lo.logic - How do we construct (in a vector space) a chain of countable dimensional subspaces that can only be bounded by an subspace of uncountable dimension?

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 a_W 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 V_α is defined and countable dimensional, let V_α+1 be the space spanned by V_α and the element a_Vα. If λ is a limit ordinal, let V_λ be the union of all earlier V_α. It is easy to see that { a_Vβ | β < α} is a basis for V_α. Thus, the dimension of each V_α is exactly the cardinality of α. In particular, if ω₁ is the first uncountable ordinal, then V_ω1 will have uncountable dimension, yet be the union of all V_α 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 A_n, where each A_n is countable. (The irritating difficulty is that although each A_n 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 V_n be the space spanned by A₁ U ... U A_n, then we can still in each case make finitely many choices to witness the countability and conclude that each V_n is countable dimensional, but the union of all V_n is all of R, which is not countable dimensional.