I'm reading a proof about orthonormal basis in Hilbert spaces and got confused about a small part.
Let $\mathcal{H}$ be a Hilbert space, $\{ x_i : i \in I \}$ an orthonormal basis, $I$ is an uncountable set.
Denote $\mathcal{H_i} := \text{span }x_i$ and let $\oplus_{i \in I}\mathcal{H_i} $ be their orthogonal direct sum, i.e
$$x \in \oplus_{i \in I}\mathcal{H_i} \implies x = (\lambda_i x_i)_{i \in I}$$
and $\{ \|\lambda_ix_i\|^2 : i \in I \}$ is summable in $\mathbb{R}$.
Let $F$ denote the linear subspace of all $(\lambda_ix_i)_{i \in I}$ with $\lambda_i \neq 0$ for at most finite number of indices $i \in I$. Then $F$ is by definition dense in $\oplus_{i \in I} \mathcal{H}_i$
I couldn't convince myself of that. My reasoning is, since $\{\|\lambda_ix_i\|^2\}_{i \in I}$ is summable and the orthogonal direct sum of $1$-dimensional hilbert spaces is a hilbert space, then $\{\lambda_ix_i\}_{i \in I}$ is summable in $\oplus_{i \in I}\mathcal{H_i}$. Then $\lambda_ix_i \neq 0$ for at most countable number of indices.
So somehow, the elements where $\lambda_ix_i \neq 0$ for at most finite number of indices can approximate all those elements? How exactly?
Using definition of summability, we can say that $\big\{\sum_{i \in J} \lambda_ix_i\big\}$ where $J \subset I$ finite, approximate our actual sum $\sum_{i \in I} \lambda_ix_i$ arbitrarily well, but I can't see clearly how this implies that $F$ is dense.
Best Answer
The partial sums of the series are in $F$ and they converge to $x$. So every $x \in \oplus_{i \in I} \mathcal H_i$ is in the closure of $F$.