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Prove this function on metric space X is onto! 
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#1
Mar712, 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 oneone. (2) => f is uniformly continuous. I tried to proceed by assuming the existence of y in X such that y has no preimage. That, and the fact that f is 11, 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 nonempty 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 oneone. 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. 


#2
Mar712, 11:37 AM

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P: 18,019

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 N1 sets. Can you prove that f(X) can be covered with N1 sets?? Can you deduce that X can be covered with N1 sets of radius smaller than ε?? 


#3
Mar712, 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 N1 sets. I'm a bit confused about what to do next... 


#4
Mar712, 12:01 PM

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P: 18,019

Prove this function on metric space X is onto!
Take the preimage of the open sets. Do these preimages have size <epsilon?



#5
Mar712, 12:13 PM

P: 85

The preimages are open and have diameter less than e because of condition (2).
And if p is in X, it must be contained in the preimage of whatever e/2 ball its image is in. So the preimages are an open cover of X with only N1 sets. But these preimages 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 N1 sets. 


#6
Mar712, 12:19 PM

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P: 18,019




#7
Mar712, 12:22 PM

P: 85

Yes, that would fix it....
Thank you! 


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