Can a non-zero vector have zero image under every linear functional? Let $X$ be an infinite-dimensional vector space, and let $x_0$ be an element of $X$ such that $f(x_0)=0$ for every linear functional $f$ defined on $X$. Then can we prove that $x_0$ is the zero vector in $X$? Or, is it possible to find a non-zero $x_0$ whose image under every linear functional is zero? 
In the case of $X$ being finite-dimensional, we can prove that if $f(x_0)=0$ for every linear functional $f$ on $X$, then $x_0$ must be the zero vector. What is the situation in the infinite-dimensional case? 
 A: Let $x_0\in X$ be a nonzero vector. If $X$ has a [Hamel] basis then it has one in which $x_0$ is a member, this follows from Steinitz exchange lemma. Even if $X$ is not finitely generated we can still apply this lemma to the smallest subset of the basis that $x_0$ lies in its span.
Now consider $H$ to be a basis which includes $x_0$, let $\pi\colon H\to\Bbb F$ (where $\Bbb F$ is the field) be the function which maps $x_0$ to $1$ and $0$ elsewhere, then we can extend it uniquely to a linear functional which gives $x_0$ the value $1$.
Note that we do use the axiom of choice here to ensure the existence of this Hamel basis. It is consistent without the axiom of choice that there is a vector space that every linear functional is zero, in which case the answer to your question is negative. The question whether or not the existence of such functionals implies the axiom of choice is open to my knowledge.
A: If $x_0\ne 0$, the set $\{x_0\}$ is linearly independent, hence can be extended to a basis of $X$. Mapping to the coefficient of $x_0$ with respect to this basis is a linear functional with $x_0\mapsto 1$.
A: As noted, the answer is "no".  And, as also noted, proofs require the Axiom of Choice, or at least something beyond ZF.  
My favorite example here is the Banach space $l^\infty/c_0$, which is infinite-dimensional, but such that no nonzero linear functional can be proved to exist using only ZF.
