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Line 37:
: <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 ''ε'' > 0 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>
Line 56:
:<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, ''T'' is continuous, so bringing the limit inside ''T'' was justified. Lastly, ''T'' cannot have more than one fixed point in (''X'',''d''), since any pair of distinct fixed points ''p''<sub>1</sub>
:<math> d(T(p_1),T(p_2)) = d(p_1,p_2) > q d(p_1, p_2).</math>
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