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 $\ell^p$.
It seems that the proof using the Baire category theorem can be found in several places on this site, but none of those questions is an exact duplicate of this one. Therefore I'm posting a CW-answer, so that this question is not left unanswered.
We assume that a Banach space $X$ has a countable basis $\{v_n; n\in\mathbb N\}$. Let us denote $X_n=[v_1,\dots,v_n]$.
Then we have:
So we see that $\operatorname{Int} \overline{X_n} = \operatorname{Int} X_n=\emptyset$, which means that $X_n$ is nowhere dense. So $X$ is a countable union of nowhere dense subsets, which contradicts the Baire category theorem.
Some further references:
Other questions and answers on MSE
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Books
- Corollary 5.23 in Infinite Dimensional Analysis: A Hitchhiker's Guide by Charalambos D. Aliprantis, Kim C. Border.
- A Short Course on Banach Space Theory By N. L. Carothers, p.25
- Exercise 1.81 in Banach Space Theory: The Basis for Linear and Nonlinear Analysis by Marián Fabian, Petr Habala, Petr Hájek, Vicente Montesinos, Václav Zizler
Best Answer
Take any infinite dimensional vector space. Take a set $S$ of countably infinitely many linearly independent vectors from that vector space. Be $V$ the subspace spanned by $S$ (using finite linear combinations). Then the vectors in $S$ form a Hamel basis of $V$.
Thus $V$ has a countably infinite Hamel basis.
Note however, that in a vector space that has an uncountable Hamel basis, all other Hamel bases are also uncountable, as all Hamel bases of the same vector space have the same cardinality.