Let $\{\mathscr{H}_{i}\}_{i\in I}$ be a family of Hilbert spaces. Their direct sum, denoted by:
$$\mathscr{H} = \bigoplus_{i} \mathscr{H}_{i},$$
is defined as the space of all functions $\psi: I \to \bigcup_{i}\mathscr{H}_{i}$, with $\psi(i) \equiv \psi_{i} \in \mathscr{H}_{i}$ for each $i$ such that:
$$\sum_{i\in I}\|\psi_{i}\|^{2} < +\infty. \tag{1}\label{1}$$
It is a Hilbert space when equipped with the inner product:
$$\langle\psi,\phi\rangle = \sum_{i\in I}\langle \psi_{i},\phi_{i}\rangle. \tag{2}\label{2}$$
As far as I understand, the sum $\sum_{i\in I}x_{i}$ of real or complex numbers being convergent to, say, $x$ which is used in (\ref{1}) and (\ref{2}) means that given $\varepsilon > 0$ there exists a finite subset $F_{\varepsilon} \subset I$ such that:
$$|x-\sum_{i\in F_{\varepsilon}}x_{i}| < \varepsilon.$$
I have two questions about this definition, however.
(1) It is known that if an uncountable sequence of nonnegative real numbers is summable, then at most a countable number of its elements is nonzero. This is what happens in (\ref{1}). Does this imply that $\bigoplus_{i}\mathscr{H}_{i}$ is precisely the space of all sequences $\psi = (\psi_{i})_{i\in I}$ with at most a countable number of nonzero entries? And in this case, the inner product (\ref{2}) also becomes a (at most) countable sum? In other words, it is probably not standard, but can I define $\bigoplus_{i}\mathscr{H}_{i}$ as the space of sequences $\{\psi_{i}\}_{i\in I}$, $\psi_{i} \in \mathscr{H}_{i}$, with at most a countable number of nonzero entries with (\ref{2}) as its inner product?
(2) If each $\mathscr{H}_{i}$ is a closed subspace of a Hilbert space $\mathscr{H}$, then in particular each $\mathscr{H}_{i}$ is a Hilbert space itself and we can consider its direct sum as before. If $\mathscr{H}$ is separable, is it true that the set of indices $I$ can be at most countable? It seems natural because, in this case, $\mathscr{H}$ has a countable orthonormal basis, but I am not completely sure.
Best Answer
"as far as I understand, the sum $\sum_{i\in I}$..." Not exactly. This means that given $\varepsilon>0$, there exists a finite set $F_\varepsilon\subset I$ such that, for any finite set $F\subset I$ with $F_\varepsilon\subset F$, we have that $|x-\sum_{i\in F}x_i|<\varepsilon$.
(1) Definitely not: what if $I=\mathbb{N}$ already, say all Hilbert spaces $H_n$ are the same space $H$ and you take $(\psi_n)_{n\ge1}=(h)_{n\ge1}$? wouldn't the condition $\sum_{n\in\mathbb{N}}\|\psi_n\|^2<\infty$ fail? It is true that when an uncountable sum converges then at most countably many summands are non-zero, but this does not change anything. Why not define $\bigoplus_{i\in I}H_i$ as it is already defined?
(2) Again no, but for a stupid reason: just take any uncountable set $I$ and for each $i\in I$ set $H_i:=H$. This is an uncountable collection of Hilbert subspaces of $H$ and you can just as well form their direct sum. Now there is the inner direct sum construction, which refers to taking direct sums of mutually orthogonal Hilbert subspaces of a given Hilbert space. If you are confused with this and the question is "given a separable Hilbert space $H$ and a family $\{H_i\}_{i\in I}$ of mutually orthogonal subspaces, does it follow that $I$ is countable?", then the answer is yes: pick $E_i\subset H_i$ an orthonormal basis of $H_i$. Then take $E:=\bigcup_iE_i\subset H$. Then $E$ is an orthonormal set, which can thus be extended to an orthonormal basis of $H$. But then $|I|\le|E|\le|\mathbb{N}|$, so $I$ is countable.