# Separable Hilbert space has a countable orthonormal basis

I 'm studying about Hilbert Spaces this semester, and the following is a Proposition from yesterday's class which I can't completely understand.

"Obviously,the closed linear span of $V\;$ coincides with $H$."

It doesn't seem so obvious to me. It might be really silly, but how do I know that the closed linear span of a dense subset of $H$ is also dense in $H$? I have the feeling that it's quite elementary but I'm new to Functional Analysis.

I would appreciate any help. Thanks in advnace!!

• Hint: linear span of $V$ contains $V$. – Wojowu Mar 21 '17 at 20:46

For any topological space $X$ and subspaces $A\subset B\subset X$ we have that $\overline{A}\subset\overline{B}$. Therefor if $A$ is dense then $B$ will be dense as well.
For this specific case, since $V$ is dense and $V\subset span(V)$ we get that $span(V)$ is dense.
The linear span of $V$ contains $V$, so its closure must contain the closure of $V$. But the closure of $V$ is all of $H$, so the closure of the linear span is also all of $H$.