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If given set $V = \mathbb{C}\setminus\{0\}$ and field $\mathcal{F} = \mathbb{C}$ with operations defined for all $\vec{x}, \vec{y} \in V, \vec{x} = x \in \mathbb{C}\setminus\{0\}, \vec{y} = y \in \mathbb{C}\setminus\{0\}$ and $\alpha \in \mathbb{C}$ as: $$\vec{x} \oplus \vec{y} = x\cdot y$$ $$\alpha \odot \vec{x} = \alpha\cdot x$$ Is it really enough to take $\alpha = 0$, which gives $\alpha\cdot \vec{x} \notin V$ to prove that this is not a vector space over $\mathbb{C}$? I.e. just one counterexample is enough, right?

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  • $\begingroup$ Since $\alpha\odot x$ is not defined for $\alpha=0$, the problem doesn't even start up. Note, to the contrary, that $\mathbb{R}_+$ (the positive reals) is a vector space over $\mathbb{R}$ with $x\oplus y=xy$ and $\alpha\odot x=x^\alpha$. This cannot be done over the complex numbers, because complex exponentiation cannot be defined as a single-valued operation. $\endgroup$ – egreg Oct 15 '16 at 11:07
  • $\begingroup$ @egreg so this is a correct disproof? :-) $\endgroup$ – Accelerate to the Infinity Oct 15 '16 at 11:08
  • $\begingroup$ In my opinion there is nothing to disprove. $\endgroup$ – egreg Oct 15 '16 at 11:10
  • $\begingroup$ @egreg: I don't really understand your comment. $\alpha\odot x$ is defined to be $0$. As the OP points out, this is not in $V$, so $\{V,\mathcal F,\oplus,\odot\}$ is not a vector space. What's wrong with this? $\endgroup$ – TonyK Oct 15 '16 at 11:21
  • $\begingroup$ @TonyK Since $0\notin V$, there is no definition. $\endgroup$ – egreg Oct 15 '16 at 12:15
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Yes, that is enough to show that the definition doesn't work.

$\odot$ is supposed to be $\mathcal F\times V\to V$, but the definition shown here produces values outside $V$ for at least one combination of $a\in\mathcal F$ and $\vec x\in V$. This is enough to disqualify the purported thing from being a vector space.

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The answer is yes. The set $\mathbb{C}\backslash \{0\}$ with usual multiplication is an abelian group, so $V$ with $\oplus$ is an abelian group. The problems arise with "scalar multiplication", and although one example is enough, it's also true that distributive laws fail for (almost) any $\alpha,x,y$

$$\alpha\odot (x\oplus y) \neq \alpha \odot x \oplus \alpha\odot y$$

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