Is there a more elementary proof of the fact that $L^2(\mathbb{R})$ is a separable Hilbert Space? The one I am aware of uses the fact that the set of polynomials is dense in $C([0,1])$ (the Stone-Weierstrass Theorem), but I wonder if something simpler can be done.
[Math] Proof of separability of $L^2(\mathbb{R})$ without Stone-Weierstrass Theorem
functional-analysishilbert-spaceslp-spacesseparable-spaces
Best Answer
If $S$ is the set of all functions of the form $$f=\sum_1^n\alpha_j\chi_{(a_j,b_j)}$$where $\alpha_j$, $a_j$ and $b_j$ are all rational then $S$ is countable and it's not hard to show $S$ is dense in $L^2$.