y)$ for every $x,y\in A$.
Define $(x_n)_{n\in\mathbb{N}}$ by $x_{n+1}= f(x_n)$ for an arbitrary starting point $x_0\in A$.
Then exists exactly one $a\in A$ such that $f(a)=a$ and $\lim_{n\to\infty} x_n=a$ for an arbitrary $x_0$.
Obviously you can look up a proof easily, but I have a theorem, which states the following, and I am curious if you can use it to proof Banach's fixed-point theorem.
Let $(X,d)$ be a complete metric space. Let $A_0\supset A_1\supset\dotso$ be a decreasing sequence of nonempty closed subsets with $\operatorname{diam}(A_n)\stackrel{n\to\infty}{\longrightarrow} 0$, where $\operatorname{diam}(A_n):=\sup\{d(x,y)|x,y\in A\}$.
Then exists exactly one unique point in $\bigcap_{n\in\mathbb{N}} A_n$
The theorems are kinda familiar and my guess is, that you can construct such a decreasing sequence of sets, such that $\bigcap_{n\in\mathbb{N}} A_n=\{a\}$. Where $a$ is the desired fixed point.
If it is possible to use this efficiently, I would appreciate a hint, to try it myself.
Thanks in advance.
Best Answer
The suggestion $A_0=A$, $A_{n+1}=\overline{f(A_n)}$ only works if $A$ has finite diameter.
You could derive the first theorem from the second by setting $$A_n=\overline{\{x_n,x_{n+1},\dots\}}.$$
I'm not suggesting that's the right way to prove Banach's theorem! It's certainly not the "efficient" argument you asked for, since working out the details seems like more work then just proving Banach's theorem by the method you know. But it does serve to show how the two results are related, even if $A$ has infinite diameter.