Thread: 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.

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 ? To see that it's Cauchy merely note that . Now you can easily prove that for sufficiently large that last term is smaller than any given . So, is cauchy but not convergent since