725: For Vectors Space, Generator of Space, and Linearly Independent Subset Contained in Generator, Generator Can Be Reduced to Be Basis with Linearly Independent Subset Retained

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description/proof of that for vectors space, generator of space, and linearly independent subset contained in generator, generator can be reduced to be basis with linearly independent subset retained

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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: Natural Language Description
• 3: Note
• 4: Proof

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Starting Context

• The reader knows a definition of generator of module.
• The reader knows a definition of basis of module.
• The reader admits Zorn's lemma.

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Target Context

• The reader will have a description and a proof of the proposition that for any vectors space, any generator of the space, and any linearly independent subset contained in the generator, the generator can be reduced to be a basis with the linearly independent subset retained.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$S^{\prime }$: $\in \{\text{ the generators of }V\}$
$S$: $\subseteq S^{\prime }$, $\in \{\text{ the linearly independent subsets of }V\}$
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Statements:
$\mathrm{\exists }B\subseteq V(S\subseteq B\subseteq S^{\prime }\wedge B\in \{\text{ the bases of }V\})$
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2: Natural Language Description

For any field, $F$, any $F$ vectors space, $V$, any generator of $V$, $S^{\prime }$, and any linearly independent subset, $S\subseteq V$, such that $S\subseteq S^{\prime }$, there is a basis, $B\subseteq V$, such that $S\subseteq B\subseteq S^{\prime }$.

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3: Note

$V$ does not need to be finite-dimensional.

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4: Proof

Whole Strategy: Step 1: define the subset of $Pow(S^{\prime })$, $S^{″}$, such that each element contains $S$ and is linearly independent; Step 2: apply Zorn's lemma to $S^{″}$ and get any maximal element as $B$; Step 3: see that $B$ spans $S^{\prime }$.

Step 1:

Let us define $S^{″}:=\{p\in Pow(S^{\prime })|S\subseteq p\wedge p\in \{\text{ the linearly independent subsets of }V\}\}$.

Step 2:

Let us see that $S^{″}$ satisfies the conditions for Zorn's lemma.

$S^{″}$ is nonempty because $S\in S^{″}$. Let $S^{‴}\subseteq S^{″}$ be any nonempty chain, which means that for each $S_1,S_2\in S^{‴}$, $S_1\subseteq S_2$ or $S_2\subseteq S_1$. $\cup S^{‴}\in S^{″}$? $\cup S^{‴}=\{p\in S^{\prime }|\mathrm{\exists }{p}^{\prime }\in S^{‴}(p\in p^{\prime })\}$, so, $\cup S^{‴}\in Pow(S^{\prime })$. $S\subseteq \cup S^{‴}$, obviously. $\cup S^{‴}$ is linearly independent, because for each finite subset, $\{p_1,...,p_n\}\subseteq \cup S^{‴}$, there are some $S_1\in S^{‴},...,S_n\in S^{‴}$ such that $p_1\in S_1,...,p_n\in S_n$, but as $S^{‴}$ is a chain, there is an $S_k$ such that $S_j\subseteq S_k$ for each $j\in \{1,...,n\}$, so, $\{p_1,...,p_n\}\subseteq S_k$, and as $S_k$ is linearly independent, $\sum _{j\in \{1,...,n\}}{c}^j{p}_j=0$ has only all zero $c^j$ s. So, $\cup S^{‴}\in S^{″}$.

By Zorn's lemma, there is a maximal element, $B\in S^{″}$.

$S\subseteq B\subseteq S^{\prime }$.

$B$ is linearly independent as it is an element of $S^{″}$.

Step 3:

Let us see that $B$ spans $S^{\prime }$.

Let $p\in S^{\prime }\setminus B$ be any.

$B\cup \{p\}$ is linearly dependent, because otherwise, $B$ would not be maximal, because it would be $B\cup \{p\}\in S^{″}$ while $B\subset B\cup \{p\}$.

So, there is a finite subset, $\{b_1,...,b_n,p\}\subseteq B\cup \{p\}$, such that $\sum _{j\in \{1,...,n\}}{c}^j{b}_j+cp=0$ has a not-all-zero $(c^1,...,c^n,c)$. In fact, $c\ne 0$, because $\sum _{j\in \{1,...,n\}}{c}^j{b}_j=0$ would imply that all the $c^j$ s were $0$, because $B$ was linearly independent. So, $p=c^{-1}\sum _{j\in \{1,...,n\}}-c^j{b}_j$.

As $S^{\prime }$ spans $V$, $B$ spans $V$.

So, $B$ is a basis of $V$.

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References

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