I want to understand better the structure of the vector space $\mathbb{R}$ over $\mathbb{Q}$. I know that it is an infinite dimensional vector space with a non countable Hamel basis, and it is cited in many works just for this properties. But, searching online, I don't find any satisfactory treatment of the properties of such space. E.g. :

What is its dual space? and the dual of the dual?

There is some way to characterize his subspaces? there is some equivalence relation between them (as for the dimension in countable vector spaces) and, in general, we can define a dimension of the subspaces?

Is it possible to define an inner product on this space?

So: I don't want here an answer to all this question, but I ask if there is some accessible resurse where I can find a specific and exhaustive study of this space that seem to me very interesting.

  • $\begingroup$ The usual inner product is also one when the space is regarded over the rationals. $\endgroup$ – Tobias Kildetoft Jun 4 '15 at 19:11
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    $\begingroup$ Just a bit of a grammar tip, in English if you are referring to something that's not alive (like a vector subspace) the possessive is "its" instead of "his" (i.e. "What is his dual space?" $\rightarrow$ "What is its dual space?"). It's clear what you mean, I just thought you might like to know. $\endgroup$ – Matt Samuel Jun 4 '15 at 19:16
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    $\begingroup$ @Matt :Thank a lot ! My English is very primitive :) $\endgroup$ – Emilio Novati Jun 4 '15 at 19:22

The reason that you can't find any references on this is that there's nothing particularly special about $\mathbb{R}$ as a vector space over $\mathbb{Q}$ -- it's isomorphic to any other vector space over $\mathbb{Q}$ whose basis has the same cardinality.

In particular, define the dimension of a vector space to be the cardinality of any basis for the vector space. It is not hard to see that $\dim(V) = |V|$ for any infinite-dimensional vector space over $\mathbb{Q}$, and thus $\mathbb{R}$ is a vector space of dimension $|\mathbb{R}|$. Two vectors spaces over $\mathbb{Q}$ are isomorphic if and only if they have the same dimension.

If $B$ is a basis for $\mathbb{R}$ over $\mathbb{Q}$, then linear functionals from $\mathbb{R}$ to $\mathbb{Q}$ are in one-to-one correspondence with functions $B\to \mathbb{Q}$. In particular, the dual space is isomorphic to $\mathbb{Q}^\mathbb{R}$, the vector space of all functions $\mathbb{R}\to\mathbb{Q}$, whose cardinality is the same as that of the power set of $\mathbb{R}$.

In general, if $V$ is any infinite-dimensional vector space over $\mathbb{Q}$, then $|V^*| = |\mathcal{P}(V)|$. By Cantor's theorem, it follows that $$ |\mathbb{R}| \;<\; |\mathbb{R}^*| \;<\; |\mathbb{R}^{**}| \;<\; |\mathbb{R}^{***}| \;<\; \cdots. $$

As for subspaces, every subspaces of $\mathbb{R}$ has a dimension as defined above, which must be a cardinal less than or equal to $|\mathbb{R}|$. Thus there are finite-dimensional subspaces of every possible dimension, as well as countable-dimensional subspaces and subspaces isomorphic to $\mathbb{R}$. If you believe the continuum hypothesis then these are the only possibilities, but if you don't believe the continuum hypothesis then there are subspaces of other intermediate cardinalities.

Every subspace $S$ also has a codimension, which is the dimension of $\mathbb{R}/S$, or equivalently the dimension of any complementary subspace to $S$. If $\dim(S) < |\mathbb{R}|$, then $\mathrm{codim}(S) = |\mathbb{R}|$, but if $\dim(S) = |\mathbb{R}|$, then $\mathrm{codim}(S)$ may be any cardinal less than or equal to $|\mathbb{R}|$. If $S$ and $S'$ are subspaces of $\mathbb{R}$, then there exists a $\mathbb{Q}$-linear bijection $\mathbb{R}\to\mathbb{R}$ taking $S$ to $S'$ if and only if $S$ and $S'$ have the same dimension and the same codimension.

It is possible to put a $\mathbb{Q}$-valued inner product on $\mathbb{R}$, but it isn't very meaningful. Basically, for any basis $B$ of $\mathbb{R}$, there exists a $\mathbb{Q}$-valued inner product on $\mathbb{R}$ under which that basis is orthonormal. This is basically defined by taking the "dot product" of the coefficient vectors of two real numbers with respect to this basis. More generally, it is possible to put a $\mathbb{Q}$-valued inner product on any vector space over $\mathbb{Q}$ by choosing a basis to be orthonormal.

  • $\begingroup$ thank you! Do you know some reference useful to better understand such kind of vector spaces over $\mathbb{Q}$? $\endgroup$ – Emilio Novati Jun 4 '15 at 20:08
  • $\begingroup$ @EmilioNovati There's really not much to know. Most advanced linear algebra or advanced abstract algebra books prove most of the basic theory of vector spaces without assuming finite-dimensionality -- just look for a book that uses Zorn's lemma to prove that every vector space has a basis. $\endgroup$ – Jim Belk Jun 4 '15 at 20:17
  • $\begingroup$ I've same problem to figure how we can define a dot product for a non countable basis. How can I find the coefficents of a real number with repect to basis if I can not know all the elements of the basis? $\endgroup$ – Emilio Novati Jun 4 '15 at 20:18
  • $\begingroup$ @EmilioNovati You can't. This is all abstract -- you can't actually find a basis for $\mathbb{R}$ over $\mathbb{Q}$ without the axiom of choice, which means that practically speaking you can't make one. However, it is true that any set of $\mathbb{Q}$-linearly independent real numbers extends to a basis. For example, if you want a basis that contains $\{1,\pi,\sqrt{2}\}$, you can have it, in which case $1$, $\pi$, and $\sqrt{2}$ will be orthonormal. $\endgroup$ – Jim Belk Jun 4 '15 at 20:23

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