Proof of The Subspaces Spanned by an Orthonormal Sequence and Exhausting Sequences

For each $n\in\mathbb{N}$ the tuple $e^{(n)}$ is orthonormal, since its components are components of the orthonormal sequence $(e_{k})$ and distinct indices in $[n]$ are distinct in $\mathbb{N}$. Write $P^{(n)}$ for the projection onto $H_{n}=\operatorname{span}(e^{(n)})$ of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace, so that $P^{(n)}x=\sum_{k=1}^{n}\langle x,e_{k}\rangle e_{k}$.

Claim 1. By The Span of a Finite Family is the Smallest Subspace Containing It the set $H_{n}$ is a linear subspace of $H$, and it is closed in $H$ by claim 4 of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace, hence a closed linear subspace. Since $|e_{1}|=1$ we have $e_{1}\ne0_{H}$, so $H_{n}\ne\{0_{H}\}$; apply claim 1 of Gram-Schmidt Orthonormalisation in a Real Inner Product Space to the tuple $e^{(n)}$, whose span is $H_{n}$. Since $e^{(n)}$ is itself an orthonormal $n$-tuple with $\operatorname{span}(e^{(n)})=H_{n}$, it is a tuple of the kind to which the second sentence of that claim refers, so the corresponding tuple in $H_{n}$ is a basis of $H_{n}$; hence $H_{n}$ is finite-dimensional.

For the inclusion, let $u\in H_{n}$, say $u=\sum_{k=1}^{n}c_{k}e_{k}$ with $c\in\mathbb{R}^{n}$. Let $c'\in\mathbb{R}^{n+1}$ have components $c'_{k}=c_{k}$ for $k\in[n]$ and $c'_{n+1}=0$. By the recursion of Finite Sum Notation in a Vector Space and claim 1 of Properties of Finite Sums of Vectors,

$$\sum_{k=1}^{n+1}c'_{k}e_{k}=\Bigl(\sum_{k=1}^{n}c_{k}e_{k}\Bigr)+0\,e_{n+1}=u+0_{H}=u,$$

using claim 3 of Elementary Identities in a Vector Space for $0\,e_{n+1}=0_{H}$. Hence $u\in H_{n+1}$ and $H_{n}\subseteq H_{n+1}$.

Finally, $H$ is a real Hilbert space and $H_{n}$ is closed, so by claim 5 of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace the map $P^{(n)}$ is the orthogonal projection $P_{H_{n}}$, which is the asserted formula.

Claim 2. Suppose first that $(e_{k})_{k\in\mathbb{N}}$ is an orthonormal basis of $H$. Conditions 1 and 2 of Exhausting Sequence of Finite-Dimensional Subspaces of a Real Hilbert Space hold by claim 1. For condition 3, let $x\in H$ and let $\varepsilon$ be positive. By claim 3 of Orthonormal Expansions in a Real Hilbert Space the series $\sum_{k=1}^{\infty}\langle x,e_{k}\rangle e_{k}$ converges to $x$, so there is $n\in\mathbb{N}$ with $|x-P_{H_{n}}x|<\varepsilon$, and $P_{H_{n}}x\in H_{n}$. Hence $(H_{n})_{n\in\mathbb{N}}$ is an exhausting sequence for $H$.

Conversely, suppose $(H_{n})_{n\in\mathbb{N}}$ is an exhausting sequence for $H$, and let $x\in H$ satisfy $\langle x,e_{k}\rangle=0$ for every $k\in\mathbb{N}$. By claim 3 of Properties of Finite Sums of Vectors, applied with the factor $0$, and claim 1 above,

$$P_{H_{n}}x=\sum_{k=1}^{n}0\,e_{k}=0_{H}\qquad\text{for every }n\in\mathbb{N}.$$

Let $\varepsilon$ be positive. By condition 3 of Exhausting Sequence of Finite-Dimensional Subspaces of a Real Hilbert Space there are $n\in\mathbb{N}$ and $y\in H_{n}$ with $|x-y|<\varepsilon$. By claim 5 of Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace the point $P_{H_{n}}x$ is a nearest point of $H_{n}$ to $x$, so

$$|x|=|x-0_{H}|=|x-P_{H_{n}}x|\le|x-y|<\varepsilon .$$

As $\varepsilon$ was an arbitrary positive real number, Comparison of Real Numbers with Arbitrary Positive Slack gives $|x|\le0$; since $0\le|x|$, we get $|x|=0$ and hence $\langle x,x\rangle=0$, so $x=0_{H}$ by condition (d) of Real Inner Product Spaces (i). Thus $(e_{k})_{k\in\mathbb{N}}$ is an orthonormal basis of $H$.