1106: For Separable Hilbert Space and Subset, Closure of Subspace Generated by Subset Is Double-Orthogonal Complement of Subset

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description/proof of that for separable Hilbert space and subset, closure of subspace generated by subset is double-orthogonal complement of subset

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of Hilbert space.
• The reader knows a definition of topology induced by metric.
• The reader knows a definition of separable topological space.
• The reader knows a definition of sub-'vectors space' generated by subset of vectors space.
• The reader knows a definition of closure of subset of topological space.
• The reader knows a definition of orthogonal complement of subset of vectors space with inner product.
• The reader admits the proposition that for any complete metric space, any closed subspace is complete.
• The reader admits the proposition that for any separable topological space induced by any metric, any topological subspace is separable.
• The reader admits the proposition that any separable Hilbert space has an orthonormal Schauder basis.
• The reader admits the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
• The reader admits the proposition that for any vectors space with the topology induced by the metric induced by the norm induced by any inner product and any subspace, the closure of the subspace is a subspace.
• The reader admits the proposition that for any separable Hilbert space and any orthonormal subset, the subset can be expanded to be an orthonormal Schauder basis.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any separable Hilbert space and any subset, the closure of the subspace generated by the subset is the double-orthogonal complement of the subset.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\mathbb{R},\mathbb{C}\}$, with the canonical field structure
$(V^{\prime },dist)$: $=\{\text{ the separable Hilbert spaces }\}$, with any inner product, $⟨\bullet ,\bullet ⟩$, with the topology induced by $dist$
$S$: $\subseteq V^{\prime }$
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Statements:
$\overline{(S)}={S^{\perp }}^{\perp }$
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2: Note

It is not "$\bar{S}={S^{\perp }}^{\perp }$": the subspace generated by $S$ is required.

For example, when $S=\{s\}$ for a single $s$, $\bar{S}=\{s\}$, which does not equal ${S^{\perp }}^{\perp }$: while ${S^{\perp }}^{\perp }$ is a vectors subspace, $\bar{S}$ is not even any vectors subspace at all.

Of course, when $S$ is already a vectors subspace, $(S)=S$, and $\bar{S}={S^{\perp }}^{\perp }$ holds.

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3: Proof

Whole Strategy: Step 1: see that $S^{\perp }=(S{)}^{\perp }$; Step 2: see that for each closed subspace of $V^{\prime }$, $V\subseteq V^{\prime }$, $V={V^{\perp }}^{\perp }$; Step 3: see that ${\overline{(S)}}^{\perp }={(S)}^{\perp }$ and ${{\overline{(S)}}^{\perp }}^{\perp }={{(S)}^{\perp }}^{\perp }$; Step 4: conclude the proposition.

Step 1:

Let us see that $S^{\perp }=(S{)}^{\perp }$.

$(S)=Span(S)$, by Note for the definition of sub-'vectors space' generated by subset of vectors space.

For each $v\in S^{\perp }$, $v\in (S{)}^{\perp }$, because for each $c^1{s}_1+...+c^n{s}_n\in (S)$, $⟨v,c^1{s}_1+...+c^n{s}_n⟩=\overline{c^1}⟨v,s_1⟩+...+\overline{c^n}⟨v,s_n⟩=0+...+0=0$.

For each $v\in (S{)}^{\perp }$, $v\in S^{\perp }$, because $S\subseteq (S)$.

Step 2:

Let us see that for each closed subspace of $V^{\prime }$, $V\subseteq V^{\prime }$, $V={V^{\perp }}^{\perp }$, which is in fact a special case of this proposition, because $\overline{(V)}=V$, because $(V)=V$ as $V$ is already a vectors subspace and $\bar{V}=V$ as $V$ is already closed.

$V$ is a separable Hilbert space, by the proposition that for any complete metric space, any closed subspace is complete and the proposition that for any separable topological space induced by any metric, any topological subspace is separable.

$V$ has an orthonormal Schauder basis, $B=\{b_1,b_2,...\}$, by the proposition that any separable Hilbert space has an orthonormal Schauder basis.

