Abstract Nonsense

Crushing one theorem at a time

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 \mathbb{R} is a vector space over \mathbb{Q}, that there must be a set \mathcal{H}\subseteq\mathbb{R} (the letter “H” chosen for historical reasons) such that every real number r there is finitely many x_1,\cdots,x_n\in\mathcal{H} and q_1,\cdots,q_n\in\mathbb{Q} such that r=q_1x_1+\cdots+q_nx_n. But, moreover the elements of \mathcal{H} 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 \mathcal{H}. 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, \text{Hamel Basis}-1,\text{ Student}-0. 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 \dim_{\mathbb{Q}}\mathbb{R}?

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 \dim_{\mathbb{Q}}\mathbb{R}?

So, as the preface promised I will give a few proofs as to why \mathbb{R} cannot be a finite-dimensional vector space over \mathbb{Q}. 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 \pi instead of \tau (see below) and given with the caveat that “I won’t prove for you why \pi is transcendental.”  Here, we somewhat circumvent this.

Theorem: Let \mathbb{R} be a vector space over \mathbb{Q}, then \mathbb{R} is infinite dimensional.

Proof (1): By a previous post we know that there exists transcendental numbers. So, let \tau be such a transcendental number. Consider the set

T=\left\{\tau^n:n\in\mathbb{N}\cup\{0\}\right\}

Clearly this is set is infinite. Furthermore suppose that \{\tau^{m_1},\cdots,\tau^{m_n}\} is any finite subset of T and suppose there existed \displaystyle \frac{p_1}{q_1},\cdots,\frac{p_n}{q_n}\in\mathbb{Q} such that

\displaystyle \frac{p_1}{q_1}\tau^{m_1}+\cdots+\frac{p_n}{q_n}\tau^{m_n}=0

Then, by simple multiplication we find that

\displaystyle \prod_{j\ne1}\frac{p_j}{q_j}\tau^{m_1}+\cdots+\prod_{j\ne n}\frac{p_j}{q_j}\tau^{m_n}=0

Or, put differently p(\tau)=0 where

\displaystyle p(x)=\sum_{i=1}^{n}\prod_{j\ne i}\frac{p_j}{q_j}x^{m_i}

Note though that p(x)\in\mathbb{Z}[x] (this is just notation for polynomials with integer coefficients) and since \tau is transcendental it follows that

\displaystyle \frac{p_1}{q_1}=\cdots=\frac{p_n}{q_n}=0

(otherwise we’d have that \tau is algebraic!). Thus, since \tau^{m_1},\cdots,\tau^{m_n} was arbitrary it follows that T is a linearly independent set of vectors. But, note that T is infinite, and so any basis for \mathbb{R} cannot be finite. The conclusion follows. \blacksquare

By far the shortest proof, although not the most insightful is:

Proof (2): Suppose that \mathbb{R} was finite dimensional, say \dim_{\mathbb{Q}}\mathbb{R}=n. Then, by an elementary theorem in linear algebra we’d have that

\mathbb{R}\cong\mathbb{Q}^n

In particular, we’d have that \text{card }\mathbb{R}=\text{card }\mathbb{Q}^n. But, this is clearly impossible and so \mathbb{R} cannot be finite dimensional. \blacksquare

 

Even though we  have proved in several ways why \mathbb{R} cannot be a finite-dimensional vector space over \mathbb{Q}, 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 \dim_{\mathbb{Q}}\mathbb{R}=\square

Theorem: Let \mathscr{V} be an infinite-dimensional vector space over a countable  field F. Then, \dim_{F}\mathscr{V}=|\mathscr{V}|.

Proof: We know that since \mathbb{R} is a vector space over \mathbb{Q} it must admit some basis, call it \mathcal{H}.  We will show that given such a basis that |\mathcal{H}|\leqslant|\mathscr{V}| and |\mathscr{V}|\leqslant|\mathcal{H}|, the rest will then follow from the Schroeder-Bernstein Theorem and the Dimension Theorem.

Luckily, the first of our goals, namely that |\mathcal{H}|\leqslant |\mathscr{V}| is completely evident since \mathcal{H}\subseteq\mathscr{V} and so the inclusion map

i:\mathcal{H}\hookrightarrow\mathscr{V}:x\mapsto x

is an injection.

Thus, to prove that |\mathscr{V}|\leqslant |\mathcal{H}| we notice that by definition

\displaystyle \mathbb{R}=\bigcup_{S\subseteq\mathcal{H}:|S|<\infty}\text{span }S

But, notice that for each S\subseteq\mathcal{H} such that |S|<\infty we have that

\text{span }S\cong F^{|S|}

and since |S| is finite we may conclude that

\left|\text{span }S\right|=\left|F^{|S|}\right|\leqslant \aleph_0

Thus, we may conclude that

\displaystyle \left|\mathscr{V}\right|=\text{card }\bigcup_{S\subseteq\mathcal{H}:|S|<\infty}\text{span }S\leqslant \left|\left\{S\subseteq\mathcal{H}:|S|<\infty\right\}\right|\cdot\aleph_0

Notice though, that by inspecting this post we can see that we may rephrase this as

|\mathscr{V}|\leqslant\left|\left\{S\subseteq\mathcal{H}:|S|<\infty\right\}\right|=\left|\mathcal{P}_0\left(\mathcal{H}\right)\right|\cdot\aleph_0=|\mathcal{H}|\cdot\aleph_0=|\mathcal{H}|

where we’ve implicitly used the fact that since |\mathcal{H}| is infinite that

|\mathcal{H}|\cdot\aleph_0=\text{max}\left\{|\mathcal{H}|,\aleph_0\right\}=|\mathcal{H}|

since \aleph_0 is in a sense the “smallest” infinite cardinal number. It follows by previous discussion that

|\mathscr{V}|=|\mathcal{H}|=\dim_F\mathscr{V}

which is what was to be proven. \blacksquare

So we can finally apply to this theorem to the immensely satisfying corollary:

Corollary: Consider the vector space \mathbb{R} over \mathbb{Q}. Since \mathscr{V} is infinite and \mathbb{Q} countable we see that

\dim_{\mathbb{Q}}\mathbb{R}=|\mathbb{R}|=2^{\aleph_0}

 

References:

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.

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October 22, 2010 - Posted by | Algebra, Linear Algebra, Uncategorized | , , ,

2 Comments »

  1. […] Since is a subfield of we know that is a -vector space in the obvious way. We have proven before on this blog that , and so we see that . Thus, and so as -vector spaces. In particular though, this implies […]

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  2. […] that infinite dimensional vector spaces aren’t isomorphic to their dual since, as we have proven before on this blog since is countable and lambda,kappa infinite we have that and . So, if the two are […]

    Pingback by Q and Q^2 are not Isomorphic as Groups « Abstract Nonsense | January 31, 2012 | Reply


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