I'm aware that the vector space of $\mathbb{R}$ over $\mathbb{Q}$ has a basis, and that this basis is uncountable (as we would otherwise have a contradiction.)

as is pointed out in the comment section, the basis must have cardinality $|\mathbb{R}|$.

[[for every $r \in \mathbb{R}$, there exists a finite collection of basis elements $b_2,...,b_n$ so that $r=q_1+q_2b_2+...+q_nb_n$ for $q_{i} \in \mathbb{Q}$ for all $i$.]]

Moreover, for every linearly independent set $A \subseteq \mathbb{R}$ there exists a basis so that it contains $A$.

My question is: when does an uncountable collection of linearly independent elements [over $\mathbb{Q}$] form a basis for $\mathbb{R}$ ?

While it is difficult to specify a basis, it seems that it wouldn't be too difficult to tell whether or not something specified is indeed a basis (namely that every $r \in \mathbb{R}$ is a linear combination of finitely many basis elements.) But if I pick an arbitrary set of uncountably many linearly independent elements, how can I tell if it's a basis?

edit: The reason I ask is that given, say, $\mathbb{R}^2$ as a vector space over the real numbers, $\mathbb{R}^2$ has dimension 2 and so if we show that any $2$ vectors are linearly independent, then they form a basis for $\mathbb{R}^2$.

However, $\mathbb{R}$ over the rationals has at least uncountably many basis elements. to prove that that an uncountable set is a basis, does it suffice to show their linear independence? The answer is clearly "no," given that we may remove a single basis element, and what remains will be an uncountable linearly independent set.

so, how can we tell when a linearly independent set is a basis?

  • $\begingroup$ hmm. Regarding the close vote, I figured the question was clear, but if there is some part in particular that is unclear, I'd be glad to edit. $\endgroup$ Mar 21, 2016 at 1:54
  • $\begingroup$ You can do better than saying that a basis is uncountable. In fact, it must have size $|\mathbb R|$. Moreover, there are ``simple'' $\mathbb Q$-linearly independent subsets of $\mathbb R$ of size $|\mathbb R|$, see here. On the other hand, no basis for $\mathbb R$ over $\mathbb Q$ is simple, in fact, they must be pathological and their existence uses the axiom of choice in an essential way. $\endgroup$ Mar 21, 2016 at 2:57
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    $\begingroup$ It bothers me that the question you close your post with is trivial: Suppose $A$ is independent and uncountable, and $a\in A$. Then $A\setminus\{a\}$ is also independent and uncountable, and does not span $\mathbb R$, so the answer to the question is obviously no. $\endgroup$ Mar 21, 2016 at 2:58
  • $\begingroup$ what an absolutely brilliant answer! And clearly, it has $|\mathbb{R}|$ but is not a basis for $\mathbb{R}$. I was aware that it depended on the axiom of choice. Could you direct me to a proof that no basis is simple? $\endgroup$ Mar 21, 2016 at 3:00
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    $\begingroup$ You may want to see one of the answers to this question for an example of how any basis is complicated (it cannot be Borel or even analytic, that is, it cannot be the continuous image of a Borel set). On the other hand, another answer points out that a basis does not need to be entirely hopeless (it can be measurable, albeit its measure must be 0. $\endgroup$ Mar 21, 2016 at 3:19

1 Answer 1


The problem is what the input to your criterion would be. Any candidate for a $\mathbb{Q}$-basis for $\mathbb{R}$ can't be explicitly constructed, so how exactly do we articulate our criterion? It doesn't appear we can say anything simpler than just stating the definition of a basis.

You ask in the comments if we can define the reals to be the vector space generated by an uncountable basis over $\mathbb{Q}$. If you replace "uncountable" (which means the cardinality of the continuum or higher) with "continuum," then this is true, provided that by "the reals" we mean only the rational vector space -- you will not be able to recover multiplication of arbitrary reals.

  • $\begingroup$ Yeah, I think there was a fundamenal misunderstanding on my part regarding how bases could be defined. I was wondering about a simple solution, but it appears as though this isnt possble $\endgroup$ Mar 21, 2016 at 20:48
  • $\begingroup$ Uncountable most certainly does not mean "of cardinality at least continuum." $\endgroup$ Mar 22, 2016 at 21:47
  • $\begingroup$ Fair enough -- I guess there are people out there who reject the continuum hypothesis. $\endgroup$ Mar 22, 2016 at 21:49

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