1104: For Vectors Space with Topology Induced by Metric Induced by Norm Induced by Inner Product, if Space Is Separable, It Has No Uncountable Orthonormal Subset

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description/proof of that for vectors space with topology induced by metric induced by norm induced by inner product, if space is separable, it has no uncountable orthonormal 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 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 separable topological space.

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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, if the space is separable, it has no uncountable orthonormal 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
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$ with any inner product, $⟨\bullet ,\bullet ⟩:V\times V\to F$, with the topology induced by the metric induced by the norm induced by the inner product
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Statements:
$V\in \{\text{ the separable topological spaces }\}$
$⟹$
$\mathrm{\neg }\mathrm{\exists }S\in \{\text{ the uncountable orthonormal subsets of }V\}$
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2: Note

Typically, although not necessarily, $V$ is a Hilbert space.

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3: Proof

Whole Strategy: Step 1: suppose that there was an $S$; Step 2: for each $s\in S$, take an open ball, $B_{s,\delta }\subseteq V$, such that such open balls were disjoint, and see that any dense subset of $V$ would need to have a point in each open ball; Step 3: conclude the proposition.

Step 1:

Let us suppose that there was an uncountable orthonormal subset, $S$.

Step 2:

For each $s\in S$, let us take an open ball, $B_{s,\delta }\subseteq V$, such that such open balls were disjoint, as follows.

For each $s,s^{\prime }\in S$ such that $s\ne s^{\prime }$, $dist(s,s^{\prime })=\sqrt{⟨s-s^{\prime },s-s^{\prime }⟩}=\sqrt{⟨s,s⟩+⟨s,-s^{\prime }⟩+⟨-s^{\prime },s⟩+⟨-s^{\prime },-s^{\prime }⟩}=\sqrt{1+0+0+1}=\sqrt{2}$.

Let $p\in B_{s,\delta }$ be any, which means that $dist(s,p)<\delta$.

$dist(s,s^{\prime })\le dist(s,p)+dist(p,s^{\prime })$, so, $dist(s,s^{\prime })-dist(s,p)\le dist(p,s^{\prime })$.

So, $\sqrt{2}-\delta <dist(p,s^{\prime })$.

So, we can take $\delta =\sqrt{2}-\delta$, which means that $\delta =\sqrt{2}/2$, then, $\delta <dist(p,s^{\prime })$, which implies that $p\notin B_{s^{\prime },\delta }$.

So, the open balls are disjoint.

As $S$ had some uncountable points, there would be some uncountable open balls.

Then, any dense subset of $V$ would need to have a point in each $B_{s,\delta }$, because otherwise, $s$ would not be in the closure of the dense subset.

Then, the dense subset would need to have some uncountable points, because there were those uncountable open balls, a contradiction against $V$'s being separable (having a countable dense subset).

Step 3:

So, the supposition was wrong, and there is no uncountable orthonormal subset of $V$.

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References

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