Hilbert dimension

Remarks

Example: in $\ell^2$, the unit vectors $e_n$, $n \ge 1$, form an orthonormal basis. They are orthonormal, and if $\ip{x}{e_n} = x_n = 0$ for all $n$ then $x = 0$. So the Hilbert dimension of $\ell^2$ is $\aleph_0$, the cardinality of $\N$.

Non-example: the family $(e_n)_{n \ge 2}$ in $\ell^2$ is orthonormal but not maximal, since $e_1$ is orthogonal to all of it. So it is not an orthonormal basis, though its cardinality is also $\aleph_0$.

For every orthonormal basis, each $x \in H$ is the sum $x = \sum_{i \in I} \ip{x}{e_i} e_i$, with only countably many nonzero terms. For finite-dimensional $H$, the Hilbert dimension is the usual dimension. For $H = L^2([0, 2\pi])$, the functions $e^{inx} / \sqrt{2\pi}$, $n \in \Z$, form an orthonormal basis, so $L^2([0, 2\pi])$ and $\ell^2$ are isometrically isomorphic.