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I want to find an example of a vector space with no base if we assume that axiom of choice is incorrect. This question might be duplicate so please alert me. Thanks.

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    $\begingroup$ I can't prove it, but I'd be surprised if one could establish a basis for the space of all (real- or complex-valued) sequences without the axiom of choice. $\endgroup$ – Daniel Fischer Aug 6 '13 at 22:06
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    $\begingroup$ @DanielFischer DERP. $\endgroup$ – Pedro Tamaroff Aug 6 '13 at 22:09
  • $\begingroup$ @PeterTamaroff Nope, COFFEE! $\endgroup$ – Daniel Fischer Aug 6 '13 at 22:09
  • $\begingroup$ @DanielFischer Why no both? $\endgroup$ – Pedro Tamaroff Aug 6 '13 at 22:10
  • $\begingroup$ Inspired by the comment of @ZhenLin below: If the axiom of choice is incorrect, the only thing we know is that there exists a setting where it fails. But we don't know which one, which probably makes it impossible to pin down a concrete vector space without a base. So probably, you should reformulate this part to something like "if we cannot invoke the axiom of choice". $\endgroup$ – azimut Aug 6 '13 at 22:24
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The proof is not as constructive as one would expect.

I suggest that you take a look at Andreas Blass' paper in which he proved that the existence of bases implies the axiom of choice. From the proof it is easy to construct a counterexample.

The proof idea is to take a family of non-empty sets, and to show that there exists a choice function (not quite exactly, Blass goes through another equivalence first). Assuming the axiom of choice fails, there is a family of non-empty sets which doesn't have a choice function. From this you can construct a counterexample to the principle that Blass is using in his equivalence, and then you can easily construct the vector space which doesn't have a basis.

On the other hand, it is not very hard to construct specific models in which there are specific vector spaces without bases. $\ell_2$ doesn't have a basis in Solovay's model, and generally in models of $\sf ZF+DC+BP$ (where $\sf BP$ is the statement that every set of real numbers have the Baire property).


See also:

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  • $\begingroup$ I was waiting for you! Thanks. What can we say about classification? $\endgroup$ – Dahlersit Aug 6 '13 at 22:27
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    $\begingroup$ I was just watching a movie, sorry it took so long! ;-) What do you mean by classification? $\endgroup$ – Asaf Karagila Aug 6 '13 at 22:27
  • $\begingroup$ :) - Classification of all vector spaces with this property. I'm not good at that subject obviously so this question might be nonsense. $\endgroup$ – Dahlersit Aug 6 '13 at 22:30
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    $\begingroup$ Ah. I don't think that this sort of thing exists. We know how to generate counterexamples, and we know to prove that certain counterexamples exist. But I don't think we have any idea what sort of wild creatures are out there. $\endgroup$ – Asaf Karagila Aug 6 '13 at 22:33
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I think for any field $F$, the $F$-vector space $F^{\mathbb{N}}$ of all sequences over $F$ provides a suitable example. (Note that the unit vectors only generate the subspace consisting of all vectors with finitely many non-zero entries.)

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  • $\begingroup$ Okay, can you classify all vector spaces with this property? $\endgroup$ – Dahlersit Aug 6 '13 at 22:10
  • $\begingroup$ @Dahlersit: This sounds like a new and much harder question. $\endgroup$ – azimut Aug 6 '13 at 22:16
  • $\begingroup$ Yes, can I edit/change my question? Is this appropriate? $\endgroup$ – Dahlersit Aug 6 '13 at 22:19
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    $\begingroup$ Just because we can't describe the basis doesn't mean it doesn't exist! For this particular example, if the axiom of choice holds up to a sufficiently large cardinal, then $F^{\mathbb{N}}$ would have a basis. But the axiom of choice could still fail higher up. $\endgroup$ – Zhen Lin Aug 6 '13 at 22:21
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    $\begingroup$ @Dahlersit: No, you shouldn't change your question in something different. Post a new one instead. $\endgroup$ – azimut Aug 6 '13 at 22:27

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