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I was reading up on the difference between countable and uncountable sets, and was wondering if there was a basis of uncountable size. I now know there are, however they all seem to be covering rather high level areas in math. So I was wondering; what is the simplest example of a basis of uncountable size for a vector space over a field?

Asaf Karagila
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Nik
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2 Answers2

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Pick $F$ to be some field, and $X$ to be some uncountable set. Consider now $V$ to be the vector space whose elements are functions $f\colon X\to F$ such that all but finitely many $x\in X$ satisfy $f(x)=0$.

The addition is pointwise addition, and scalar multiplication is pointwise multiplication. One can easily check that this is a vector space.

Now it is not hard to verify that $\{\delta_x\mid x\in X\}$ where $\delta_x$ is the function $$\delta_x(y)=\begin{cases}1 & x=y\\ 0 & x\neq y\end{cases}$$ is a basis for $V$, and it is clearly uncountable ($x\mapsto\delta_x$ is a bijection).

Perhaps it might be worth adding that using a set theoretic axiom known as The Axiom of Choice, or more often its useful equivalent Zorn's Lemma, we can prove that every vector space has a basis. And then we can conclude under certain conditions that this basis must be uncountable.

However the axiom of choice only assures us the existence of certain objects, it does not supply a description. In particular some naturally occurring vector spaces can be shown to have an uncountable basis using the axiom of choice. And we can show that this use of the axiom of choice is in fact necessary.

If we're mentioning the axiom of choice, perhaps it should also be pointed out that the axiom of choice is in fact equivalent to the statement "Every vector space has a basis".

Asaf Karagila
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  • Wow, the second part is really fascinating. Since the axiom of choice cannot be proven or disproven in the ZF system, I guess we also can't prove whether or not every vector space has a basis? Is there any vector space with an infinite or uncountable basis in the natural world? If there were, would that essentially "prove" the Axiom of Choice? – Rasputin Feb 08 '17 at 16:33
  • What is "the natural world"? The basis of $\Bbb R$ over $\Bbb Q$ is necessarily uncountable, and it is consistent that the axiom of choice fails and such basis does not exist. – Asaf Karagila Feb 08 '17 at 16:36
  • What exactly is it consistent with? Is it not also "consistent" that the axiom of choice is true and such a basis does exist? As for "the natural world", I don't know, maybe it's possible that DNA or black holes or something exist in a vector space that necessarily has an infinite basis. Maybe Hausdorff dimension could come into play here . . . – Rasputin Feb 17 '17 at 11:11
  • The axiom of choice implies that a Hamel basis exists. Again, the axiom of choice is equivalent to "Every vector space has a basis". – Asaf Karagila Feb 17 '17 at 11:13
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    Hmm ok, as for "the natural world", I don't know, maybe it's possible that DNA or black holes or something exist in a vector space that necessarily has an infinite basis. Maybe Hausdorff dimension could come into play here . . . – Rasputin Feb 17 '17 at 11:17
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    Riiiiiiiiight... – Asaf Karagila Feb 17 '17 at 11:19
  • In Quantum Mechanics, an application of functional analysis that was very influential in its development, the model of a particle moving in a continuous space requires the use of an uncountably infinite basis (the orthogonal eigenvectors of the position operator). In fact, any quantum system that has an observable continuous variable requires an uncountably infinite basis. Moreover, the basis given in this answer is commonly used for just that purpose. – Andrew Nov 20 '20 at 03:55
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Consider $\Bbb R$ as a vector space over the rational scalars $\Bbb Q$. As mentioned in Asaf's answer, this vector space must have a basis (at least if you accept the Axiom of Choice). However a countable subset of $\Bbb R$ will give rise to only countably many linear combinations with rational coefficients; therefore we need an uncountable set in order to form a basis.

David
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  • But what about the set {1}? It is trivially linearly independent and the span (multiplication by a scalar) will give rise to every real number, thus forming a basis. – Coward Jul 14 '15 at 18:59
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    @Coward Note that I said $\Bbb R$ is a vector space over the rationals. That is, multiplication by a scalar means multiplication by a rational scalar, and the span of ${1}$ gives rationals only, not all reals. – David Jul 15 '15 at 04:56
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    right! That bit of technicality.. should've paid more attention instead of embarrassing myself. – Coward Jul 23 '15 at 16:58