A Hamel basis for $\ell^p$?

I am looking for an explicit example for a Hamel basis for $\ell^{p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one can express any element of the vector space as a finite linear combination of these. After some trying I could not write one explicitly. A quick google search did not reveal anything useful except for the proof of uncountability of a an infinite Hamel basis. Maybe I am being a bit silly but I don't think the answer is as obvious as for a Schauder basis for the same case.

So, what is an explicit example for a Hamel basis for $\ell^{p}$??

Solution 1:

The existence of a Hamel basis for $\ell^p$ cannot be proved without some of the axiom of choice, which in modern terms usually means that we cannot write it explicitly.

It is consistent with ZF+DC (a weak form of the axiom of choice which is sufficient to do a lot of the usual mathematics) that all sets of real numbers have Baire property, and in such model we have that every linear function from $\ell^p$ to itself is continuous.

It is also true (in ZF) that $\ell^p$ is separable for $1\leq p<\infty$. It is a known fact that continuous endomorphisms are determined completely by the countable dense set.

If there exists a Hamel basis then its cardinality is at least $\frak c$ (or rather exactly that), and therefore it has $2^\frak c$ many permutations, each extends uniquely to a linear automorphism, which is continuous.

Now, note that $\ell^p$ has size $\leq\frak c$ itself, since it is a separable metric space (and again, this is in fact $\frak c$) and therefore it has only $\frak c$ many continuous endomorphisms.

Cantor's theorem tells us that $2^\frak c\neq c$, and therefore in Shelah's model where every set of real numbers have the Baire property there is no Hamel basis for $\ell^p$.