Let $(H,\langle\cdot,\cdot\rangle))$ be a (real or complex) Hilbert space. Let $Y,Z\subseteq H$ be two closed linear subspaces. Let $\alpha(Y,Z)\in[0,\pi/2]$ be the unique number such that \begin{align} \cos \alpha(Y,Z) = \sup \{ |\langle y,z \rangle| : y\in Y, z\in Z, \|y\| = \|z\| = 1 \}. \end{align}
Construct an example of $H,Y,Z$ such that \begin{align} Y\cap Z = \{0\} \quad \text{ and }\quad \alpha(Y,Z) = 0. \end{align}
I know that $H$ must be infinite dimensional since this is impossible in finite dimensions. Any hints?
Let $H$ be $l_2$, $Y$ be spanned by $e_{2i}$ (vector that has $1$ on $2i$-th coordinate and $0$ on rest), so it consists of vectors that have $0$ on all odd coordinates, and $Z$ be spanned by $\cos\frac{1}{i} e_{2i} + \sin\frac{1}{i} e_{2i + 1}$ - vectors s.t. if $z^{2i} \neq 0$ then $\frac{z^{2i}}{z^{2i + 1}} = \cot \frac{1}{i}$.
Their intersection is $\{0\}$: if $y \in Y$, $y^{2k} \neq 0$ and $Z \ni z_n \to_n y$, then $z_n^{2k} \to_n y^{2k}$, thus $z_n^{2k + 1} = \tan \frac{1}{2k + 1} \cdot z_n^{2k} \to_n \tan \frac{1}{2k + 1} \cdot y^{2k} \neq 0$, but $z_n^{2k + 1} \to_n y^{2k + 1} = 0$ - contradiction.
But angle between $Y$ and $Z$ is zero, as $\langle e_{2i}, \cos\frac{1}{i} e_{2i} + \sin\frac{1}{i} e_{2i + 1}\rangle = \cos\frac{1}{i}$, which can be arbitrary close to $1$.