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.

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    $\begingroup$ 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. $\endgroup$ – GEdgar Jan 10 '18 at 12:50
  • $\begingroup$ @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. $\endgroup$ – Andrei Kh Mar 13 '18 at 23:26

In the article

Wright, J. D. Maitland. All operators on a Hilbert space are bounded. Bull. Amer. Math. Soc. 79 (1973), no. 6, 1247--1250.

it is shown that in ZF + DC + BP the following is a theorem:

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.

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