Take the 2-minute tour ×  If $V$ is given to be a vector space
that is not finite-dimensional, it doesn't seem to be possible
to exhibit an explicit non-zero linear functional on $V$ without
further information about $V$. The existence of a non-zero
linear functional can be shown by taking a basis of $V$ and
specifying the values of the functional on the basis.  To find a
basis of $V$, the axiom of choice (AC) is needed, and indeed, it
was shown by Blass in 1984 that in Zermelo-Fraenkel set theory
(ZF) it is equivalent to the axiom of choice that any vector
space has a basis. However, it's not clear to me that the
existence of a non-zero element of $V^*$ really needs the full
strength of AC. I couldn't find a reference anywhere, so here is
my question:  Consider the following statement:  (D) For any
vector space $V$ that is not finite-dimensional, $V^*\neq
\{0\}$.  Is (D) equivalent to AC in ZF? If not, is there some
known axiom that is equivalent to (D) in ZF?  Note that this
question is about the algebraic