Basis and cardinality

Question

Suppose that $B$ is a separable Hilbert space and $\{b_i\}_{i \in \mathbb{N}}$ is a countable linearly independent set. Then is $\{b_i\}_{i \in \mathbb{N}}$ necessarily a Schauder basis for $B$?

Answer 1

Obviously not, because for example any subset is also independent, but a proper subset of a basis cannot be a basis.

Answer 2

No. For example, the sequence $$\begin{align} b_0 &= (1,0,0,0,0,\dots)\\ b_1 &= (0,0,1,0,0,\dots)\\ b_2 &= (0,0,0,0,1,\dots)\\ \vdots &= \vdots \end{align}$$

is linearly independent in $\ell^2$, but not a Schauder basis. ($b_n$ has a $1$ in position $2n$ and zeros elsewhere.)

Answer 3

Besides taking a proper subset of a Schauder basis, you can also take a superset that is still (algebraically) linearly independent. For instance, if $(e_n)_{n\in\mathbb{N}}$ is an orthonormal basis, then $\{e_n\}_{n\in\mathbb{N}}\cup\{v\}$ is still linearly independent where $v=\sum_n e_n/2^n$, but is not a Schauder basis since the representation of $v$ as an infinite linear combination of basis elements is not unique.