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A separable metric space and surjective, continuous function

  1. Sep 19, 2013 #1
    The problem statement, all variables and given/known data.

    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?
     
  2. jcsd
  3. Sep 19, 2013 #2

    Office_Shredder

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    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.
     
  4. Sep 19, 2013 #3
    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.
     
  5. Sep 19, 2013 #4

    Office_Shredder

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    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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