Fixed points on compact spaces

[l888l888l888]

Homework Statement

Let X be a compact metric space. if f:X-->X is continuous and d(f(x),f(y))<d(x,y) for all x,y in X, prove f has a fixed point.

Homework Equations

The Attempt at a Solution

Assume f does not have a fixed point. By I problem I proved before if f is continuous 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.

 

[micromass]

Hi l888l888l888!

Try to use that the function

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

is continuous. What can you infer about this function using that X is compact?

 

[l888l888l888]

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

 

[micromass]

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))&lt;d(x,y))

 

[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)?

 

[micromass]

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

 

[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?

 

[micromass]

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?

Yes!

 

[l888l888l888]

Great. thanks!