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Let X be a vector space of infinite dimension (possibly) and $x_0 \in X$.

I would like to show that if $f(x_0)=0 \forall f \in X^*$ then $x_0=0$. So the first intuition would be to pick a function such as $Id_X$ but this is not even into the dual $X^*$.

Edit :

Actually $X^*$ was the topological dual, so containing the continuous functions (Sorry, I totally forgot that this could be confused with the algebraic dual, but actually, the answers still helped me for this, so thanks).

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    $\begingroup$ Hint: Hahn-Banach $\endgroup$ – Severin Schraven Jan 12 at 14:18
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    $\begingroup$ Do you know that for each $x_0 \in X$, there is an $f \in X^*$ such that $f(x_0) = \Vert x_0 \Vert?$ $\endgroup$ – MisterRiemann Jan 12 at 14:19
  • $\begingroup$ Well, if $X$ is normed then $X^*$ refers to the continuous dual space and thus we need something like Hahn-Banach. But if there is no norm on $X$ then $X^*$ refers to the algebraic dual space, in which case we just find a Hamel basis $\mathcal{B}$ and let $f(b_0)=1$ for some $b_0$ in the nonzero basis expansion of $x_0$ and $f(b)=0$ for all other elements of $\mathcal{B}$. $\endgroup$ – Ben W Jan 12 at 14:28
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    $\begingroup$ If there is no topology, then tag (functional-analysis) should be removed. $\endgroup$ – GEdgar Jan 12 at 14:36
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    $\begingroup$ @roi_saumon Try to use the Hahn Banach theorem (It shouldn't be that hard, but this depend on what form of the Hahn Banach theorem you're using). If you fail to do so I would ask a separate question about the case where $X$ is not just a vector space but a Banach space instead. $\endgroup$ – Yanko Jan 12 at 15:09
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I think that the statment is not necessarily true without assuming the axiom of choice. But if we assume it, then $X$ has a basis. It's easy to dedudce from this that, if $x_0\neq0$, then $X$ has a basis $B$ such that $x_0\in B$. Now, consider $\varphi\in X^*$ which maps $x_0$ into $1$ and all other elements of $B$ into $0$.

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    $\begingroup$ Is it clear that $\varphi$ is continuous? Edit: I think it's not but also the OP didn't specify any topology so I guess this answer is valid. $\endgroup$ – Yanko Jan 12 at 14:23
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    $\begingroup$ Did you mention continuity in your question. Actually, did you assume anything about $X$ other than the fact that it is a vector space? $\endgroup$ – José Carlos Santos Jan 12 at 14:31
  • $\begingroup$ That's not my question. Also you're right. I suppose it's the "functional analysis"- tag and the notation of $X$ (instead of $V$) for a vector space that made me think unconsciously that the question is about Banach spaces. $\endgroup$ – Yanko Jan 12 at 14:33
  • $\begingroup$ @Yanko I'm so sorry. It was automatic for me that you were the person who asked the question. $\endgroup$ – José Carlos Santos Jan 12 at 15:06
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I take the hypothesis that $X^*$ is the algebraic dual space as you don’t mention any norm or something similar.

Suppose that $x_0 \neq 0$ and complement it with $(y_i)_{i \in I}$ to get a basis $\mathcal B=(x_0,(y_i)_{i \in I})$. This uses axiom of choice.

Then define $f \in X^*$ on that basis by $f(x_0)=1$ and $f(y_i)=0$ for $i \in I$.

You’re done.

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The question does not specify any topology, so it could well be that $X^*$ is supposed to be the algebraic dual of $X$. As a matter of interest, if $X$ is a Banach space and $X^*$ is supposed to be the space of continuous linear functionals, then the construction in the other two answers, "choose a basis and send one basis element to $1$, the others to $0$", does not necessarily give a continuous functional:

Say $\Lambda$ is an unbounded linear functional on $X$ and let $Y$ be the nullspace of $\Lambda$. Let $B_0$ be a Hamel basis for $Y$, and let $B=B_0\cup\{x_0\}$ where $\Lambda x_0=1$. Then $\Lambda$ is the functional constructed above.

So now we know: Coordinate functionals with respect to a Hamel basis need not be continuous.

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  • $\begingroup$ A question concerning your construction : If you take any unbounded linear functional $\Lambda$, how do you know there is a $b$ such that $\Lambda b = 1$? $\endgroup$ – roi_saumon Jan 12 at 14:49
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    $\begingroup$ @roi_saumon If $\Lambda=0$ then $\Lambda$ is bounded. If $\Lambda\ne0$ there exists $x_0$ with $\Lambda x_0\ne0$, so $\Lambda(x_0/\Lambda x_0)=1$. $\endgroup$ – David C. Ullrich Jan 12 at 14:52

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