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According to the book I am studying (Royden & Fitzpatrick):

We can infer from Zorn's Lemma that every linear space possesses a Hamel basis.

A Hamel Basis is defined as a subset $\mathcal{B} $ of a linear space $X$ provided each vector in $X$ is expressible as a unique $\bf{finite}$ linear combination of vectors in $\mathcal{B}$.

When I first read about the concept of Hamel bases (compared to other sets referred to as basis), I recognized that bases of finite spaces are Hamel basis, and that, for example, $\{x^k\}_{k=0}^\infty$ is a Hamel basis of $\mathbb{R}[x]$, the infinite-dimensional space of polynomials with real coefficients.

As comments have pointed out, there might not be a Hamel basis that is explicit to state for other common infinite-dimensional linear spaces ($\ell^p$ or $L^p$) - their Hamel bases would need to be uncountable.

However, even if we cannot explicitly characterize the whole basis, could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'? (For example, include $ \{x^k\}_{k=0}^\infty$ in the Hamel basis for $L^p([a,b])$). If so, what are the applications of this fact?

Asaf Karagila
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ertl
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    When the dimension is continuum, I think there is almost no hope for writing such explicit basis. Choice of one that includes a specific countable linearly independent set is possible though. – Sungjin Kim Jul 29 '18 at 05:32
  • Possible duplicate of A Hamel basis for $\ell^p$? – adfriedman Jul 29 '18 at 06:02
  • Thank you for the link! I changed the question to ask what (if any) 'benefits' there might be to being able to choose a Hamel basis including a certain linearly independent set of vectors. This is not a duplicate as far as I can tell. – ertl Jul 29 '18 at 06:26
  • Yes, an arbitrary linearly independent subset $S$ can be extended to a Hamel basis. Just apply Zorn's lemma to the family of all linearly independent sets which contain $S$, ordered by inclusion. The maximal element will be a Hamel basis. – mechanodroid Jul 29 '18 at 12:38

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could we choose one that includes a specific (countable) set of vectors as a subset, given they are 'independent'?

Yes, we can.

If so, what are the applications of this fact?

None.

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    Care to explain the "None"? – SBK Jul 29 '18 at 15:26
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    I think "none" is at least an exaggeration, if not just false. For example, a simple application of the fact that we can always extend a linearly independent set to a basis (assuming AC) is the existence of a discontinuous linear functional on any infinite dimensional normed space. – Rhys Steele Aug 01 '18 at 20:20