Proposition 9

There exists a subset C of E such that E is the Hilbert sum of the spaces $E_x$ for $x\in$ C. If E is of countable type, the set C is countable.

Let $\mathscr{O}$ be the set of subsets C of E such that the subspaces $E_x$ for $x\in C$ are pairwise orthogonal. The set $\mathscr{O}$, ordered by inclusion, is of finite character (E, III, p. 34, Def. 2) since C belongs to $\mathscr{O}$ if and only if the sets formed of two elements of C belong to $\mathscr{O}$. By E, III, p. 35, Theorem 1, there exists a maximal element C of $\mathscr{O}$.

Let F be the closed subspace of E generated by the subspaces $E_x$ for $x\in C$. It is enough to prove that $F^{\circ}$ is reduced to 0 in order to complete the proof of the proposition. Let $y$ be an element of $F^{\circ}$. For every $x\in C$ and every $a$ and $b$ in A, we have $\langle a(y)|b(x)\rangle =\langle y|a^*b(x)\rangle = 0$ since $a^*b(x)$ belongs to $E_x$, hence to F. Since the elements $a(y)$ (resp. $b(x)$) generate a dense subspace of $E_y$ (resp. $E_x$), we therefore have $E_y\subset E^{\circ}_x$. Since C is maximal in $\mathscr{O}$, this means that $E_y$ is reduced to 0, hence that $y= 0$.

Suppose that E is of countable type. Since $E_x$ is nonzero for every nonzero $x\in C$, every set C such that E is the Hilbert sum of the spaces $E_x$ for $x\in C$ is countable.