A mapping from a metric space to itself is called a contraction if there exists an such that for every we have . The Banach fixed point theorem (aka the contraction theorem) says that every contraction on a non-empty complete metric space has exactly one fixed point.

**Proof**: Suppose is a complete metric space and that is a contraction. Let be any point. Define , , and so forth. If we write in terms of and, we see that . Now consider where we assume, without loss of generality, that . By the triangle inequality we have

We can obviously make this value as small as we’d like by picking a large enough , so the sequence is Cauchy and must converge. Call the limit of this sequence . Now consider the distance between and .

By picking a large enough , we can make this as small as we’d like as well, so we have so has a fixed point. For the uniqueness of , suppose is also a fixed point.

The only way this holds is if , so . Note that the we constructed didn’t depend on on our initial value of , so every sequence we construct by picking a point in and iterating will result in the same limit.

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