1103: For Vectors Space with Topology Induced by Metric Induced by Norm Induced by Inner Product and Subspace, Closure of Subspace Is Subspace

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for vectors space with topology induced by metric induced by norm induced by inner product and subspace, closure of subspace is subspace

---

Topics

About: vectors space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of %field name% vectors space.
• 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 closure of subset of topological space.
• 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 Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.

---

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 and any subspace, the closure of the subspace is a subspace.

---

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.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$V^{\prime }$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V$: $\in \{\text{ the vectors subspaces of }{V}^{\prime }\}$
//

Statements:
$\bar{V}\in \{\text{ the vectors subspaces of }{V}^{\prime }\}$
//

---

2: Proof

Whole Strategy: see that for each $v_1^{\prime },v_2^{\prime }\in \bar{V}$, $r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }\in \bar{V}$; Step 1: take any neighborhood of $r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }$, $N_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }}\subseteq V^{\prime }$, and an open ball around $r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }$, $B_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime },ϵ}$, such that $B_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime },ϵ}\subseteq N_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }}$; Step 2: take an open ball around $v_1^{\prime }$, $B_{v_1^{\prime },\delta }$, and an open ball around $v_2^{\prime }$, $B_{v_2^{\prime },\delta }$, take any $v_1\in B_{v_1^{\prime },\delta }\cap V$ and any $v_2\in B_{v_2^{\prime },\delta }\cap V$, and choose $\delta$ such that $r_1^{\prime }{v}_1+r_2^{\prime }{v}_2\in B_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime },ϵ}$.

Step 1:

Let $v_1^{\prime },v_2^{\prime }\in \bar{V}$ and $r_1^{\prime },r_2^{\prime }\in F$ be any.

We intend to see that $r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }\in \bar{V}$, because then, 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, $\bar{V}$ will be a vectors subspace.

When $r_1^{\prime }=r_2^{\prime }=0$, $r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }=0\in V\subseteq \bar{V}$.

Let us suppose that $r_1^{\prime }\ne 0$ or $r_2^{\prime }\ne 0$ hereafter.

Let any neighborhood of $r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }$ be $N_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }}\subseteq V^{\prime }$.

There is an open ball around $r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }$, $B_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime },ϵ}\subseteq V^{\prime }$, such that $B_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime },ϵ}\subseteq N_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }}$.

Step 2:

Let us take an open ball around $v_1^{\prime }$, $B_{v_1^{\prime },\delta }\subseteq V^{\prime }$, where $\delta$ will be determined later.

Let us take the open ball around $v_2^{\prime }$, $B_{v_2^{\prime },\delta }\subseteq V^{\prime }$.

As $v_1^{\prime }\in \bar{V}$, there is a $v_1\in V$ such that $v_1\in B_{v_1^{\prime },\delta }$, because $B_{v_1^{\prime },\delta }\cap V\ne \mathrm{\varnothing }$.

As $v_2^{\prime }\in \bar{V}$, there is a $v_2\in V$ such that $v_2\in B_{v_2^{\prime },\delta }$, because $B_{v_2^{\prime },\delta }\cap V\ne \mathrm{\varnothing }$.

We intend to choose $\delta$ such that $r_1^{\prime }{v}_1+r_2^{\prime }{v}_2\in B_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime },ϵ}$, because $r_1^{\prime }{v}_1+r_2^{\prime }{v}_2\in V$ and that will mean that $B_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime },ϵ}\cap V\ne \mathrm{\varnothing }$.

