A separable metric space and surjective, continuous function

In summary: 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.
  • #1
mahler1
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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?
 
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  • #2
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.
 
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  • #3
Office_Shredder said:
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
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).
 
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1. What is a separable metric space?

A separable metric space is a mathematical concept in topology that describes a space where every point can be approximated by a sequence of points from a countable subset of the space. This means that the space has a dense subset, which is a set of points that are "close" to every point in the space.

2. How is a separable metric space different from a metric space?

A metric space is a mathematical concept that describes a space where the distance between any two points can be measured. A separable metric space is a type of metric space that has the additional property of having a countable dense subset.

3. What is a surjective function?

A surjective function is a type of function in mathematics that maps every element in the target set to at least one element in the domain set. In other words, every element in the output set has a corresponding input element.

4. How does a surjective function relate to a separable metric space?

In the context of a separable metric space, a surjective function is often used to map the dense subset to the entire space. This means that the function "covers" the entire space and can be used to approximate any point in the space with a point from the dense subset.

5. What is the significance of a continuous function in a separable metric space?

A continuous function is a type of function in mathematics that preserves the "closeness" of points. In a separable metric space, a continuous function is important because it ensures that the dense subset remains dense in the target space. This means that the approximations of points from the dense subset will still be "close" to the actual points in the space.

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