All Hilbert spaces are isometric to $l^2(E)$ - how? I was reading about Hilbert spaces and came across this line on Wikipedia:

By choosing a Hilbert basis (i.e., a maximal orthonormal subset of
$L^2$ or any Hilbert space), one sees that all Hilbert spaces are
isometric to $ℓ^2(E)$, where $E$ is a set with an appropriate
cardinality.

My questions are:

*

*What does the $E$ stands for? Is it the basis of $l^2$?

*What is meant by "appropriate cardinality"?

*Why is $l^2$ isometric to any other Hilbert space? Yes, the norm on $l^2$ is square root of sum of squares. But how do we know the isometry applies, even when some Hilbert spaces have elements with n coordinates, while $l^2$ has infinite coordinates? (Because elements are sequences and each sequence is infinite - has infinite "coordinates").

Thank you very much for your insights.

Every Hilbert space $H$ admits an orthonormal basis $\{v_i : i \in E \}$. Then $H$ is isometric to $\ell^2(E)$.
EDIT: The $E$ is dependent upon the Hilbert space. Moreover, if $E, F$ are two abstract sets, then $\ell^2(E)$ and $\ell^2(F)$ are isometric if and only if $E, F$ are of the same cardinality. The point of this question is that every Hilbert space can be represented as $\ell^2$ over an (essentially) unique set $E$.
