# disprove completeness ; fixed point for contractions

• Feb 27th 2010, 11:53 PM
derek walcott
disprove completeness ; fixed point for contractions
Space X - graph of f(x) = sin(1/x) on (0,1]
X = {(x,y): 0<x<=1, y = sin(1/x)}

Prove:
a) X is not complete
b) Every contraction h:X-->X has a fixed point

Hints for b given by my professor:
1.) X={(x,y) are elements in X: x = delta} U {(x,y) are elements in X: x >= delta}
2.) If delta is sufficiently small then diam(h(x1))<2; so h(x1) is an arc
3.) h(x2) is an arc (COMPACTNESS)
4.) therefore h(x) bounded away from the y-axis
5.) therefore h has a fixed point

Any help would be useful on any parts of these problems. He said proving part a would be simple but then he said part b would be long. Again, any help would be useful. I will post what I have for part a in a little bit ... still working on it. Thank you for reading.
• Feb 28th 2010, 08:43 AM
Drexel28
Quote:

Originally Posted by derek walcott
Space X - graph of f(x) = sin(1/x) on (0,1]
X = {(x,y): 0<x<=1, y = sin(1/x)}

Prove:
a) X is not complete
b) Every contraction h:X-->X has a fixed point

Hints for b given by my professor:
1.) X={(x,y) are elements in X: x = delta} U {(x,y) are elements in X: x >= delta}
2.) If delta is sufficiently small then diam(h(x1))<2; so h(x1) is an arc
3.) h(x2) is an arc (COMPACTNESS)
4.) therefore h(x) bounded away from the y-axis
5.) therefore h has a fixed point

Any help would be useful on any parts of these problems. He said proving part a would be simple but then he said part b would be long. Again, any help would be useful. I will post what I have for part a in a little bit ... still working on it. Thank you for reading.

I'm assuming this is with the usual metric.

What about $x_n=\left(\frac{1}{2n\pi},\sin\left(\frac{1}{\frac {1}{2n\pi}}\right)\right)$? To see that it's Cauchy merely note that $d(x_n,x_m)=\sqrt{\left(\frac{1}{2n\pi}-\frac{1}{2m\pi}\right)^2+\left(\sin\left(2n\pi\rig ht)-\sin\left(2m\pi\right)\right)^2}=\frac{1}{2\pi}\le ft|\frac{1}{n}-\frac{1}{m}\right|$. Now you can easily prove that for sufficiently large $m,n$ that last term is smaller than any given $\varepsilon>0$. So, $x_n$ is cauchy but not convergent since $x_n\to (0,0)\notin \Gamma_{f}$