Let $(X, d)$ be a metric space, let $f : X → X$ be a contraction, and let $a_o \in X$. Let $a_1 = f(a_o)$ and $a_{n+1} = f(a_n)$ for $n \geq 1$. Prove that $(a_n)$ is a Cauchy sequence.


Given contraction constant $k$$$ 0 \leq k < 1$$

First note that by definition of contraction $$kd(a_2,a_1) \geq d(f(a_2), f(a_1)) = d(a_3, a_2)$$

Then $$k^2d(a_2,a_1) \geq kd(a_3,a_2) \geq d(f(a_3), f(a_2)) = d(a_4, a_3)$$

$$k^3d(a_2,a_1) \geq k^2d(a_3,a_2) \geq kd(a_4, a_3) \geq d(a_5,a_4)$$

Therefore, $$k^n d(a_2,a_1) \geq d(a_{n+2}, a_{n+1})$$

Then to show that $(a_n)$ is Cauchy, we must show that $\forall \epsilon > 0, \exists N$ such that $\forall n, m \geq N, d(a_n, a_m) < \epsilon$

Then by triangle inequality:

$$d(a_n, a_m) \leq d(a_{n+1},a_{n}) + d(a_{n+2}, a_{n+1}) + \ldots d(a_{m-1}, a_{m})$$

Let $$k^n d(a_2,a_1) = k^n r \geq d(a_{n+2}, a_{n+1}), r = d(a_2, a_1)$$

How do I use the above to provide a bound on $d(a_n, a_m)$?

Note: some stuff maybe messed up in the above derivation, fixing ....


Like you say above $$d(a_n,a_m) \leq \sum_{j=n}^{m-1} d(a_j,a_{j+1}) \leq \sum_{j=n}^{m-1} d(a_1,a_2)\cdot k^{j-1} \leq d(a_1,a_2) \sum_{j=n}^{\infty} k^{j-1} = d(a_1,a_2) \frac{k^{n-1}}{1-k}$$

Thus given $\epsilon>0$, choose $N$ such that $d(a_1,a_2)\frac{k^{N-1}}{1-k}< \epsilon$. Then clearly, $m> n \geq N$ imply that $d(a_n,a_m) \leq d(a_1,a_2) \frac{k^{n-1}}{1-k}<\epsilon$.

  • $\begingroup$ Does the converse of this theorem true... Please give counter example if not true $\endgroup$ – Cloud JR Feb 4 at 6:04

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