It is known that without the axiom of choice, not every vector space has a basis. But I was wondering, if I don't assume the axiom of choice, and I choose a vector space $V$ which does have a basis (assume it is infinite, otherwise the following question is trivial - it might also be trivial if it is infinite but I can't see why just yet). Does every subspace of $V$ have a basis ?
Bases in vector spaces without $AC$
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linear-algebra
vector-spaces
set-theory
axiom-of-choice
axioms
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2I suspect that this is not the case!! Very interesting. – 2017-02-13
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1I would think it's false too, but I was wondering whether (1) my intuition was false, (2) i assuming it's true, do we get full $AC$ or less ? (3) depending on the answer to (2), what does it imply ? – 2017-02-13
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3I think this is answered here: http://mathoverflow.net/questions/80765/if-v-is-a-vector-space-with-a-basis-w-subseteq-v-has-to-have-a-basis-too – 2017-02-13
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0Ok thanks, I'll look it up ! – 2017-02-13
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1The link by @CZiegler contains a counterexample. At the time I worked with Martin Goldstern a bit more, but we didn't get any substantial results beyond a mild generalization of his answer to me. – 2017-02-13
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0I see. But in the link CZiegler gave, Goldstern uses the negation of $CC(2)$ (countable choice for $2$-element sets). Did you get any results concerning the least amount of choice one needs to have the property I described ? – 2017-02-13
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0Nothing substantial. You can replace Countable Choice for Pairs by something along the lines of "well-orderable families of finite sets" or maybe slightly more general than that. Admittedly, I don't remember anymore. – 2017-02-13
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0@AsafKaragila. The Teichmuller-Tukey lemma (equivalent to AC) comes to mind. – 2017-02-22
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0@user254665: I don't understand your remark. – 2017-02-22