Why is $\{\sin(n\pi x)\}_{n \in \mathbb N}$ dense in $L^{2}([0,1])$

Question

Background to the question: I want to find dense subsets $U, V$ of the Hilbert Space $L^{2}([0,1])$ so that $U\cap V=0$. One candidate is obviously $U:=$the space of polynomials on $[0,1]$. And now I have seen that the other candidate would be $V=(f_{n})_{n}$

where $f_{n}(x)=\sin{(nx)}$ and $x \in [0,1]$. But I struggle to see why in this case $U \cap V=0$ and further why $V$ is in any case dense in $L^{2}([0,1])$

Any ideas?

Answer 1

Consider instead your $U$ and let $V$ be the space of nonconstant simple functions on $[0,1]$. Then $V$ is dense in $L^2[0,1]$ and $U \cap V = \emptyset$.

Answer 2

$V$ is not dense. A better solution to finding disjoint dense sets would be to consider your $U$ and the set $\{e^{x}p(x):p \,\text {is a polynomial} \}$. Can you verify both the facts in this case?

Answer 3

You could also let $U$ be the set of polynomials with rational coefficients, $V$ the set of polynomials with irrational coefficients.