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This question already has an answer here:

Given an arbitrary nonzero vector space $V$, is there a nonzero linear functional on $V$, without assuming axiom of choice? I know that by assuming existence of a basis for $V$, we can consider the dual basis for a subspace of $V^*$, which justifies the existence of nonzero linear functional, but this argument fails without axiom of choice. I guess there may not always be a nonzero linear functional, but my knowledge on axiom of choice and infinite-dimensional vector spaces is lacking. A quick search on Google fails to give an answer.

Note that I am not talking about normed spaces or continuous linear functionals, just plain vector spaces with no additional structure.

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marked as duplicate by Asaf Karagila axiom-of-choice Jul 9 '17 at 12:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The answer is negative. In other words, there are models os set theory (without the axiom of choice) for which there are vector spaces $V\neq\{0\}$ such that $V^*=\{0\}$. See here, for instance.

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  • $\begingroup$ No need to go to MathOverflow for a reference. This was covered in many many threads on this site as well. $\endgroup$ – Asaf Karagila Jul 9 '17 at 12:17
  • $\begingroup$ @AsafKaragila I should have suspected that, yes, but should I abstain from making a link to MathOvrflow. $\endgroup$ – José Carlos Santos Jul 9 '17 at 12:19
  • $\begingroup$ No, don't worry about it... I just made a remark. $\endgroup$ – Asaf Karagila Jul 9 '17 at 12:23

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