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==Proof==
Let ''x''<submath>0x_0\in X</submath> ∈ ''X'' be arbitrary and define a sequence {''x<submath>n\{x_n\}</submath>''} by setting ''x<submath>n</sub>'' x_n= ''T''(''x''<sub>''x_{n''−1-1})</submath>). 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<submath>n\{x_n\}</submath>''} 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 ε <math>\varepsilon> 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<submath>n\{x_n\}</submath>''} 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<submath>1p_1</submath>'' and ''p<submath>2p_2</submath>'' 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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