# Explicit Hamel basis for an infinite dimensional Banach space

Does there exist an explicit example of an algebraic basis in a Banach space with infinite dimension? By "explicit", I mean a basis that can be constructed without using the axiom of choice for chosing the basis elements.

I know that such a basis must be uncountable and that for some Banach spaces it can be shown that there is no such "explicit" basis.

• I suspect no such example is known. Indeed, I suspect there is no example (established in ZF) of an infinite-dimensional Banach space that has a Hamel basis. – GEdgar Jan 10 '18 at 12:50
• @GEdgar This really blows my mind. This just means that if we have any explicit uncountable set (like $\mathbb{R}$) and take the free vector space $F$ over this set, then we can't ever define a norm on $F$ such that $F$ is complete w.r.t this norm. – Andrei Kh Mar 13 '18 at 23:26

Let $$X$$ be a F-space. Then $$X^* = X'$$, i.e. the algebraic dual is the topological dual.
If we could find a set $$S \subseteq X$$ and prove that $$S$$ is an algebraic basis in ZF, then I think we could obtain in ZF + DC an unbounded linear functional by taking $$T\frac{s_n}{||s_n||} = n \quad \text{for}\quad n \in \mathbb{N}\,,$$ where $$\{s_n \mid n \in \mathbb{N}\} \subseteq S$$ is a countable subset, which would contradict the above theorem.