All Questions
Tagged with axiom-of-choice vector-spaces
12
questions
4
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1
answer
328
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Automorphisms of vector spaces and the complex numbers without choice
In Zermelo-Fraenkel set theory without the Axiom of Choice (AC), it is consistent to say that there are models in which:
there are vector spaces without a basis;
the field of complex numbers $\mathbb{...
- 4,503
3
votes
0
answers
66
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Field automorphisms of projective spaces without the axiom of choice
Suppose P is a projective space over the field $k$. If P has finite dimension $n$, we can fix a base. Relative to this base, the full automorphism group of P can be described by the action on the ...
- 4,503
16
votes
1
answer
876
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Examples of vector spaces with bases of different cardinalities
In this question Sizes of bases of vector spaces without the axiom of choice it is said that "It is consistent [with ZF] that there are vector spaces that have two bases with completely different ...
- 163
0
votes
0
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175
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Eigenvalues without the axiom of choice
Without the Axiom of Choice (AC), we can find models of ZF set theory in which some vector spaces have no base, and also models in which some vector spaces have bases of different cardinalities.
The ...
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0
votes
0
answers
180
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Eigenbases without the Axiom of Choice
I understand that in ZF set theory without the Axiom of Choice (AC), it is consistent to have models in which there exist vector spaces over some (unspecified) field $k$ without a basis.
So in ...
- 4,503
0
votes
0
answers
133
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Axiom of Choice and bases of $k$-vector spaces, $k$ fixed
I know that from ZF + the Axiom of Choice (AC) follows that every vector space has a basis.
And, conversely, Blass proved that in ZF set theory, the assumption that every vector space has a basis ...
- 4,503
43
votes
2
answers
2k
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Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein?
If $V \hookrightarrow W$ and $W \hookrightarrow V$ are injective linear maps, then is there an isomorphism $V \cong W$?
If we assume the axiom of choice, the answer is yes: use the fact that every ...
- 62.6k
13
votes
1
answer
938
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Axiom(s) of choice and bases of vector spaces
I'm not sure this question is more suitable for MO or for MSE, so feel free to move it to MSE if necessary.
I work here in ZF theory. Consider the following statements:
$(C)$ Axiom of choice: for ...
- 1,661
8
votes
2
answers
394
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Relation between well-orderings of $\mathbb{R}$, and bases over $\mathbb{Q}$
The following question arose from a discussion about the definability of bases of $\mathbb{R}$ as a $\mathbb{Q}$-vector space.
(ZF without AC) something we can note is that the existence of a (...
- 14.4k
3
votes
2
answers
1k
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Linear space with (Hamel) basis and the axiom of choice
It is true that the axiom of choice is equivalent to the statement that every linear space has a Hamel basis. There are some linear spaces which definitely don't need axiom of choice to possess (...
- 9,240
7
votes
1
answer
353
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Sets of cardinalities of bases without choice
For a vector space $V$, let $BS(V)$ be the set of cardinalities (not necessarily $\aleph$s) of bases of $V$. Of course, in ZFC each $BS(V)$ is a singleton, but supposing the axiom of choice fails, it ...
- 20.7k
6
votes
2
answers
1k
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Properties of vector spaces without AC
With AC, it is easy to see that any vector space is injective, and free, therefore alse flat and projective.
Without AC, vector spaces can be not free. Are they must be projective modules? Flat ...
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