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?

\(\)

**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.

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