Is The Banach Fixed Point Theorem True?

Let $(M,d)$ be a complete metric space. A contraction is a function $f: M \to M$ for which there exists some constant $0 < c < 1$ such that $d\big(f(x), f(y)\big) < c \cdot d(x,y)$ for all $x, y \in M$.

Answer the following yes-no questions:

• If $f: M \to M$ is a contraction, does $f$ have a fixed point? $($I.e., is there some $x\in M$ such that $f(x) = x?)$

• If $f: M \to M$ has a fixed point, is $f$ a contraction?

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Hint: The first question is much harder than the second. In fact, the answer is yes, and this extremely important result is known as the Banach fixed point theorem.

To prove it, choose an arbitrary $x_0 \in M$ and set $x_n = T(x_{n-1})$ for $n\ge 1$. Then, show that $x_n$ converges to some $x\in M$ and that $x$ is the desired fixed point.