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Conventionally, a vector space over a field $F$ is a set $V$ together with two operations $+:V\times V\to V$ and $\cdot:F\times V\to V$ satisfying

  1. $(a+b)+c = a+(b+c)$ for all $a,b,c\in V$
  2. $a+b=b+a$ for all $a,b\in V$
  3. There exists an element $\vec 0\in V$ so that $a+\vec 0=a$ for all $a\in V$
  4. For all $a\in V$ there exists $-a\in V$ satisfying $a+(-a) = \vec 0$ (with $\vec 0$ satisfying axiom 3)
  5. $\alpha\cdot (a+b)=\alpha\cdot a+\alpha\cdot b$ for all $a,b\in V$ and $\alpha\in F$
  6. $(\alpha+\beta)\cdot a=\alpha\cdot a+\beta\cdot a$ for all $a\in V$ and $\alpha,\beta\in F$
  7. $(\alpha\beta)\cdot a=\alpha\cdot (\beta \cdot a)$ for all $a\in V$ and $\alpha,\beta\in F$
  8. $1\cdot a=a$ for all $a\in V$.

Then we prove that $\vec 0$ is unique, which justifies the sloppiness in stating axiom 4.

I recently made a careless quantifier error, defining the following:

A Q-space over a field $F$ is a set $V$ together with two operations $+:V\times V\to V$ and $\cdot:F\times V\to V$ satisfying for all $a,b,c\in V$ and $\alpha,\beta \in F$

  1. $(a+b)+c = a+(b+c)$
  2. $a+b=b+a$
  3. There exists an element $\vec 0_a\in V$ so that $a+\vec 0_a=a$
  4. There exists $-a\in V$ satisfying $a+(-a) = \vec 0_a$ (with $\vec 0_a$ satisfying axiom 3)
  5. $\alpha\cdot (a+b)=\alpha\cdot a+\alpha\cdot b$
  6. $(\alpha+\beta)\cdot a=\alpha\cdot a+\beta\cdot a$
  7. $(\alpha\beta)\cdot a=\alpha\cdot (\beta \cdot a)$
  8. $1\cdot a=a$

I suspect that a Q-space is different than a vector space. My question is, can anyone give an example of a Q-space that isn't a vector space? I do know that if axiom 4 of a Q-space is changed to

  1. There exists $-a\in V$ so that $b+a+(-a)=b$

we can then prove $\vec 0_a=\vec 0_b$, making the modified Q-space the same as a vector space.

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    $\begingroup$ There are standard analyses of the notion of "group". If each element has a left identity and a left inverse, then is it a group. Things like that. I guess your question belongs in math.se instead? $\endgroup$
    – GEdgar
    Jan 12, 2018 at 2:14

2 Answers 2

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Let $\mathbf{V}$ be any vector space. Let $\infty$ be something not in $\mathbf{V}$, and let $\mathbf{W}=\mathbf{V}\cup\{\infty\}$. Define addition on $\mathbf{W}$ to be the same as in $\mathbf{V}$ if both summands are in $\mathbf{V}$, and to be $\infty$ if either (or both) summands are $\infty$. Define scalar multiplication in $\mathbf{W}$ to be the same as in $\mathbf{V}$ for elements of $\mathbf{V}$, and $\alpha\cdot\infty=\infty$ for all scalars $\alpha$. Then $\mathbf{0}_{\mathbf{v}}=\mathbf{0}$ for all $\mathbf{v}\in\mathbf{V}$, and $\mathbf{0}_{\infty}=\infty$ satisfy your axioms 3 and 4. It is easy to verify that if any of the vectors in axioms 1, 2, 5, 6, 7, or 8 are equal to $\infty$, then both sides are $\infty$; and if they are all in $\mathbf{V}$, then the axioms hold since they hold in $\mathbf{V}$.

This amounts to taking your underlying abelian group $(\mathbf{V},+)$, and adding an element to turn it into a commutative semigroup with zero, then extending the scalar multiplication to be zero at zero.

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I think the following works: Let V be $\mathbb{R}_{\geq 0}$ with $v+w$ being given by ordinary multiplication and let $F$ be $\mathbb{R}$ with $\alpha \cdot v$ being given by ordinary exponentiation $v^\alpha$, and $\alpha \cdot 0 = 0$ for all $\alpha$. Then:

1 & 2 are obvious

3 If $a$ isn't $0$, then $0_a$ can just be $1$, otherwise $0_a$ is $0$

4 If $a$ isn't $0$ then $-a$ is $1/a$ otherwise $-a$ is $0$

5,6,7,8 Are usual rules of exponents for $a,b \neq 0$ and otherwise in each case both sides are $0$

Nevertheless $\mathbb{R_{\geq 0}}$ with multiplication is not a group and hence not a vector space

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    $\begingroup$ Note that this is isomorphic via exponentiation to the example given by Arturo above with V = $\mathbb{R}^1$ and $-\infty$ instead of $\infty$ $\endgroup$
    – danamansana
    Jan 12, 2018 at 2:32

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