Prove this function on metric space X is onto

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Homework Help Overview

The discussion revolves around proving that a function \( f: X \to X \) defined on a compact metric space \( (X, d) \) is onto, given that \( d(f(x), f(y)) = d(x, y) \) for all \( x, y \in X \). Participants explore properties of compactness, continuity, and the implications of the function being one-to-one.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of compactness and the properties of the function, including its injectivity and uniform continuity. There are attempts to construct sequences and cover sets to analyze the image of \( f \). Questions arise about the nature of open covers and the relationship between the function's image and the compact space.

Discussion Status

The discussion is ongoing, with participants providing insights and raising questions about the properties of the function and the implications of their assumptions. Some guidance is offered regarding the use of open covers and the relationship between the function's pre-images and the compact space, but no consensus has been reached.

Contextual Notes

Participants note the constraints of the problem, including the compactness of \( X \) and the implications of the function being one-to-one. There is also mention of specific assumptions regarding covers and the nature of limit points in relation to the image of the function.

Oster
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(1) (X,d) is a COMPACT metric space.

(2) f:X->X is a function such that
d(f(x),f(y))=d(x,y) for all x and y in (X,d)

Prove f is onto.

Things I know:

(2) => f is one-one.
(2) => f is uniformly continuous.

I tried to proceed by assuming the existence of y in X such that y has no pre-image.
That, and the fact that f is 1-1, implies that the sequence y(n)={f applied to y n times} is a sequence of distinct points. X is compact and hence y(n) has a convergent subsequence.

Also, X, f(X), f(f(X)),...and so on are all closed and nested (because f is continuous and X is compact?). Their intersection is non-empty because y(n) has a limit point which should be in the intersection? So, f restricted to the intersection is a continuous bijection.

Note: the case where X is finite can be solved by using the pigeonhole principle to show that Image(f) =/= X implies f is not one-one. And, loosely, compactness can be thought of as a generalization of finiteness...so...??

I really don't think I'm getting anywhere...
WHERE ARE YOU CONTRADICTION?

Please help. This is really bugging me.
 
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Let d(x,f(X))=ε. We know that ε>0 (why?).

We can cover X with sets of radius smaller than ε. Let's say we cover it with N sets and that we can't cover it with N-1 sets.
Can you prove that f(X) can be covered with N-1 sets?? Can you deduce that X can be covered with N-1 sets of radius smaller than ε??
 
f(X) is closed. So, if x is not in f(X), it is not a limit point and distance from f(X) is more than 0. Say the distance is 'e'.
{B(z,e/2) / z in X} is an open cover for X. Let {B(z_i,e/2) / i=1,2...N} be a minimal subcover. Say, x is in B(z_k,e/2). This open ball is contained in B(x,e) and hence it doesn't intersect f(X). So we can chuck this from the finite subcover and still be left with an open cover for f(X) with N-1 sets.
I'm a bit confused about what to do next...
 
Take the pre-image of the open sets. Do these pre-images have size <epsilon?
 
The pre-images are open and have diameter less than e because of condition (2).
And if p is in X, it must be contained in the pre-image of whatever e/2 ball its image is in.
So the pre-images are an open cover of X with only N-1 sets.
But these pre-images are not necessarily sets in the infinite cover I started with...
I used the assumption that my particular infinite subcover did not admit a finite subcover of less than N-1 sets.
 
Oster said:
The pre-images are open and have diameter less than e because of condition (2).
And if p is in X, it must be contained in the pre-image of whatever e/2 ball its image is in.
So the pre-images are an open cover of X with only N-1 sets.
But these pre-images are not necessarily sets in the infinite cover I started with...
I used the assumption that my particular infinite subcover did not admit a finite subcover of less than N-1 sets.

You started with a very specific subcover. What if you start with a subcover with the smallest number of sets??
 
Yes, that would fix it...
Thank you!
 

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