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In more rigorous language: " V: a vector space having an uncountable base S: The set of subspaces of V that have countable dimension. Can we construct explicitly a chain in the poset S (ordered by inclusion), such that this chain has NO upper bound in S? "

Apparently, this chain must have uncountable terms. Also,because S doesn't satisfy Zorn's lemma, we know such chain must exist in S.

But how do we construct it?

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It's very rare that something proven to exist by Zorn's lemma can be constructed – Harry Gindi Feb 15 '10 at 1:41
Others have given solutions that involve well ordering, presumably for 'explicitness', but I'm not sure that uncountable ordinals are really 'explicit'. Note that it doesn't make sense to order a random set by inclusion, so I assume that you mean to order subsets of $\mathbf S$. If that is so, then why not just let the chain be indexed by the countable subsets (under inclusion), with the subspace indexed by $\mathbf S'$ being its span? – L Spice Nov 17 '14 at 9:34
up vote 7 down vote accepted

The other answers asked you first to well-order the whole vector space (or a basis for it), and those answers are perfectly correct, but perhaps you don't like well order the whole space. So let me describe a construction that appeals directly to the Axiom of Choice.

Let V be your favorite vector space having uncountable dimension. For each countable dimension subpace W, let aW be an element of V that is not in W. Such a vector exists, since W is countable dimensional and V is not, and we choose such elements by the Axiom of Choice.

Having made these choices, the rest of the construction is now completely determined. Namely, we construct a linearly ordered chain of countable dimensional spaces, whose union is uncountable dimension. Let V0 be the trivial subspace. If Vα is defined and countable dimensional, let Vα+1 be the space spanned by Vα and the element aVα. If λ is a limit ordinal, let Vλ be the union of all earlier Vα. It is easy to see that { aVβ | β < α} is a basis for Vα. Thus, the dimension of each Vα is exactly the cardinality of α. In particular, if ω1 is the first uncountable ordinal, then Vω1 will have uncountable dimension, yet be the union of all Vα for α < ω1, which all have countable dimension, as desired.

If you forbid one to use the Axiom of Choice, then it is no longer true that every vector space has a basis (since it is consistent with ZF that some vector spaces have no basis), and the concept of dimension suffers in this case. But some interesting things happen. For example, it is consistent with the failure of AC that the reals are a countable union of countable sets. R = U An, where each An is countable. (The irritating difficulty is that although each An is countable, one cannot choose the functions witnessing this uniformly, since of course R is uncountable.) But in any case, we may regard R as a vector space over Q, and if we let Vn be the space spanned by A1 U ... U An, then we can still in each case make finitely many choices to witness the countability and conclude that each Vn is countable dimensional, but the union of all Vn is all of R, which is not countable dimensional.

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Some subset of a basis of $V$ can be put into bijection with the first uncountable ordinal $\omega_1$. Consider the subspaces spanned by each initial segment of $\omega_1$, all of which have countable dimension.

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I'm not sure if this will satisfy your sense of "construct" but order your uncountable base in some well-ordering and let your chain be {first basis element}, {first two basis elements}, etc up through all countable ordinals. Then this will not have an upper bound.

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Thank you James! – QuackQuack Feb 15 '10 at 2:18

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