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When defining a basis of a linear space
my textbook gives two definitions:

  • finite basis (linear space of finite measure/dimension $n$)
  • infinite (but countable) basis e.g. when the linear space
    consists of the the polynomials of a single variable P(x)
    the basis can be this one: $1, x, x^2, x^3, ... $

I wonder if there are linear spaces with
uncountable basis, and if they are useful in some cases?
What are some examples of such spaces and their bases?

peter.petrov
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  • The continuous functions on a compact real interval are a linear space with uncountable basis. – ViktorStein Sep 02 '21 at 13:07
  • It is not so much about if they are useful, but that they just naturally arise in mathematics. Likwise, finite, countably infinite and uncountably infinite sets arise in mathematics. – Dietrich Burde Sep 02 '21 at 13:08
  • @DietrichBurde Yeah, OK... that's probably what I meant to ask. – peter.petrov Sep 02 '21 at 13:09
  • @Ramanujan What could be one basis of that space? – peter.petrov Sep 02 '21 at 13:11
  • Every vector space has a basis. But one may not be able to find one explicitly. The proof uses the axiom of choice, so it is only an existence proof. Nobody has seen yet an explicit basis of the $\Bbb Q$-vector space $\Bbb R$ - see this post for example. – Dietrich Burde Sep 02 '21 at 13:19

2 Answers2

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The space of real-valued sequences $ℝ^ℕ$ has an uncountable (Hamel/Algebraic/Finitary)-basis.

Note that $B=\{e_k ∣ k∈ℕ\}$ is not a Hamel-basis as not every element of $ℝ^ℕ$ can be written as a finite linear combination of elements of $B$.

I'd love to give an example of a basis of $ℝ^ℕ$, but it is one of these axiom-of-choice kind situations.

Hyperplane
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A thing that comes to my mind is the following: all Banach spaces over $\mathbb R$ and $\mathbb C$ (except finite dimensional ones) have an uncountable Hamel basis (but this basis is not usually explicit). See the answers to the following question: Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$

Lorenzo Pompili
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