Separable Hilbert space has a countable orthonormal basis

Question

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!!

Answer 1

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.

Answer 2

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 $H$.

Answer 3

There are some details missing for fully understand why that holds:

As has been mentioned; $V\subset span(V)$ and this implies $\overline{V}\subset\overline{span(V)}$. On the other hand, $span(V)\subset \overline{V}$ (Because $\overline{V}=H$ is a vector space), then $\overline{span(V)}\subset \overline{\overline{V}}=\overline{V}$.

Then, since $\overline{V}=H$; $$ \overline{span(V)} \subset H\subset \overline{span(V)} \longrightarrow H=\overline{span(V)}. $$