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The sequence

$\displaystyle a_{0},a_{1},a_{2},\ldots$ | (1) |

in a metric space $(X,d)$ is called contractive, iff there is a real number $r\in(0,1)$ such that for any positive integer $n$ the inequality

$\displaystyle d(a_{n},a_{{n+1}})\leqq r\!\cdot\!d(a_{{n-1}},a_{n})$ | (2) |

is true.

We will prove the

Theorem. If the sequence (1) is contractive, it is
a Cauchy sequence^{}.

Proof. Suppose that the sequence (1) is contractive. Let $\varepsilon$ be an arbitrary positive number and $m,n$ some positive integers from which e.g. $n$ is greater than $m,$ $n=m+\delta$.

Using repeatedly the triangle inequality^{} we get

$\displaystyle d(a_{m},a_{n})$ | $\displaystyle\leqq d(a_{m},a_{{m+1}})+d(a_{{m+1}},a_{{m+\delta}})$ | ||

$\displaystyle\leqq d(a_{m},a_{{m+1}})+d(a_{{m+1}},a_{{m+2}})+d(a_{{m+2}},a_{{m% +\delta}})$ | |||

$\displaystyle\ldots$ | |||

$\displaystyle\leqq d(a_{m},a_{{m+1}})+d(a_{{m+1}},a_{{m+2}})+d(a_{{m+2}},a_{{m% +3}})+\ldots+d(a_{{n-1}},a_{n}).\par$ |

Now the contractiveness gives the inequalities

$d(a_{1},a_{2})\leqq rd(a_{0},a_{1}),$ |

$d(a_{2},a_{3})\leqq rd(a_{1},a_{2})\leqq r^{2}d(a_{0},a_{1}),$ |

$d(a_{3},a_{4})\leqq rd(a_{2},a_{3})\leqq r^{3}d(a_{0},a_{1}),$ |

$\ldots$ |

$d(a_{m},a_{{m+1}})\leqq r^{m}d(a_{0},a_{1}),$ |

$\ldots$ |

$d(a_{{n-1}},a_{n}})\leqq r^{{n-1}}d(a_{0},a_{1}),$ |

by which we obtain the estimation

$\displaystyle d(a_{m},a_{n})$ | $\displaystyle\leqq d(a_{0},a_{1})(r^{m}+r^{{m+1}}+\ldots+r^{{m+\delta-1}})$ | ||

$\displaystyle=d(a_{0},a_{1})r^{m}(1+r+r^{2}+\ldots+r^{{\delta-1}})$ | |||

$\displaystyle=d(a_{0},a_{1})r^{m}\frac{1-r^{\delta}}{1-r}$ | |||

$\displaystyle<d(a_{0},a_{1})\frac{r^{m}}{1-r}.$ |

The last expression tends to zero as $m\to\infty$. Thus there exists a positive number $M$ such that

$d(a_{m},a_{n})<\varepsilon\mbox{ for each }m>M$ |

when $n>m$. Consequently, (1) is a Cauchy sequence.

Remark. The assertion of the Theorem cannot be reversed. E.g. in the usual metric of $\mathbb{R}$, the sequence $1,\frac{1}{2},\frac{1}{3},\ldots$ converges to 0 and hence is Cauchy, but for it the ratio

$|a_{n}-a_{{n+1}}|:|a_{{n-1}}-a_{n}}|\;=\;1-\frac{2}{n+1}$ |

tends to 1 as $n\to\infty$.

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