• Support PF! Buy your school textbooks, materials and every day products Here!

A separable metric space and surjective, continuous function

  • Thread starter mahler1
  • Start date
  • #1
222
0
Homework Statement .

Let X, Y be metric spaces and ##f:X→Y## a continuous and surjective function. Prove that if X is separable then Y is separable.

The attempt at a solution.

I've tried to show separabilty of Y by exhibiting explicitly a dense enumerable subset of Y:

X is separable → ##\exists## ##E\subset X## such that E is a dense enumerable subset. Let's prove that f(E) is a dense enumerable subset of Y. Let ##y\in Y## and let ##ε_y>0##, f is surjective so there is ##x \in X## such that f(x)=y; and f is continuous, so ##f^{-1}B(f(x),ε_y)## is an open subset of X. By definition of open subset, there exists ##δ_x>0## such that ##B(x,δ_x) \subset f^{-1}B(f(x),ε_y)##. E is dense in X, then ##\exists## ##e \in E## : ##e \in B(x,δ_x)##. But this means that ##f(e) \in B(f(x),ε_y)##, which implies that f(E) is a dense subset of Y.

Here is my doubt: providing that what I've proved up to now is correct, I haven't got the slightest idea of how to prove that f(E) is enumerable. As a matter of fact, I am not at all convinced that this is even true. Couldn't be the case that the function sends all the elements of E to one single element in Y? Then f(E) would consist of only one element, I suppose that this being the case, f(E) wouldn't be enumerable in Y. Maybe what I have to prove is that the function can't send the domain E to a finite subset in Y. And as E is enumerable and f is surjective, then I would conclude f(E) is not uncountable, so the only thing f(E) can be enumerable. Am I correct in all of these?
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
If f(E) is finite, most people would qualify that as being countable. At any rate you are correct that the definition of separable has to allow for your dense set to be finite - for example if X is any separable space and Y is just a point.
 
  • #3
222
0
If f(E) is finite, most people would qualify that as being countable. At any rate you are correct that the definition of separable has to allow for your dense set to be finite - for example if X is any separable space and Y is just a point.
Maybe I am wrong and confused: I thought that a metric space was separable if it contained a dense enumerable subset, I mean, not only countable, but also not finite. I will check this in my textbook. If we only need to ask for the dense subset to be countable, then proving that a surjective function from a countable set to another implies that the codomain has to be countable is enough.
 
  • #4
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
Enumerable typically just means countable - I think your textbook intended to allow finite subsets. At any rate the common usage of separable allows for a finite dense subset (for example you can check wikipedia).
 

Related Threads on A separable metric space and surjective, continuous function

  • Last Post
Replies
2
Views
1K
Replies
4
Views
1K
Replies
0
Views
832
Replies
2
Views
3K
Replies
14
Views
5K
Replies
6
Views
2K
Replies
0
Views
839
  • Last Post
Replies
1
Views
5K
Replies
9
Views
415
  • Last Post
Replies
1
Views
5K
Top