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Prove this function on metric space X is onto!

by Oster
Tags: function, metric, prove, space
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Oster
#1
Mar7-12, 10:24 AM
P: 85
(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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micromass
#2
Mar7-12, 11:37 AM
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P: 18,299
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??
Oster
#3
Mar7-12, 11:58 AM
P: 85
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...

micromass
#4
Mar7-12, 12:01 PM
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P: 18,299
Prove this function on metric space X is onto!

Take the pre-image of the open sets. Do these pre-images have size <epsilon?
Oster
#5
Mar7-12, 12:13 PM
P: 85
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.
micromass
#6
Mar7-12, 12:19 PM
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P: 18,299
Quote Quote by Oster View Post
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??
Oster
#7
Mar7-12, 12:22 PM
P: 85
Yes, that would fix it....
Thank you!


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