This question is about vector space $V$ over arbitrary field $k$ equipped with a billinear form $\langle \cdot,\cdot \rangle$.
The space $V$ is said to be Symplectic if $\langle \cdot,\cdot \rangle$ alternates. That id $\forall v \in V \; . \; v,v \rangle = 0$.
The space $V$ is said to be nonsingular, if for any $v \in V$, such that $v \neq 0$, there is $u \in V$ such that $\langle v,u \rangle \neq 0$.
A 2-dimesnsional subspace $h$ of $V$ is Hyperbolic plane, if can be spanned by a pair of vectors $u,v$ such that $\langle u, v \rangle =\pm1 $ and $\langle v,v \rangle = \langle u,u\rangle =0.$ Then, The space $V$ is hyperbolic if $V = \bigoplus_{i \in I} h_i$ with all $h_i$ being orthogonal hyperbolic planes.
I know that every finite-dimensional symplectic nonsingular vector space is hyperbolic. Is the sane result holds in in infinite dimensional case too?
The proof of finite-dimensional case used existence of maximal hyperbolic subspace $H$ of $V$, that must be equal to $V$ itself. The finite dimensionality was only used to prove that $\mathcal{H}$ is non-empty. I think that in the infinite dimensional case it is possible to use Zorn lemma to show that the maximal hyperbolic subspace exists.
That is, I want to show that $H_i$ is increasing chain in $\mathcal{H}$, then $\bigcup_i H_i$ is also hyperbolic. But this claim seems fishy in case of the infinite chain. However, if modify respective relation to mean $H_i \le H_j$ iff $H_j = H_i \oplus \bigoplus_{l \in L} h_l$, where $h_l$ are some orthogonal hyperbolic planes. I see no harm in doing so.
Is my reasoning correct? Otherwise, I would like to see an example of symplectic nonsingular vector space which is not hyperbolic due to its huge dimension.