$⟨r_1^{\prime }{v}_1+r_2^{\prime }{v}_2-(r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime }),r_1^{\prime }{v}_1+r_2^{\prime }{v}_2-(r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime })⟩=⟨r_1^{\prime }(v_1-v_1^{\prime })+r_2^{\prime }(v_2-v_2^{\prime }),r_1^{\prime }(v_1-v_1^{\prime })+r_2^{\prime }(v_2-v_2^{\prime })⟩=|r_1^{\prime }{|}^2⟨v_1-v_1^{\prime },v_1-v_1^{\prime }⟩+|r_2^{\prime }{|}^2⟨v_2-v_2^{\prime },v_2-v_2^{\prime }⟩+r_1^{\prime }\overline{r_2^{\prime }}⟨v_1-v_1^{\prime },v_2-v_2^{\prime }⟩+r_2^{\prime }\overline{r_1^{\prime }}⟨v_2-v_2^{\prime },v_1-v_1^{\prime }⟩\le |r_1^{\prime }{|}^2⟨v_1-v_1^{\prime },v_1-v_1^{\prime }⟩+|r_2^{\prime }{|}^2⟨v_2-v_2^{\prime },v_2-v_2^{\prime }⟩+|r_1^{\prime }\overline{r_2^{\prime }}⟨v_1-v_1^{\prime },v_2-v_2^{\prime }⟩|+|r_2^{\prime }\overline{r_1^{\prime }}⟨v_2-v_2^{\prime },v_1-v_1^{\prime }⟩|=|r_1^{\prime }{|}^2⟨v_1-v_1^{\prime },v_1-v_1^{\prime }⟩+|r_2^{\prime }{|}^2⟨v_2-v_2^{\prime },v_2-v_2^{\prime }⟩+|r_1^{\prime }||r_2^{\prime }||⟨v_1-v_1^{\prime },v_2-v_2^{\prime }⟩|+|r_2^{\prime }||r_1^{\prime }||⟨v_2-v_2^{\prime },v_1-v_1^{\prime }⟩|=|r_1^{\prime }{|}^2⟨v_1-v_1^{\prime },v_1-v_1^{\prime }⟩+|r_2^{\prime }{|}^2⟨v_2-v_2^{\prime },v_2-v_2^{\prime }⟩+2|r_1^{\prime }||r_2^{\prime }||⟨v_1-v_1^{\prime },v_2-v_2^{\prime }⟩|$.

By the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, $|⟨v_1-v_1^{\prime },v_2-v_2^{\prime }⟩|\le \sqrt{⟨v_1-v_1^{\prime },v_1-v_1^{\prime }⟩}\sqrt{⟨v_2-v_2^{\prime },v_2-v_2^{\prime }⟩}$.

So, $⟨r_1^{\prime }{v}_1+r_2^{\prime }{v}_2-(r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime }),r_1^{\prime }{v}_1+r_2^{\prime }{v}_2-(r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime })⟩<|r_1^{\prime }{|}^2{\delta }^2+|r_2^{\prime }{|}^2{\delta }^2+2|r_1^{\prime }||r_2^{\prime }|{\delta }^2={\delta }^2(|r_1^{\prime }{|}^2+|r_2^{\prime }{|}^2+2|r_1^{\prime }||r_2^{\prime }|)$.

So, let us choose $\delta$ such that ${\delta }^2(|r_1^{\prime }{|}^2+|r_2^{\prime }{|}^2+2|r_1^{\prime }||r_2^{\prime }|)<{ϵ}^2$, so, ${\delta }^2<{ϵ}^2/(|r_1^{\prime }{|}^2+|r_2^{\prime }{|}^2+2|r_1^{\prime }||r_2^{\prime }|)$, which is possible.

Then, $⟨r_1^{\prime }{v}_1+r_2^{\prime }{v}_2-(r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime }),r_1^{\prime }{v}_1+r_2^{\prime }{v}_2-(r_1^{\prime }{v}_1^{\prime }+r_2^{\prime }{v}_2^{\prime })⟩<{ϵ}^2$, which means that $r_1^{\prime }{v}_1+r_2^{\prime }{v}_2\in B_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime },ϵ}\subseteq N_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }}$, which means that $r_1^{\prime }{v}_1+r_2^{\prime }{v}_2\in N_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }}\cap V$, which means that $N_{r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }}\cap V\ne \mathrm{\varnothing }$.

So, $r^{\prime 1}{v}_1^{\prime }+r^{\prime 2}{v}_2^{\prime }\in \bar{V}$.

So, 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, $\bar{V}$ is a vectors subspace of $V^{\prime }$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>