The Dimension of R over Q
Point of post: Often in intermediate linear algebra courses it is taught that every vector space has a basis. Immediately after this most teachers will point out the mind-boggling fact that since is a vector space over , that there must be a set (the letter “H” chosen for historical reasons) such that every real number there is finitely many and such that . But, moreover the elements of are “spread out enough” that no one is a linear combination of the others.
At this point any inquisitive math student worth their weight in sawdust will then proceed to find such a set . The student may search for five minutes, maybe ten, scribbling violently on their paper while the lecturer lectures. But, the end of the class quickly comes and the student is defeated, . This set has eluded the student’s best tricks, his most clever maneuvers. It’s at this point that he/she may ask: “What could such a damned set look like?”
Here sits such a student, writing this post. When, I first heard (actually, proved when prompted by a book) that such a basis existed, I was flabbergasted. I honestly could not conceive such a set. In fact, it’s almost by definition unconceivable. That said, I grew curious as to what this set looked like. I wanted to know things about it: was it measure zero, was it dense, was it a subgroup? The first and foremost, in my mind at least, is how big must such a set be. In other words, what is ?
This post is the realization of that question. Within it I will begin by describing some of the methods shown in intermediate level courses as to half-answer this question, and end it by proving a slightly more general theorem which as a corollary will answer the question at hand; what is ?
So, as the preface promised I will give a few proofs as to why cannot be a finite-dimensional vector space over . This will partially sate a student’s interest, for it was intuitively clear that this is true.
This first proof is one that is sometimes cited, but usually using instead of (see below) and given with the caveat that “I won’t prove for you why is transcendental.” Here, we somewhat circumvent this.
Theorem: Let be a vector space over , then is infinite dimensional.
Proof (1): By a previous post we know that there exists transcendental numbers. So, let be such a transcendental number. Consider the set
Clearly this is set is infinite. Furthermore suppose that is any finite subset of and suppose there existed such that
Then, by simple multiplication we find that
Or, put differently where
Note though that (this is just notation for polynomials with integer coefficients) and since is transcendental it follows that
(otherwise we’d have that is algebraic!). Thus, since was arbitrary it follows that is a linearly independent set of vectors. But, note that is infinite, and so any basis for cannot be finite. The conclusion follows.
By far the shortest proof, although not the most insightful is:
Proof (2): Suppose that was finite dimensional, say . Then, by an elementary theorem in linear algebra we’d have that
In particular, we’d have that . But, this is clearly impossible and so cannot be finite dimensional.
Even though we have proved in several ways why cannot be a finite-dimensional vector space over , we are still left with the nagging question “Ok, it’s not finite dimensional. So what dimension is it?” This question is not as easily answered but with some work, can be dug out with a little set-theoretic know-how. So, we prove a (not much more) general theorem, which as a corollary will finally enable is to fill in the box
Theorem: Let be an infinite-dimensional vector space over a countable field . Then, .
Proof: We know that since is a vector space over it must admit some basis, call it . We will show that given such a basis that and , the rest will then follow from the Schroeder-Bernstein Theorem and the Dimension Theorem.
Luckily, the first of our goals, namely that is completely evident since and so the inclusion map
is an injection.
Thus, to prove that we notice that by definition
But, notice that for each such that we have that
and since is finite we may conclude that
Thus, we may conclude that
Notice though, that by inspecting this post we can see that we may rephrase this as
where we’ve implicitly used the fact that since is infinite that
since is in a sense the “smallest” infinite cardinal number. It follows by previous discussion that
which is what was to be proven.
So we can finally apply to this theorem to the immensely satisfying corollary:
Corollary: Consider the vector space over . Since is infinite and countable we see that
1. Jech, Thomas J. Set Theory. Berlin: Springer, 2003. Print.
2. Halmos, Paul R. Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print
3. Roman, Steven. Advanced Linear Algebra. New York: Springer, 2005. Print.