# Basis for the linear spacer $\ell^p$ [duplicate]

Is there any well known basis (Hamel basis) for the vector space $\ell^p$? And what about the cardinality of such basis? Is it countable?

• The Hammel basis is uncountable and it is impossible to see one since it requires the axiom of choice. – azarel Jun 10 '15 at 1:39
• Thanks for the cardinality. How can we deduce that? I know that using the axiom of choice we can prove the existance of a basis, but that doesn't imply that we need the A.C. to build one, does it? – Chilote Jun 10 '15 at 1:45
• To see that the cardinality is uncountable you can use the Baire category theorem. Regarding the other question, there are models of set theory where there are no Hammel basis for $\ell^p$, so although it is not obvious you really need AC. – azarel Jun 10 '15 at 1:54
• @Chilote: You may be right: The axiom of choice does not exclude the existence of an explicit Hamel basis for certain spaces, or? As another example the existence of ONB for Hilbert spaces requires the axiom of choice. However there are spaces that admit explicit ONB's anyway even in the inseparable case. As an example consider $\ell^2(\mathbb{R})$. It has the explicit ONB $e_s:=\chi_{\{s\}}$ for $s\in\mathbb{R}$. – C-Star-W-Star Jun 10 '15 at 8:28
• @azarel: Exploiting the axiom of choice does not exclude the existence of an explicit Hamel basis. As an example for inseparable Hilbert spaces the proof for ONB requires the axiom of choice, in general. However, there are examples of inseparable Hilbert spaces where there is a well-known explicit ONB; e.g. $\ell^2(\mathbb{R})$ has ONB $e_s:=\chi_{\{s\}}$ for $s\in\mathbb{R}$. – C-Star-W-Star Jun 10 '15 at 8:29