cauchy sequence, contractive sequence

Hello,

Let X1 be in E. For n>=1, let Xn+1 = f(Xn)
show {Xn} is contractive

I have attempted several things:
1. |f(X2) - f(X1)|
= |X3-X2|
<= |X2-X1| + |X1 - X3| (by triangle inequality)
< |X2-X1|
By definition, a sequence is contractive if |Pn+1 -Pn|<= b|Pn-Pn-1| or |Pn+1-Pn|<= b^n-1|p2-p1|

So, I know |X3-X2| < |X2-X1|, but how do i find the b?

2. Show {Xn} is cauchy
to show that, I have to find an N s.t. |Pn -Pm|< epsilon for all n,m>=N.
So, |f(Xn-1)-f(Xm-1)|
= |Xn-Xm|
<= |Xn -p| + |p-Xm| (let |Xn - p| < epsilon/2 and |Xm -p| < epsilon/2)
< epsilon/2 + epsilon/2
=epsilon

But what is N?

Would either work?

Quote:

---

Originally Posted by inthequestofproofs

Let X1 be in E. For n>=1, let Xn+1 = f(Xn)
show {Xn} is contractive

---

You have not given enough information here. What is E? How is the function f defined?

This is the whole question:
Let E be a subset of R. If E is closed, and f: E -> E is contractive, prove there exists a unique point x0 in E s.t. f(x0) = x0 (a fixed point)

On the back of the book, it is stated:
"Let X1 be in E. For n>=1, let Xn+1 = f(Xn)
show {Xn} is contractive"

I don't understand how doing the latter proves the former.

Thanks

If f is contractive, then it's a quick step to show that the sequence f(x), f(f(x)), f(f(f(x))), etc. is Cauchy (hint: geometric series). Since E is closed and R is complete, that sequence has its limit L in E. Use the sequence definition of continuity to show that f(L)=L. To show uniqueness, suppose x and y are both fixed points. Then |f(x)-f(y)|=|x-y| is a contradiction unless |x-y|=0.