The empty set fails to satisfy only one of the axioms of vector spaces. Which one? ("Linear Algebra Done Right" by Sheldon Axler.)

Question

I am reading "Linear Algebra Done Right" by Sheldon Axler.

This book contains the following exercise:

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?

This exercise says that the empty set fails to satisfy only one of the axioms of vector spaces.

I think that the empty set fails to satisfy the following two axioms:

There exists an element $0\in V$ such that $v+0=v$ for all $v\in V$.

For every $v\in V$, there exists $w\in V$ such that $v+w=0$.

Does the empty set really fail to satisfy only one of the axioms of vector spaces?

There doesn't exist an element $0\in \emptyset$ such that $v+0=v$ for all $v\in \emptyset$.
But the following axiom uses $0$.
No problem?

For every $v\in V$, there exists $w\in V$ such that $v+w=0$.

Answer 1

The first axiom you cite is the one you are looking for.

The second one is satisfied by the empty set. It states that if you take some $v\in\emptyset$, then there is some $w\in\emptyset$ for which $v+w=0$. This is an implication whose antecedent (the part including the "if") never holds, hence the implication itself is true (since "false implies true" and "true implies true" are implications which are true).

Answer 2

The first axiom is stronger than the axiom you get from dropping the condition which just says there exists an element $0 \in V$. If it fails to satisfy that, then it fails the stronger axiom as well. And the empty set fails on both accounts.

The second one you give has a universal quanification at the beginning. But because $V$ is empty, saying something holds for all $v \in V$ is a vacuous condition. There is nothing it has to hold for so it is vacuously true. That axiom is satisfied.

The first one is an existential and the second is universal and that makes the difference about which the empty set satisfies.

Answer 3

The second statement could be rephrased as "if $v \in V$ then there exists $w \in V$ such that $v + w = 0$". That statement is vacuously true if $V$ is the empty set.