Editing Banach fixed-point theorem (section)

==Proof==

Let <math>x_0 \in X</math> be arbitrary and define a [[sequence]] <math>(x_n)_{n\in\mathbb N}</math> by setting <math>x_n = T(x_{n-1})</math>. We first note that for all <math>n \in \N,</math> we have the inequality

:<math>d(x_{n+1}, x_n) \le q^n d(x_1, x_0).</math>

This follows by [[Principle of mathematical induction|induction]] on <math>n</math>, using the fact that <math>T</math> is a contraction mapping. Then we can show that <math>(x_n)_{n\in\mathbb N}</math> is a [[Cauchy sequence]]. In particular, let <math>m, n \in \N</math> such that <math>m > n </math>:

: <math>\begin{align}
d(x_m, x_n) & \leq d(x_m, x_{m-1}) + d(x_{m-1}, x_{m-2}) + \cdots + d(x_{n+1}, x_n) \\[5pt]
& \leq q^{m-1}d(x_1, x_0) + q^{m-2}d(x_1, x_0) + \cdots + q^nd(x_1, x_0) \\[5pt]
& = q^n d(x_1, x_0) \sum_{k=0}^{m-n-1} q^k \\[5pt]
& \leq q^n d(x_1, x_0) \sum_{k=0}^\infty q^k \\[5pt]
& = q^n d(x_1, x_0) \left ( \frac{1}{1-q} \right ).
\end{align}</math>

Let <math>\varepsilon > 0</math> be arbitrary. Since <math>q \in [0,1)</math>, we can find a large <math>N \in \N</math> so that

:<math>q^N < \frac{\varepsilon(1-q)}{d(x_1, x_0)}.</math>

Therefore, by choosing <math>m</math> and <math>n</math> greater than <math>N</math> we may write:

:<math>d(x_m, x_n)  \leq q^n d(x_1, x_0) \left ( \frac{1}{1-q} \right ) < \left (\frac{\varepsilon(1-q)}{d(x_1, x_0)} \right ) d(x_1, x_0) \left ( \frac{1}{1-q} \right ) = \varepsilon.</math>

This proves that the sequence <math>(x_n)_{n\in\mathbb N}</math> is Cauchy. By completeness of <math>(X, d)</math>, the sequence has a limit <math>x^* \in X.</math> Furthermore, <math>x^*</math> must be a [[Fixed point (mathematics)|fixed point]] of <math>T</math>:

:<math>x^*=\lim_{n\to\infty} x_n = \lim_{n\to\infty} T(x_{n-1}) = T\left(\lim_{n\to\infty} x_{n-1} \right) = T(x^*). </math>

As a contraction mapping, <math>T</math> is continuous, so bringing the limit inside <math>T</math> was justified. Lastly, <math>T</math> cannot have more than one fixed point in <math>(X, d)</math>, since any pair of distinct fixed points <math>p_1</math> and <math>p_2</math> would contradict the contraction of <math>T</math>:

:<math> d(T(p_1),T(p_2)) = d(p_1,p_2) > q d(p_1, p_2).</math>