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
Topics
About: vectors space
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
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.
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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(S'\): \(\in \{\text{ the generators of } V\}\)
\(S\): \(\subseteq S'\), \(\in \{\text{ the linearly independent subsets of } V\}\)
//
Statements:
\(\exists B \subseteq V (S \subseteq B \subseteq S' \land B \in \{\text{ the bases of } V\})\)
//
2: Natural Language Description
For any field, \(F\), any \(F\) vectors space, \(V\), any generator of \(V\), \(S'\), and any linearly independent subset, \(S \subseteq V\), such that \(S \subseteq S'\), there is a basis, \(B \subseteq V\), such that \(S \subseteq B \subseteq S'\).
3: Note
\(V\) does not need to be finite-dimensional.
4: Proof
Whole Strategy: Step 1: define the subset of \(Pow (S')\), \(S''\), 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 \(S'\).
Step 1:
Let us define \(S'' := \{p \in Pow (S') \vert S \subseteq p \land p \in \{\text{ the linearly independent subsets of } V\}\}\).
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 \(S_2 \subseteq S_1\). \(\cup S''' \in S''\)? \(\cup S''' = \{p \in S' \vert \exists p' \in S''' (p \in p')\}\), so, \(\cup S''' \in Pow (S')\). \(S \subseteq \cup S'''\), 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, \(B \in S''\).
\(S \subseteq B \subseteq S'\).
\(B\) is linearly independent as it is an element of \(S''\).
Step 3:
Let us see that \(B\) spans \(S'\).
Let \(p \in S' \setminus B\) 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 \(\sum_{j \in \{1, ..., n\}} c^j b_j + c p = 0\) has a not-all-zero \((c^1, ..., c^n, c)\). In fact, \(c \neq 0\), 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 \(S'\) spans \(V\), \(B\) spans \(V\).
So, \(B\) is a basis of \(V\).
