Cardinality of Hamel basis of a separable incomplete inner product space

Question

I know that cardinality of Hamel basis of an infinite dimensional separable Hilbert space is always equal to the cardinality of the continuum. But if an infinite dimensional separable inner product space is incomplete then is it true that the cardinality of Hamel basis may or may not be countable. Do we have examples for both the situation and how to determine the cardinality of Hamel basis under such conditions.

Answer 1

Let $c_{00}$ be the space of all sequences $(a_n)_{n\in\Bbb N}$ of real numbers such that $a_n=0$ if $n\gg1$ and consider the inner product$$\bigl\langle(a_n)_{n\in\Bbb N},(b_n)_{n\in\Bbb N}\bigr\rangle=\sum_{n=1}^\infty a_nb_n.$$If, for each $n\in\Bbb N$, $e(n)$ is the sequence $0,0,\ldots,0,1,0,0,\ldots$, whith the $1$ as the $n$^th element, then $\{e(n)\mid n\in\Bbb N\}$ is a Hamel basis of $c_{00}$.

And if $H$ is a non-closed hyperplane of $\ell^2$, then $H$ is a non-complete Hilbert space and any Hamel basis of $H$ will have the same cardinal as a Hamel basis of $\ell^2$, which is the cardinality of the continuum.