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==Proof==
Let ''x''<mathsub>x_0\in X0</mathsub> ∈ ''X'' be arbitrary and define a sequence {''x<mathsub>\{x_n\}n</mathsub>''} by setting ''x<mathsub>x_nn</sub>'' = ''T''(x_{''x''<sub>''n-1})''−1</mathsub>). We first note that for all <math>''n\in\mathbb'' ∈ '''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 {''x<mathsub>\{x_n\}n</mathsub>''} is a [[Cauchy sequence]]. In particular, let <math>''m'', ''n\in\mathbb'' ∈ '''N</math>''' such that <math>''m'' > ''n</math>'':
: <math>\begin{align}
\end{align}</math>
Let <math>\varepsilonε > 0</math> be arbitrary, since <math>''q\in'' ∈ [0, 1)</math>, we can find a large <math>''N\in\mathbb'' ∈ '''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 {''x<mathsub>\{x_n\}n</mathsub>''} 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 ''p<mathsub>p_11</mathsub>'' and ''p<mathsub>p_22</mathsub>'' 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>
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