# Cardinality of a basis of an infinite-dimensional vector space

by andytoh
Tags: basis, cardinality, infinitedimensional, space, vector
 P: 363 I am reading "The linear algebra a beginning graduate student ought to know" by Golan, and I encountered a puzzling statement: Let V be a vector space (not necessarily finitely generated) over a field F. Prove that there exists a bijective function between any two bases of V. Hint: Use transfinite induction. If V is generated by a finite set (with n elements), then I know how to prove that any basis has at most n elements, and thus all bases will have the same number of elements. But for infinite-dimensional vector spaces, I'm confused. How do I use transfinite induction to prove that there is a bijective correspondence between two bases of V if V is infinite-dimensional?
 Emeritus Sci Advisor PF Gold P: 16,101 Doesn't the same proof work?
 P: 363 Well the proof of the finite case goes like this: G = {v_1,...,v_n} is a generator and D={w_1,...,w_m} is an independent set. Then join the w's to the beginning of G and remove the v's out until we're left with all the w's with some v's left over at the end, and so m
P: 363

## Cardinality of a basis of an infinite-dimensional vector space

Let G = {v_1,...,v_n,...} be a generator and D={w_1,...,w_m,...} an independent subset of an infinite-dimensional vector space V (after well-ordering G and D). Then
{w_1,v_1, v_2,...}
is linearly dependent and so one of the v's can be removed and the resulting set still span V. Continuing,
{w_1,w_2, v_1, v_2,...}
is linearly dependent and so one of the v's can again be removed. Continue in this manner, we have by transfinite induction (which applies since D is well-ordered):
D U G' spans V (where G' is a subset of G), and |D U G'| = |G|.
Thus |D|<=|G|.

So if A and B are two bases of V, then |A|<=|B| (since B generates V) and |B|<=|A| (since A generates V). Thus |A|=|B| by the Schroeder-Bernstein Theorem, i.e. there is a bijective correspondence between any two bases of V.

Am I on track?
P: 363
Here's my formal solution. Opinions are welcomed.
Attached Files
 My Solution.pdf (12.6 KB, 58 views)
 P: 363 Additions to my proof: Some of you may question whether we can actually find such a vector v_m1 from G to delete, since not every element of G will have an index from the natural numbers using its well-ordering relation (if G is not countable). But if no such index m1 from the natural numbers exists, then that would mean that every finite subset of B1 is linearly independent, so that by definition B1 itself is linearly independent, a contradiction. The same applies to the other v_mi from G. Also, I omitted some cardinal arithmetic at the end of my proof: |DUG'| = |G| |DU(G'-D)| = |G| (since D and G' are not necessarily disjoint) |D| + |G'-D| = |G| |D| = |G| - |G'-D| <= |G|
 P: 1 I think the solution has a problem. You wrote that : We have just shown above that if, ......Thus 0 D is inductive But I think you didn't prove the above clame. Since the proof you gave only for those W_n which has a predecessor element. In fact, you can well-order, but can't always well-order like natural numbers.
 P: 17 You have most of the proof down, just a few minor tweaks are needed. Hodge is right that you need to address the limit case as well, but I also think that you mean to construct B_n by adding w_n to B'_n-1 instead of just B_n-1, which isn't clear in your initial examples. It might be easier to prove that any basis of an uncountable vector space has the same cardinality as the vector space itself. EDIT: I just looked at the date of the OP's post. Looks like this thread has been dead for a long time.

 Related Discussions Quantum Physics 30 Linear & Abstract Algebra 7 Beyond the Standard Model 1 Linear & Abstract Algebra 8 Calculus 0