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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?

Andres Mejia
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  • 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. – Andres Mejia Mar 21 '16 at 01:54
  • 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. – Andrés E. Caicedo Mar 21 '16 at 02:57
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    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. – Andrés E. Caicedo Mar 21 '16 at 02:58
  • 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? – Andres Mejia Mar 21 '16 at 03:00
  • You are right, that was just my initial motivation for asking the question. I suppose I didn't want anyone to think the question was fruitless. Actually the real origin of the question was whether or not we could define the reals to be the vector space generated by an uncountable basis over the rationals...edited so that it doesn't seem like the edit is the question I'm ultimately trying to ask. – Andres Mejia Mar 21 '16 at 03:01
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    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. – Andrés E. Caicedo Mar 21 '16 at 03:19

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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.

Daniel McLaury
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