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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?

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You've asked at least 36 questions here. So far you've accepted only 3! Maybe you don't know that when you get a satisfying answer, you're supposed to accept your favorite one. In order to accept an answer, hover your mouse below the downvote button of the desired answer and a check sign will appear. Click it, when it turns green you've accepted the answer. Please consider going through all of your questions and accepting answers that you like. – Git Gud Feb 23 '13 at 12:00
Visually impaired including color blind, I'm using the High Contrast (Black) display scheme and can't judge colors. I discovered the particular checkbox only recently and have to revise the answer to my previously asked questions in order to decide which ones I should accept. I've already accepted some. – Saaqib Mahmuud Feb 23 '13 at 12:14
Thank you.${}{}{}$ – Git Gud Feb 23 '13 at 12:16
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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 $x_0\mapsto 1$.

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Also, Hahn-Banach theorem if the OP wants continuous linear functionals. – wj32 Feb 23 '13 at 12:04
Absolutely! Thanks. – Saaqib Mahmuud Feb 23 '13 at 12:16
Well, does a vector space necessarily have a basis? I mean an infinite-dimensional vector space of course. – Saaqib Mahmuud Feb 23 '13 at 12:23
Yes, it is equivalent to the axiom of choice. – Berci Feb 23 '13 at 12:28
@wj32, to use Hahn-Banach it is necessary to ask more on the space: for example a Banach space. – Tomás Feb 23 '13 at 12:29

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.

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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.

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I think that correct term is "not finitely generated" when the vector space does not have a basis to begin with. Dimension is something we associate with the size of Hamel bases, and that size can be undefined because there might be no Hamel bases, or there might be several of distinct cardinality. – Asaf Karagila Feb 23 '13 at 16:09
The common term is, indeed, infinite-dimensional. Even if no Hamel basis exists. – GEdgar Feb 23 '13 at 17:12
I find that outrageous. – Asaf Karagila Feb 23 '13 at 17:15

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