How does one prove associativity, commutativity, etc of an empty set? In Linear Algebra Done Right by Sheldon Axler one of the exercises is as follows:
"The empty set is not a vector space. The empty set fails to satisfy only one of the requirements listed in 1.19. Which one?"
1.19 lists the requirements of a vector space. I.e. associativity, commutativity, additive identity, additive inverse, multiplicative inverse, and distributive property.
The answer to the question is that the empty set doesn't satisfy additive identity due to the fact that there is no element $\mathbf{0}$ in vector space $\mathbf{V}$.
How does one going about proving that the empty set meets the other requirements of associativity, commutativity, additive inverse, multiplicative inverse, and the distributive property?
I'm confused about how the elements of an empty set can be shown to follow these properties when they don't exist.
 A: To add to the other answers, one way I like to look at this this situation is as follows:
All the of the axioms you list: associativity, commutativity, additive inverse, etc... all start with the quantifier "for all". In other words, if you wrote them out with heavy notation they would have the form $\forall x P(x)$ where $P(x)$ is some further property of $x$. To say that a space fails to satisfy such a statement amounts to saying that the space satisfies the negation of the statement, which has the form $\exists x Q(x)$ where $Q(x)$ is the negation of $P(x)$. So right away we see that an empty space cannot satisfy such a statement since it's asserting that there is some element of the space.
A: maybe you can understand it by example. Example 1:
prove that my favorite set $\{1,4,9,12\}$ meets the requirements of associativity.
Approach: Pick elements and start testing:
test #1:    $ (1 + 4) + 9 = 1 + (4 + 9)$. Obviously true.
test #2:    $ (12 + 4) + 9 = 12 + (4 + 9)$. Obviously true.
to finish the proof, you'd have to walk through all combinations -- a finite number $k$ of tests to pass in this example.
Example 2:
prove that your favorite set $\{\,\}$ meets the requirements of associativity.
Again pick elements and start testing... but we cannot form even one test without any elements. So we have no tests to conduct ($k=0$). Loosely speaking, "Since we did not fail, we have succeeded." More properly "A implies B" is true, with A being: $a,b,c$ are any elements of $\{\,\}$ , and B representing the associativity equation. "A implies B" is equivalent to "(not A) or B", and A never holds.
A: or maybe you'd prefer to look at it this way:
Example 3:  Prove that the system of 2 by 2 matrices does not enjoy the commutative property. We are done if there exists one pair of elements where commutativity fails. We can give this explicitly:
$$
 \left[ \begin{array}{c}
 0 & 1 \\ 1 & 0
\end{array}  
\right]\left[ \begin{array}{c}
 1 & -1 \\ 0 & 1
\end{array}  
\right] \neq \left[ \begin{array}{c}
 1 & -1 \\ 0 & 1
\end{array}  
\right]\left[ \begin{array}{c}
 0 & 1 \\ 1 & 0
\end{array}  
\right],\,\,\,\text{q.e.d.} 
$$
Example 4:  Prove that the system $\{\,\}$ does not enjoy the commutative property. We are done if there exists one pair of elements where commutativity fails. But we cannot find a pair of elements, period. It seems the claim is false.
Example 5:  Prove that the system $\{\,\}$ does enjoy the commutative property. We must show "A implies B", with A = (L and M are elements of $\{\,\}$) and B = (LM = ML). Here A can never be true, so the implication arrow is never "activated". More formally, "A implies B" = "(not A) or B" holds, because "not A" holds always.
