# A lemma for the idea in the modern proof of Banach's Contraction Mapping Theorem

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?

• Looks good..... – N. S. Sep 30 '18 at 21:19

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.

• Thank you for your feedback, I will correct the mistake. – G the Stackman Sep 30 '18 at 21:20

Looks good to me!

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

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