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I have a question about the vector space of the real numbers $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Is this vector space separable?

I found the assertion, that every separable vector space has a countable basis, and also the assertion, that the aforementioned vector space does not have a countable basis. But $\mathbb{Q}$ is dense in that vector space and it is also countable, so the vector space would be separable.

Where am I making a mistake?

Thank you for your help!

Best,

Luke

Peter Wacken
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  • $\mathbb{R}$ has a countable basis, namely the intervals $(q_1,q_2)$ for $q_1,q_2\in\mathbb{Q}$ form a basis for the standard topology on $\mathbb{R}$. – Levent Dec 13 '16 at 19:30
  • I might be wrong, but it's possible two different versions of the word "basis" are being used here? One in the topological sense, the other in the vector space sense. – TomGrubb Dec 13 '16 at 19:32
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    @bburGsamohT Three, actually - basis in the topological sense, and two meanings of basis in the vector sense - Hamel and Schauder. – Noah Schweber Dec 13 '16 at 19:32
  • @NoahSchweber Looked at your answer seconds after I commented haha; had not heard of the Schauder basis before, thanks for the reference. – TomGrubb Dec 13 '16 at 19:33
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    What does it mean for a vector space to be separable? I know what it means for a topological space to be separable, but an abstract vector space doesn't carry a topology. – Nate Eldredge Dec 13 '16 at 19:51
  • You could add to your post where you found the assertion that every separable vector space has a countable basis. That might help clarify in which sense you are using the words separable and basis. – Martin Sleziak Dec 14 '16 at 13:52

2 Answers2

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You are confusing topological separability and metric separability. While $\mathbb{R}$ is a separable topological space, as witnessed by $\mathbb{Q}$, it is not separable as a $\mathbb{Q}$-vector space. However, this isn't because every (Hamel) basis is uncountable; every Hamel basis of every infinite-dimensional Banach space is uncountable, but there are still infinite-dimensional separable vector spaces. The relevant notion of basis for separability is a Schauder basis.

Incidentally, $\mathbb{R}$ as a $\mathbb{Q}$-vector space has something vaguely like a Schauder basis - every element of $\mathbb{R}$ can be approximated arbitrarily well by a rational multiple of $1$, so in some sense $\{1\}$ is an "approximate Schauder basis" for $\mathbb{R}$ (and in a previous version of this answer, I mistook this for a Schauder basis).

Noah Schweber
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A Hilbert space is a separable topological space (contains a countable dense subset) if and only if it has a countable orthonormal Hilbert basis. See this question.

Note the difference between a Hilbert basis and a Hamel basis. In a Hamel basis, every vector is a finite linear combination of basis vectors, but in a Hilbert basis, every vector is uniquely expressible as an infinite linear combination of basis vectors, defined as the limit of partial sums. This requires the Hilbert space to have a metric to define limits, so does not apply in an abstract vector space.

For example, if $V=\ell^2(\mathbb{N})$ is the space of functions $f:\mathbb{N}\to\mathbb{R}$ with finite $\ell^2$ norm, modulo functions with vanishing $\ell^2$ norm, equipped with a topology coming from the $\ell^2$ norm, then the characteristic functions of singleton sets (i.e. Kronecker delta functions) form a countable Hilbert basis (not a Hamel basis), and their $\mathbb{Q}$-span (i.e. the finite rational linear combinations of them, which are functions $f:\mathbb{N}\to\mathbb{Q}$ with finite support) is a countable dense subset (isomorphic as an abstract vector space to $\mathbb{Q}^{\oplus\mathbb{N}}$).

Hilbert spaces are complete (i.e. every Cauchy sequence converges) and defined over $\mathbb{R}$ or $\mathbb{C}$, so that automatically bars $\mathbb{R}$ from being a Hilbert space over $\mathbb{Q}$. There are more general notions of topological vector spaces than Hilbert spaces as well (e.g. Banach spaces as in Noah's answer).

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