Is there a set which satisfies none of the vector space axioms?

Question

I'm taking a linear algebra course, and while studying vector spaces, I've come up with the following question: Is there a (preferably nontrivial) set which satisfies none of the vector space axioms? That is, for some set $V$ and suitable definitions of addition and scalar multiplication,

• $\exists\ u, v \in V : u+v \notin V $
• $\exists\ u, v \in V : u+v \neq v+u $
• $\exists\ u, v, w \in V :\ (u+v)+w \neq u+(v+w) $
• $\nexists\ 0 \in V : \forall v \in V,\ v+0 = v \text{ and }0+v = 0 $

• $\forall\ u \in V,\ u+v = 0 \implies v \notin V$

• $\exists\ u \in V,\ c \in \mathbb{R} : c\cdot u \notin V$

• $\exists\ u \in V,\ c,d \in \mathbb{R} : (cd)\cdot u \neq c(d\cdot u) $
• $\exists\ u \in V,\ c,d \in \mathbb{R} : (c+d)\cdot u \neq c\cdot u + d\cdot u $
• $\exists\ u, v \in V,\ c \in \mathbb{R} : c\cdot (u+v) \neq c\cdot u + c\cdot v $
• $\exists\ u \in V : 1\cdot u \neq u $

No examples come to mind easily. If we try $V = \{x \in \mathbb{R}: x\neq 0\},\ $ I have tried things like defining addition by:

$$ a+b = ae^{bi} = a(\cos b + i\sin b)$$

This fails all the addition axioms. However, I am struggling to find a definition of multiplication that fails all the multiplication axioms. It is especially hard to work with the ninth axiom; since it distributes a scalar across the sum of two vectors, I must use the convoluted addition while working on it. Is there any way to make it work? I've also tried playing around with the new axioms and trying to see if they are self-contradictory, but haven't gotten anywhere with that. Would it be better to look outside the reals and consider sets of matrices or something like that?

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Here is my work for the addition axioms:

If we let $a, b \in \mathbb{R}, a+b = ae^{bi}$, then by Euler's identity,

$$a+b = a(\cos b + i\sin b)$$

Which has the possibility to be a complex number, not a real number. So the set is not closed under addition (defined this way).

$$ae^{bi} \neq be^{ai}$$

So addition is not commutative.

$$(a+b)+c = ae^{bi}+c = ae^{bi}e^{ci} = ae^{bci^2} = ae^{-bc}$$ $$a+(b+c) = a+ be^{ci} = ae^{be^{ci}} = ae^{b(\cos c + i\sin c)}$$

In general, $-c \neq \cos c + i\sin c $, so in general $(a+b)+c \neq a + (b+c)$. So addition is not associative.

$e^z \neq 0$ for any z, so there is no $b$ such that $a+b = ae^{bi} = 0$. However, there obviously is $a=0$ such that $ae^{bi} = 0$. But this does not work the other way. So there is no additive identity in general such that $a+0 = a $ and $0+a = a$. But I excluded $0$ from $V$ anyway, just in case a one-way identity would cause problems.

Since there is no zero element, $a+b \neq 0$.

Answer 1

Just dare to be wild enough, e.g., let $V=\{1,2,3\}$ and $u\color{red}+v=u+v^2$, $c\color{red}\cdot v=c^2+v$

Answer 2

The question is meaningless; a set does not satisfy (nor not satisfy) the vector space axioms.

What is needed to test the axioms is a set together with two binary operations: addition of vectors, and multiplication of vectors by scalars (for you $\mathbb{R}$, but in general other fields could work).

There are plenty of binary operations one could impose on a set that would fail all of the axioms. For example, suppose our set is $\{a,b\}$, and we define $a+b=c$ and $b+a=d$, and multiplication by scalars as $3*a=e$ while $2*a=f$, and $x*a=g$, for all other real $x$.

But why on earth would anyone be interested in such a weird construction?