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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 $v+w=0$.

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  • $\begingroup$ Any existentially quantified proposition fails for an empty set. $\endgroup$
    – Kaz
    Jun 21, 2021 at 15:12
  • $\begingroup$ Then again, the complete axiomatization of the concept of vector spaces needs only one axiom: "A vector space over a field $F$ is an abelian group together with an action by $F$" :) $\endgroup$ Jun 21, 2021 at 15:13
  • $\begingroup$ Every element in the empty set has an additive inverse. True fact. You'll never find a counter-example. $\endgroup$
    – user121330
    Jun 21, 2021 at 15:34

3 Answers 3

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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).

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  • $\begingroup$ 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 $v+w=0$." $\endgroup$
    – tchappy ha
    Jun 21, 2021 at 6:55
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    $\begingroup$ I see what you mean. This problem doesn't occur if we consider the empty set as a subset of some vector space (so that we at least have a candidate for the zero vector). Otherwise, I guess you are right to say that both axioms indeed fail to hold in this case. $\endgroup$
    – Zuy
    Jun 21, 2021 at 7:01
  • $\begingroup$ Thank you very much for your answer and comment. $\endgroup$
    – tchappy ha
    Jun 21, 2021 at 7:04
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    $\begingroup$ @tchappyha To decide that, one would need to come up with a purely logical formulation of the second axiom, not some inherently ambiguous English sentence. For example, how does the symbol "$0$" even have a meaning in the secon daxiom? We might introduce a constant symbol $0$ to our language and in the first axiom have "$0\in V\land\forall v\in V\colon v+0=v$" and in the second axiom "$\forall v\in V\exists w\in V\colon v+w=0$" is true even though $0$ occurs because we do not claim $0\in V$ here. $\endgroup$ Jun 21, 2021 at 15:18
  • $\begingroup$ @tchappyha Get rid of the math notation and convert the proposition into English. Let V be the set of all moons in the solar system made from green cheese. And let $0$ be the king of the USA. Now, "For every moon $v$ made from green cheese, there is a moon $w$ made from green cheese, such that the sum of the two moons is the king of the USA". There is nothing logically wrong with that assertion. None of the things it mentions exist. $\endgroup$
    – alephzero
    Jun 21, 2021 at 15:26
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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.

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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.

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