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

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?

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).

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

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

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, $a+b \neq 0$.

• Your definition of addition and multiplication for this question has no meaning. Basically you are asking for two binary operations on a set which is contained in some bigger set such that the vector space axiom "fails". But you don't even have a defined linear structure so how is this even a linear algebra problem? – Jacky Chong Oct 14 '16 at 19:35
• I don't understand what restrictions on addition or multiple you are assuming. Let X {babar,pink honk honk}. babar + babar = jumbo. babar+ pink honk honk = pink honk honk while pink honk honk + babar = babar, etc. kinda silly, isn't it. – fleablood Oct 14 '16 at 20:51

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

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?

• It sounds like your first sentence is responding to the title moreso than the question. The question does say "and suitable definitions of addition and scalar multiplication". – Mark S. Oct 14 '16 at 19:34
• If a+b=c then you dont have a set with a binary operation on it unless c is equal to a or b. – Mark S. Oct 14 '16 at 19:35
• @marks what does it even mean to be suitable if you want your addition to be non commutative. – Jacky Chong Oct 14 '16 at 19:45
• What on earth does "suitable" mean? The only definition for suitable that I can tell would be that the satisfy the axioms. Why wouldn't 1+325=2 while 325+1=7 be "suitible"? – fleablood Oct 14 '16 at 20:54
• @fleablood I assumed they intended "corresponding"; I suspect they were just used to seeing the word "suitable" in that spot in math texts and used it without thinking much. Whatever apoapsis meant by "suitable", or whether it was reasonable, my only point was that since they mentioned definitions of "addition" and "scalar multiplication", I took that as evidence that they knew that an underlying set alone wasn't enough for consideration. Sorry if I was unclear. I definitely did not intend to say "hey your definition isn't 'suitable'!", whatever that would mean. – Mark S. Oct 14 '16 at 21:26