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Let $(X,d)$ be a metric space and $T: X \rightarrow X$ a contraction mapping with a fixed point $w$. Suppose that $x_0 \in X$ and we define $x_n$ inductively by $x_{n+1}= Tx_n$. Show that $d(x_n,w) \rightarrow 0$ as $n \rightarrow \infty$.

$Proof:$ We observe that since $T$ is a contraction mapping on $X$, there exists a positive number $K<1$ with $d(Tx,Ty) \leq Kd(x,y)$ for all $x,y \in X$. Therefore we have that $$d(x_n,w)=d(Tx_{n-1},Tw) \leq Kd(x_{n-1},w),$$ and we see that $$d(x_n,w) \leq K^{n}d(x_0,w),$$ and hence $d(x_n,w) \rightarrow 0$ as $n \rightarrow \infty$.

Could you please provide some feedback regarding the correctness of the above proof?

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  • $\begingroup$ Looks good..... $\endgroup$ – N. S. Sep 30 '18 at 21:19
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Note that in the first line, you should have the inequality $d(T x_{n-1}, Tw ) \leq K d(x_{n-1}, w)$, but besides this things seem correct.

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  • $\begingroup$ Thank you for your feedback, I will correct the mistake. $\endgroup$ – G the Stackman Sep 30 '18 at 21:20
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Looks good to me!

Might want to double-check that the problem lets you assume the existence of a fixed point though.

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    $\begingroup$ Since the metric space is not given to be complete, that needs to be an assumption... $\endgroup$ – N. S. Sep 30 '18 at 21:20
  • $\begingroup$ Mm, fair. I've only seen it on complete spaces so wanted to make sure that wasn't an accidental assumption. $\endgroup$ – Henry Swanson Sep 30 '18 at 22:49

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