Proof of Exhausting Sequences of Finite-Dimensional Subspaces in a Separable Real Hilbert Space, and Their Projections

We use Exhausting Sequence of Finite-Dimensional Subspaces of a Real Hilbert Space, the facts about $P_{N}$ in Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space, and the identities of Elementary Identities in a Real Inner Product Space and The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity. Convergence of a sequence $(z_{n})$ to $z$ in $(H,d)$ means that for every $\varepsilon>0$ there is $N$ with $|z_{n}-z|<\varepsilon$ for $n\ge N$ (Convergent Sequence in a Metric Space).

Claim 1. Let $D\subseteq H$ be a countable set that is dense in $H$, which exists by Separable Metric Space. $D$ is nonempty: $H\ne\varnothing$ as $0_{H}\in H$, and $\varnothing$ is not dense, since by Closure of a Subset of a Topological Space no point lies in the closure of $\varnothing$ (every open set containing a point fails to meet $\varnothing$; and $H$ itself is open). Hence by Countable Set there is a sequence $(y_{k})_{k\in\mathbb{N}}$ in $H$ whose set of terms is $D$. For $n\in\mathbb{N}$ let $y^{(n)}\in H^{n}$ be the $n$-tuple with components $y_{1},\dots,y_{n}$ (the restriction of $k\mapsto y_{k}$ to $[n]$) and put $H_{n}=\operatorname{span}(y^{(n)})$, a linear subspace of $H$ by claim 1 of The Span of a Finite Family is the Smallest Subspace Containing It. We verify the three conditions of Exhausting Sequence of Finite-Dimensional Subspaces of a Real Hilbert Space and closedness.

Closed and finite-dimensional. By Gram-Schmidt Orthonormalisation in a Real Inner Product Space §orthonormalisation, either $H_{n}=\{0_{H}\}$ or $H_{n}=\operatorname{span}(e)$ for an orthonormal tuple $e$, in which case $e^{\ast}$ is a basis of $H_{n}$. In the second case $H_{n}$ is closed by Projection onto the Span of an Orthonormal Tuple, and Coordinates on a Finite-Dimensional Subspace §closed and finite-dimensional by Finite-Dimensional Vector Space. In the first case it is finite-dimensional by that definition, and closed because $\{0_{H}\}=H^{\perp}$ (by Elementary Identities in a Real Inner Product Space §zero and Elementary Identities in a Real Inner Product Space §vanishing: $x\in H^{\perp}$ forces $\langle x,x\rangle=0$) and $H^{\perp}$ is closed by Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space §complement, $H$ being a closed linear subspace of itself (its complement $\varnothing$ is open).

Nondecreasing. By the recursion and restriction rules (claim 1 of Properties of Finite Sums of Vectors), $\sum_{k=1}^{n}c_{k}y_{k}=\sum_{k=1}^{n+1}c'_{k}y_{k}$ where $c'$ extends $c$ by $c'_{n+1}=0$ (the last summand being $0_{H}$ by claim 3 of Elementary Identities in a Vector Space), so $H_{n}\subseteq H_{n+1}$.

Density. Let $x\in H$ and $\varepsilon>0$. Since $x$ lies in the closure of $D$, Sequential Characterization of the Closure in a Metric Space provides a sequence $(a_{j})$ in $D$ converging to $x$, hence some $j$ with $|x-a_{j}|<\varepsilon$; and $a_{j}=y_{k}$ for some $k$, with $y_{k}\in H_{k}$ by claim 1 of The Span of a Finite Family is the Smallest Subspace Containing It. So $n=k$ and $y=y_{k}$ serve.

Claim 2. $P_{n}$ is linear by Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space §linear, hence so is $Q_{n}$ (differences of linear maps are linear, by the conditions of Linear Map and claim 5 of Elementary Identities in a Vector Space). By Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space §pythagoras, $|P_{n}x|\le|x|$ and $|Q_{n}x|=|x-P_{n}x|\le|x|$. By Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space §characterisation with $z=P_{n}x$, $Q_{n}x\in H_{n}^{\perp}$, and as $P_{n}x\in H_{n}$, $\langle P_{n}x,Q_{n}x\rangle=\langle Q_{n}x,P_{n}x\rangle=0$. Finally $P_{n}x\in H_{n}\subseteq H_{n+1}$, so the nearest-point property Nearest-Point Projection onto a Nonempty Closed Convex Subset of a Real Hilbert Space §existence of $P_{n+1}x$ in $H_{n+1}$ gives $|Q_{n+1}x|=|x-P_{n+1}x|\le|x-P_{n}x|=|Q_{n}x|$.

Claim 3. Let $x\in H$ and $\varepsilon>0$. 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 $|x-y|<\varepsilon$. For $n\ge N$ we have $H_{N}\subseteq H_{n}$ (by condition 1 of Exhausting Sequence of Finite-Dimensional Subspaces of a Real Hilbert Space: the set of $n\in\mathbb{N}$ with $n<N$ or $H_{N}\subseteq H_{n}$ contains $1$ and is closed under successor, so it is $\mathbb{N}$ by Principle of Induction for the Natural Numbers), so $y\in H_{n}$ and the nearest-point property gives $|Q_{n}x|=|x-P_{n}x|\le|x-y|<\varepsilon$. Since $d(Q_{n}x,0_{H})=|Q_{n}x-0_{H}|=|Q_{n}x|$ (Real Inner Product Space §distance, the additive inverse of $0_{H}$ being $0_{H}$ by claim 4 of Elementary Identities in a Vector Space), $(Q_{n}x)$ converges to $0_{H}$; and $P_{n}x=x-Q_{n}x$ converges to $x-0_{H}=x$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §linear-limits.

Claim 4. By linearity of $Q_{n}$, The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle and claim 2, $|Q_{n}x_{n}|\le|Q_{n}(x_{n}-x)|+|Q_{n}x|\le|x_{n}-x|+|Q_{n}x|$. Given $\varepsilon>0$, choose $N_{1}$ with $|x_{n}-x|<\varepsilon/2$ for $n\ge N_{1}$ and, by claim 3, $N_{2}$ with $|Q_{n}x|<\varepsilon/2$ for $n\ge N_{2}$; for $n\ge\max(N_{1},N_{2})$ (claim 1 of Elementary Properties of the Maximum of Two Elements) we get $|Q_{n}x_{n}|<\varepsilon$ (claim 8 of Elementary Order Arithmetic in an Ordered Field). Thus $(Q_{n}x_{n})$ converges to $0_{H}$, and $P_{n}x_{n}=x_{n}-Q_{n}x_{n}$ converges to $x-0_{H}=x$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §linear-limits.