Prove this function on metric space X is onto

In summary, the conversation discusses proving that a function f:X->X, on a compact metric space (X,d), is onto (surjective) if it is one-one (injective) and uniformly continuous. The proof involves using the compactness of X and the properties of f to construct an open cover for X and f(X) with a finite number of sets, ultimately showing that f(X) can be covered with a smaller number of sets than X, which leads to a contradiction.
  • #1
Oster
85
0
(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.
 
Last edited:
Physics news on Phys.org
  • #2
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??
 
  • #3
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...
 
  • #4
Take the pre-image of the open sets. Do these pre-images have size <epsilon?
 
  • #5
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.
 
  • #6
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??
 
  • #7
Yes, that would fix it...
Thank you!
 

1. What does it mean for a function to be onto?

A function is onto if every element in the range of the function has at least one preimage in the domain. In other words, every element in the codomain has at least one corresponding element in the domain that maps to it.

2. How do you prove that a function is onto?

To prove that a function is onto, you must show that for every element in the codomain, there exists at least one element in the domain that maps to it. This can be done by either using the definition of an onto function or by finding the inverse of the function and showing that it is well-defined.

3. What is a metric space?

A metric space is a set of elements that have a notion of distance or similarity between them. This distance is defined by a function called a metric, which satisfies certain properties such as non-negativity, symmetry, and the triangle inequality.

4. Can a function on a metric space be onto but not one-to-one?

Yes, a function on a metric space can be onto but not one-to-one. In this case, there may be multiple elements in the domain that map to the same element in the codomain, but every element in the codomain still has at least one preimage in the domain.

5. Why is proving that a function is onto important in mathematics?

Proving that a function is onto is important in mathematics because it ensures that every element in the codomain is being mapped to by at least one element in the domain. This allows us to make conclusions about the structure of the function and its relationship to other mathematical concepts.

Similar threads

  • Calculus and Beyond Homework Help
Replies
19
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
560
  • Calculus and Beyond Homework Help
Replies
4
Views
635
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
785
  • Calculus and Beyond Homework Help
Replies
8
Views
348
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
457
Back
Top