Fixed points on compact spaces

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Homework Help Overview

The problem involves a compact metric space X and a continuous function f mapping X to itself, with the condition that the distance between images under f is less than the distance between the original points. The goal is to prove that f has a fixed point.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to prove the existence of a fixed point by assuming the contrary and seeking a contradiction. They consider the implications of a minimum distance between points and their images under f. Other participants suggest examining the continuity of a derived function and its properties due to the compactness of X.

Discussion Status

Participants are exploring the implications of the continuity of the function defined by the distance from points to their images. There is a productive discussion about the existence of a minimum value and the potential contradictions arising from it, although no consensus has been reached yet.

Contextual Notes

There is an assumption that the function does not have a fixed point, which is being used to derive contradictions. The discussion also involves the properties of compact spaces and continuous functions.

l888l888l888
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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.
 
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Hi l888l888l888! :smile:

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?
 
you can infer that it attains its bounds on X. ?
 
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))
 
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)?
 
What is d(f(f(a)),f(a))?
 
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?
 
l888l888l888 said:
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!
 
Great. thanks!
 

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