# Example of a case that shows that a Hilbert lattice of infinite dimension is not modular.

Let $H$ be a separable Hilbert space. Let $C(H)$ be the set of closed subspaces of the Hilbert space $H$.
Let $\le$ be the inclusion $\subseteq$. For $a,b \in C(H)$ define

• $a^\perp$ as the othogonal closed space of $a$,

• $a\wedge b = a \cap b$,

• $a \vee b= \{$ the smallest closed subspace that contains $a$ and $b \}$.

Using these definitions, $C(H)$ is an orthomodular lattice, and it will be modular if and only if $H$ has finite dimension.

N.B. A lattice $\mathcal L$ is said to be modular iff $$\forall a,b \in \mathcal L: a\le b \quad\Rightarrow \quad\forall c \in \mathcal{L}: a \vee (c \wedge b) = (a \vee c) \wedge b.$$

I'd like help with finding an example of three closed subspaces $a, b, c$ of an infinite dimensional Hilbert space $H$ such that $a\le b$ doesn't imply $a \vee (c \wedge b) = (a \vee c) \wedge b$.

• You can find an example in Redei's Quantum logic in Algebraic Approach, Proposition 4.4, pg. 49. More specifically, let $(e_n)$ be an orthonormal basis of $H$, $a>1$ and define a sequence $(\xi_n)$ as $\xi_n=e_{2n}+a^{-n}e_1+a^{-2n}e_{2n+1}$. Then set $F=[\xi_n: n\in \mathbb{N}]$, $G=[e_1, \xi_n: n\in \mathbb{N}]$ and $H=[e_{2n}: n\in\mathbb{N}].$ – tree detective Aug 14 '17 at 22:02