- #1

mahler1

- 222

- 0

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?