If I have a seperable Hilbert space does any countably infinite set of linearly independent vectors span the space? If I have a seperable Hilbert space does ANY countably infinite set of linearly independent vectors span the space?
Span in the sense that the set of all linear combinations of these vectors is dense in the Hilbert space.
My intuition says yes, but I am concerned there is some subtlety regarding convergence that I am missing.
 A: This is for sure not true. Take any separable Hilbert space with a countable Hilbert basis $\mathcal B = \{b_i \mid i \in I\}$ and $ j \in I$. Then $\mathcal B \setminus \{b_j\}$ can't be a basis while it is a countable set of linearly independent vectors.
A: Here is another answer, similar in spirit but actually holds for any Banach space, to complement the already accepted (and perfectly fine) answer given.
Let $H$ be a separable Hilbert space and let $\mathcal{B}$ be a Hamel (i.e. algebraic) basis for $H$. It follows from the Baire category theorem that $\mathcal{B}$ is necessarily uncountable because $H$ is infinite-dimensional. Now, let $V\subset\mathcal{B}$ be a countable subset of basis vectors and let $b\in\mathcal{B}\setminus V$ (such a vector exists because $V$ is countable while $\mathcal{B}$ is uncountable). It follows that $b\notin\text{span}(V)$ because the vectors of $\mathcal{B}$ (and, of course, any Hamel basis) are linearly independent so, in conclusion, $V$ is a countable set of vectors in $H$ whose span is not all of $H$.
