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Consider an infinite dimensional vector space $E$ and define $$S:=\left\{F \subset E\biggr| F\ne 0\text{ is a subspace of }E\right\}.$$ Endow $S$ with the reverse inclusion. Is it possible to find a space $E$ for which $S$ is not inductive?

Asaf Karagila
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Polando
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    When you say inductive you mean...? – Asaf Karagila Nov 05 '12 at 23:41
  • I mean that every totally ordered subset of $S$ has an upper bound. – Polando Nov 05 '12 at 23:49
  • I never knew that was the meaning of inductive. But very well. – Asaf Karagila Nov 05 '12 at 23:50
  • I adopt the definition given in Brezis' "Functional Analysis, Sobolev spaces and Partial Differential Equation", but if you say so I'm pretty sure that is not an universal convention! – Polando Nov 05 '12 at 23:53
  • I see. I can understand the origin of this terminology, but for future reference it is best to give a definition (in particular when the definition is very short, and you are pretty sure there is no universal convention). I hope my answer is to your liking. – Asaf Karagila Nov 05 '12 at 23:58

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Assuming the axiom of choice (otherwise there is ambiguity in the term "infinite dimensional" as the dimension may not be well-defined) every vector space has a basis.

We have that if $E$ has dimension $\kappa$ it has $2^\kappa$ subspaces, simply by considering a fixed basis of size $\kappa$ and considering all possible subsets of this basis, each spans a distinct subspace.

It is not hard to see that it is possible to have a chain of subspaces whose intersection is empty, simply enumerate the basis by a limit ordinal, $B=\{b_\alpha\mid\alpha<\delta\}$ and consider $V_\alpha=\operatorname{span}\{b_\beta\mid\alpha\leq\beta\}$ as a linearly ordered set of subspaces. Since $\delta$ has no maximal element the intersection of all these spaces is $\{0\}$ which is not in $S$.

Let us consider the case where the axiom of choice fails. In fact, let us consider the worst case scenario, something which I happened to have dealt with in my thesis, a space which is not finitely generated but every subset is finitely generated.

It is not hard to see that the subspaces of this space only have finite chains in $S$, because an infinite chain would have to correspond to an infinitely dimensional subspace, which is only the entire space.

If the chain is finite, it has a maximal element which is its upper-bound. As requested.

I discuss this construction in: Axiom of choice and automorphisms of vector spaces and you can replace the base field by any field, not just $\mathbb F_2$ (which is used to generate an abelian group without nontrivial automorphisms).

I should also add that the theory of general vector spaces without choice is not too well-understood. We know some stuff, like the existence of bases implies choice; but we don't know a lot. We know it is consistent to have two bases of different cardinalities; no basis at all (as in my example); it is possible to have a basis but a subspace without a basis; and so on. In particular it is difficult to give a complete and comprehensive answer because $S$ can become very strange without the axiom of choice!

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Asaf Karagila
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