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? - MathOverflow

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?

2010-02-15T01:36:44Z

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?

Answer by Qiaochu Yuan for 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?

2010-02-15T01:57:45Z

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.

Answer by James Pfeiffer for 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?

2010-02-15T02:01:10Z

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.

Answer by Joel David Hamkins for 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?

2010-02-15T02:21:58Z

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_{V_α}. If λ is a limit ordinal, let V_λ be the union of all earlier V_α. It is easy to see that { a_{V_β} | 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.