description/proof of that for vectors space, bases have same cardinality
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of basis of module.
- The reader knows a definition of dimension of vectors space.
- The reader admits the proposition that any vectors space has a basis.
- The reader admits the proposition that for any finite-dimensional vectors space, there is no basis that has more than the dimension number of elements.
- The reader admits the absorption law of cardinal arithmetic.
Target Context
- The reader will have a description and a proof of the proposition that for any vectors space, the bases have the same cardinality.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
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Statements:
\(\forall B_1, B_2 \in \{\text{ the bases for } V\} (\vert B_1 \vert = \vert B_2 \vert)\)
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\(V\) does not need to be finite-dimensional.
2: Proof
Whole Strategy: Step 1: see that \(V\) has a basis; Step 2: suppose that \(V\) is finite-dimensional; Step 3: see that \(\vert B_1 \vert = \vert B_2 \vert\); Step 4: suppose that \(V\) is infinite-dimensional; Step 5: let \(B_1 = \{b_{1, j} \vert J_1\}\) and \(B_2 = \{b_{2, j} \vert J_2\}\), for each \(b_{1, j} \in B_1\), take a finite subset \(J_{2, j} \subseteq J_2\), such that \(b_{1, j} = \sum_{l \in J_{2, j}} {b_{1, j}}^l b_{2, l}\), and see that \(\cup_{j \in J_1} J_{2, j} = J_2\); Step 6: see that \(\vert J_2 \vert \le \vert J_1 \vert\); Step 7: conclude the proposition.
Step 1:
\(V\) has a basis, by the proposition that any vectors space has a basis.
Step 2:
Let us suppose that \(V\) is finite-dimensional with any dimension, \(d\).
Step 3:
There is a basis with some \(d\) elements, by the definition of dimension of vectors space.
\(\vert B_j \vert = d\), because \(\vert B_j \vert\) cannot be larger than \(d\), by the proposition that for any finite-dimensional vectors space, there is no basis that has more than the dimension number of elements, but cannot be smaller than \(d\), by the definition of dimension of vectors space.
So, \(\vert B_1 \vert = \vert B_2 \vert = d\).
Step 4:
Let us suppose that \(V\) is infinite-dimensional.
Step 5:
Let \(B_1 = \{b_{1, j} \vert J_1\}\) and \(B_2 = \{b_{2, j} \vert J_2\}\).
For each \(b_{1, j} \in B_1\), there is a finite subset \(J_{2, j} \subseteq J_2\), such that \(b_{1, j} = \sum_{l \in J_{2, j}} {b_{1, j}}^l b_{2, l}\), by the definition of basis of module.
\(\cup_{j \in J_1} J_{2, j} \subseteq J_2\), obviously.
\(\cup_{j \in J_1} J_{2, j}\) is linearly independent as a subset of \(J_2\).
\(\cup_{j \in J_1} J_{2, j}\) spans \(V\), because for each \(v \in V\), \(v = \sum_{j \in J_{1, v}} v^j b_{1, j}\) for a finite \(J_{1, v} \subseteq J_1\), and \(v = \sum_{j \in J_{1, v}} v^j \sum_{l_j \in J_{2, j}} {b_{1, j}}^{l_j} b_{2, l_j}\), which is a linear combination of a finite subset of \(B_2\).
If \(J_2 \setminus \cup_{j \in J_1} J_{2, j} \neq \emptyset\), supposing \(l \in J_2 \setminus \cup_{j \in J_1} J_{2, j}\), \(b_{2, l}\) would be a finite combination of \(\cup_{j \in J_1} J_{2, j}\), a contradiction against that \(J_2\) is linearly independent.
So, \(\cup_{j \in J_1} J_{2, j} = J_2\).
Step 6:
By cardinality arithmetic, \(\vert J_2 \vert = \vert \cup_{j \in J_1} J_{2, j} \vert \le \sum _{j \in J_1} \vert J_{2, j} \vert \le \sum _{j \in J_1} \vert \mathbb{N} \vert\), because \(\vert J_{2, j} \vert \lt \vert \mathbb{N} \vert\), \(= \vert J_1 \vert \vert \mathbb{N} \vert = \vert J_1 \vert\), by the absorption law of cardinal arithmetic.
Step 7:
By symmetry, also \(\vert J_1 \vert \le \vert J_2 \vert\).
So, \(\vert J_1 \vert = \vert J_2 \vert\).
So, \(\vert B_1 \vert = \vert B_2 \vert\).