$V^{\prime }$ has an orthonormal Schauder basis as an expansion of $B$, $B^{\prime }=\{b_1^{\prime },b_2^{\prime },...\}$, by the proposition that for any separable Hilbert space and any orthonormal subset, the subset can be expanded to be an orthonormal Schauder basis.

$V^{\prime }$ is nothing but the set of the elements that can be expressed as $\sum _j{v}^{\prime j}{b}_j^{\prime }$: each element of $V^{\prime }$ can be expressed such; each convergent $\sum _j{v}^{\prime j}{b}_j^{\prime }$ is a Cauchy sequence, so the convergence is in $V^{\prime }$, because $V^{\prime }$ is complete.

$V$ is nothing but the set of the elements that can be expressed as $\sum _j{v}^{\prime j}{b}_j^{\prime }$ where only $B$ elements appear as $b_j^{\prime }$ s: likewise.

$V^{\perp }$ is nothing but the set of the elements that can be expressed as $\sum _j{v}^{\prime j}{b}_j^{\prime }$ where only $B^{\prime }\setminus B$ elements appear as $b_j^{\prime }$ s, because each element of $V^{\perp }$ cannot have any $B$ element and such each $\sum _j{v}^{\prime j}{b}_j^{\prime }$ is in $V^{\perp }$.

Then, ${V^{\perp }}^{\perp }$ is nothing but the set of the elements that can be expressed as $\sum _j{v}^{\prime j}{b}_j^{\prime }$ where only $B$ elements appear as $b_j^{\prime }$ s, because each element of ${V^{\perp }}^{\perp }$ cannot have any $B^{\prime }\setminus B$ element and such each $\sum _j{v}^{\prime j}{b}_j^{\prime }$ is in ${V^{\perp }}^{\perp }$.

So, ${V^{\perp }}^{\perp }=V$.

Step 3:

Let us see that ${\overline{(S)}}^{\perp }={(S)}^{\perp }$.

${\overline{(S)}}^{\perp }\subseteq {(S)}^{\perp }$, because $(S)\subseteq \overline{(S)}$ and each element of ${\overline{(S)}}^{\perp }$ is perpendicular to each element of $(S)$, which is equal to or smaller than $\overline{(S)}$.

Let $v\in {(S)}^{\perp }$ be any. For each $v^{\prime }\in \overline{(S)}$, $⟨v,v^{\prime }⟩=0$? For each $ϵ$-'open ball' around $v^{\prime }$, $B_{v^{\prime },ϵ}\subseteq V^{\prime }$, there is a $v^{″}\in B_{v^{\prime },ϵ}\cap (S)$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset. $⟨v,v^{″}⟩=0$. $⟨v,v^{\prime }⟩=⟨v,v^{\prime }-v^{″}+v^{″}⟩=⟨v,v^{\prime }-v^{″}⟩+⟨v,v^{″}⟩=⟨v,v^{\prime }-v^{″}⟩+0=⟨v,v^{\prime }-v^{″}⟩\le \Vert v\Vert \Vert v^{\prime }-v^{″}\Vert$, by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, $<\Vert v\Vert ϵ$, but as $ϵ$ is arbitrary, $⟨v,v^{\prime }⟩=0$.

So, $v\in {\overline{(S)}}^{\perp }$, which means that ${(S)}^{\perp }\subseteq {\overline{(S)}}^{\perp }$.

So, ${\overline{(S)}}^{\perp }={(S)}^{\perp }$.

So, ${{\overline{(S)}}^{\perp }}^{\perp }={{(S)}^{\perp }}^{\perp }$.

Step 4:

By Step 2, ${{\overline{(S)}}^{\perp }}^{\perp }=\overline{(S)}$: $\overline{(S)}$ is a vectors subspace, by the proposition that for any vectors space with the topology induced by the metric induced by the norm induced by any inner product and any subspace, the closure of the subspace is a subspace.

By Step 1, ${{(S)}^{\perp }}^{\perp }={S^{\perp }}^{\perp }$.

So, $\overline{(S)}={S^{\perp }}^{\perp }$.

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References

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