1099: For Vectors Space with Topology Induced by Metric Induced by Norm Induced by Inner Product, Orthogonal Complement of Subset of Vectors Space Is Closed Vectors Subspace

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description/proof of that for vectors space with topology induced by metric induced by norm induced by inner product, orthogonal complement of subset of vectors space is closed vectors subspace

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Topics

About: vectors space
About: topological space

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The table of contents of this article

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

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Starting Context

• The reader knows a definition of norm induced by inner product on real or complex vectors space.
• The reader knows a definition of metric induced by norm on real or complex vectors space.
• The reader knows a definition of topology induced by metric.
• The reader knows a definition of orthogonal complement of subset of vectors space with inner product.
• The reader admits the proposition that for any vectors space, any nonempty subset of the vectors space is a vectors subspace if and only if the subset is closed under linear combination.
• The reader admits the proposition that for any real or complex vectors space with the topology induced by the metric induced by the norm induced by any inner product, the inner product with any 1 argument fixed is a continuous map.
• The reader admits the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.

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Target Context

• The reader will have a description and a proof of the proposition that for any vectors space with the topology induced by the metric induced by the norm induced by any inner product, the orthogonal complement of any subset of the vectors space is a closed vectors subspace.

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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
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$, with the topology induced by the metric induced by the norm induced by any inner product, $⟨\bullet ,\bullet ⟩$
$S$: $\subseteq V$
$S^{\perp }$:
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Statements:
$S^{\perp }\in \{\text{ the closed vectors subspaces of }V\}$
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Without supposing any topology, $S^{\perp }$ is a vectors subspace of $V$, because that fact does not require any topology.

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2: Proof

Whole Strategy: Step 1: see that $S^{\perp }$ is a vectors subspace of $V$; Step 2: see that $S^{\perp }$ is a closed subset of $V$.

Step 1:

Let us see that $S^{\perp }$ is closed under linear combination.

Let $v_1,v_2\in S^{\perp }$ and $r_1,r_2\in F$ be any.

For each $s\in S$, $⟨r_1{v}_1+r_2{v}_2,s⟩=r_1⟨v_1,s⟩+r_2⟨v_2,s⟩=0+0=0$, so, $r_1{v}_1+r_2{v}_2\in S^{\perp }$.

By the proposition that for any vectors space, any nonempty subset of the vectors space is a vectors subspace if and only if the subset is closed under linear combination, $S^{\perp }$ is a vectors subspace of $V$.

Step 2:

For each $s\in S$, the map, $f_s:V\to F,v\mapsto ⟨v,s⟩$ is continuous, by the proposition that for any real or complex vectors space with the topology induced by the metric induced by the norm induced by any inner product, the inner product with any 1 argument fixed is a continuous map.

As $\{0\}\in F$ is a closed subset, ${f_s}^{-1}(\{0\})\subseteq V$ is a closed subset, by the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.

$S^{\perp }={\cap }_{s\in S}{f_s}^{-1}(0)$, because for each $v\in S^{\perp }$, for each $s\in S$, $f_s(v)=0$, which means that $v\in {f_s}^{-1}(0)$, so, $v\in {\cap }_{s\in S}{f_s}^{-1}(0)$; for each $v\in {\cap }_{s\in S}{f_s}^{-1}(0)$, for each $s\in S$, $v\in {f_s}^{-1}(0)$, which means that $f_s(v)=0$, which means that $v\in S^{\perp }$.

$S^{\perp }$ is closed as the intersection of the closed subsets.

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References

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