Long time ago a student asked me what the dimension of $\mathbb{R}$ over $\mathbb{Q}$ is, and I said $$\dim_{\mathbb{Q}}\mathbb{R}=\mathfrak{c}$$ where $\mathfrak{c}$ is the cardinality of the continuum, since if there was a countable basis the real numbers would be countable too, since every real number could be written as a finite linear combination of countably many basis elements over a countable field. I added that we assume the continuum hypothesis. It was a course in linear Algebra I and I then felt that my answer ( I was a tutor) was sufficient. By now I think I should have been more explicit and may be should have given some hints how this could actually be proved. However keeping in mind it was the first semester for the students, does anyone have any suggestions how I could have answered more appropriately? Many thanks in advance!

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    $\begingroup$ Your answer was good. Sure, you could have given hints but not the answers, but this isnt a productive line of inquiry for a first semester of linear algebra so it's best not to spend too much time with it. $\endgroup$
    – Adam
    Commented Jun 25, 2023 at 0:46
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    $\begingroup$ Worth mentioning is that bases for the real numbers as a vector space over the field of rational numbers have been studied very extensively and form the basis (pun intended) for many counterexamples in various areas of analysis. See Applications of real numbers being a vector space over the rational numbers and this 2 October 1999 sci.math post. $\endgroup$ Commented Jun 25, 2023 at 19:48
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    $\begingroup$ Independently of any pedagogical issues, please note that the argument that you give for $\dim_{\mathbb{Q}} \mathbb{R} = \mathfrak{c}$ is mathematically not correct: what your argument, as currently worded, actually shows is that the dimension is uncountable - but this alone does not imply that it is equal to $\mathfrak{c}$ unless you assume the continuum hypothesis. $\endgroup$ Commented Dec 30, 2023 at 9:15
  • $\begingroup$ @Adam: Why isn't this a "productive line of inquiry"? I mean, sure, it's most likely not productive for the entire course, but if a good student asks this question it's likely that this is a productive question for them, and one can easily defer a detailed explanation to the end of the class or the office hours. $\endgroup$ Commented Dec 30, 2023 at 9:20
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    $\begingroup$ @JochenGlueck That is of course correct. If I remember it correctly ( it´s around twenty years ago) I even mentioned the continuum hypothesis. This post is related, especially the fifth answer which elaborates on this post. $\endgroup$ Commented Dec 30, 2023 at 10:23

2 Answers 2


I remember being confused by this as an undergraduate, or to be more honest, blind to the fact that I held a fundamental misconception that would take years to realise.

A student in a first linear algebra course might have had exposure to infinite series in Calculus II and maybe Fourier series in a physics class. Perhaps such a student has done some exploratory reading that planted the idea of a Hilbert Space such as $l^2$ and its norm. I know that I held the misconception that the dimension of $l^2$ was $\aleph_0$, perhaps even into graduate school. The prevalent use of infinite sums in Fourier series and $l^2$ was just too powerful for me to realise otherwise. My schema was not developed to the extent that I was bothered by the fact that the definition of a basis required finite linear combinations. Perhaps in my enthusiastic undergraduate "curiosity" readings I saw the phrase "Schauder basis." From my analysis class, I knew that polynomials were dense in $C^0[a,b]$, i.e., I knew about the Weierstrass Approximation Theorem, and this seemed consistent with my schema for Taylor series. A continuous function on $[a,b]$ has a nice Taylor series. I was not ready for my schema to upended. It felt as if everything was consistent with my belief that the countable set of monomials was a basis for $C^0[0,1]$.

Now if an inquisitive undergraduate in linear algebra were to ask me about the dimension of $\mathbb{R}$ over $\mathbb{Q}$, I would first ask probing questions to reveal their schema for polynomials, infinite series, Fourier series, Stone-Weierstrass Theorem, etc. I would test to see if their schema was flawed, as was mine. I would draw a parallel with the question at hand, namely the dimension of $\mathbb{R}$ over $\mathbb{Q}$.

On another note, a student in a first course in linear algebra class might not have a mental schema needed to appreciate the difference between countably infinite and the cardinality of $2^{\aleph_0}$. Perhaps this needs to be developed in the process of answering the question. We might also need some time to develop appreciation for the fact that the cardinality of the family of finite sets of $\aleph_0$ is still countably infinite.


It is possible that the argument

[...] since if there was a countable basis the real numbers would be countable too, since every real number could be written as a finite linear combination of countably many basis elements over a countable field.

is not correct, for there are models of $ZF(U)$ that satisfy the existence of a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ (called FORM 367. in Howard & Rubin) while falsifying the general statement

FORM 31. $UT(\aleph_0, \aleph_0, \aleph_0)$: The countable union theorem: The union of a denumerable set of denumerable sets is denumerable.

for example, the models $\mathcal{N}2$, $\mathcal{N}2(n)$, $\mathcal{N}2^*(3)$, $\mathcal{N}2(LO)$, $\mathcal{N}6$, $\mathcal{N}18$, $\mathcal{N}22(K, p)$, $\mathcal{N}35$, $\mathcal{N}41$, $\mathcal{N}43$, $\mathcal{N}50(E)$, loc. cit., but all these, being permutation models, satisfy the more specific

FORM 6. $UT(\aleph_0, \aleph_0, \aleph_0, \mathbb{R})$: The union of a denumerable family of denumerable subsets of $\mathbb{R}$ is denumerable.

so I'm not sure

  • $\begingroup$ Thanks for the input. Though we did not refer to a specific model of ZFC in this first course on linear algebra, we assumed the axiom of choice to be true and thus the existence of a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$. We certainly assumed the countable union theorem to be true, from which it follows that this basis cannot be countable. It is interesting to know there are models of ZF(U) where this is not that clear though. This is loosely related. $\endgroup$ Commented Dec 30, 2023 at 8:58
  • $\begingroup$ @PeterMelech of course I suppose you guys used ZF + full choice, which then guarantees both these theorems. I just couldn't help thinking about whether such a proof as you outlined would be possible :p $\endgroup$
    – ac15
    Commented Dec 30, 2023 at 16:12

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