# Fixed points on compact spaces

1. Jul 18, 2011

### l888l888l888

1. The problem statement, all variables and given/known data
Let X be a compact metric space. if f:X-->X is continous and d(f(x),f(y))<d(x,y) for all x,y in X, prove f has a fixed point.

2. Relevant equations

3. The attempt at a solution
Assume f does not have a fixed point. By I problem I proved before if f is continous with no fixed point then there exists an epsilon>0 st d(f(x),x)>=epsilon for all x in X. Using this I wanted to get a contradiction. i wanted to prove d(f(x),f(y))>=d(x,y) which leads to a contradiction of
d(f(x),f(y))<d(x,y) but I got stuck.

2. Jul 18, 2011

### micromass

Staff Emeritus
Hi l888l888l888!

Try to use that the function

$$\Phi:X\rightarrow \mathbb{R}:x\rightarrow d(x,f(x))$$

3. Jul 18, 2011

### l888l888l888

you can infer that it attains its bounds on X. ?

4. Jul 18, 2011

### micromass

Staff Emeritus
Indeed, so you can deduce that there is an a such that d(f(a),a) is the minimum value. Either this minimum value is 0, and then we got a fixpoint. If the minimum value is not 0, then we must obtain a contradiction, can you find one?? (Use that $d(f(x),f(y))<d(x,y)$)

5. Jul 18, 2011

### l888l888l888

ummmmm nothin comes to mind at this time...??? if o is not the min value then d(f(a),a)>0 and we must come up with a contradiction that contradicts d(f(x),f(a))<d(x,a)??????

6. Jul 18, 2011

### micromass

Staff Emeritus
What is d(f(f(a)),f(a))?

7. Jul 18, 2011

### l888l888l888

d(f(f(a)),f(a))<d(f(a),a). but then this contradicts the fact that d(f(a),a) is a min value?

8. Jul 18, 2011

### micromass

Staff Emeritus
Yes!

9. Jul 18, 2011

### l888l888l888

Great. thanks!!!!