What is the need to include the "additive identity exists" axiom in the set of vector space axioms? A vector space is a set V along with an addition on V and a scalar multiplication on V such that the following properties hold, according to S. Axler's Linear Algebra Done Right:

*

*Commutativity

*Associativity

*Existence of additive identity

*Existence of additive inverse

*Multiplicative identity

*Distributive Properties

Now, by definition, an addition of two elements in the set V should also be a member of V. Using that fact, and (4) from the list above, shouldn't (3) be provable?
Note: A few more moments of thought made me question my statement. What is $0$? Unless (3) is taken, does (4) make any sense? Am I on the right track?
Also, S. Axler says, about subspaces, that they have to satisfy three conditions for them to be considered a subspace (excluding the condition that they have to be a subset of a vector space, which would ensure that the other properties (distributive, commutative) are satisfied $\forall v \space \epsilon\space Subspace$).

*

*$0 \space\epsilon\space Subspace$

*Scalar multiplication is closed.

*Addition is closed.

In addition to this, he also says we can replace (1) here with a similar condition:

*

*The subspace is non-empty.

He says that since scalar multiplication is closed within the subspace, and that $0v = 0$ (the proof of this involves the additive inverse axiom and the fact that the "Zero" produced by the inverse CAN be added to a vector from $V$; Does this zero have to exist within $V$ for us to define "addition" between this zero and a vector from $V$?) for any $v\space\epsilon\space Subspace$, this would imply that $0\space \epsilon \space Subspace$.
So here, is this replacement possible only because (3) from the vector space axioms defines "zero"?
Note: I tried my best to organize my thoughts and questions, but something seems amiss. I know my questions aren't sequential and coherent but I am struggling to understand which part of this is the head and which one's the tail.
 A: (1) "Does ${4}$ and the addition of two vectors being in ${V}$ imply ${(3)}$?" No. The definitions are intertwined. In order to define an additive inverse, you need a definition of the identity element, since its definition is based on the identity element. By definition, if you take a vector ${v}$, its additive inverse, ${(-v)}$ is the vector ${\in V}$ such that
$${v + (-v) = 0_V}$$
where ${0_V}$ is the additive identity vector in the vector space. If we have no concept of what ${0_V}$ is, the definition for this additive inverse doesn't really make sense. It's like asking "what does blue mean?" without knowing the concept of a colour.
(2) Exactly. Given any subset of ${V}$,  (call it ${U}$. That is ${U\subseteq V}$) it's either empty or non-empty. If it's empty, it's not a subspace (since if it's empty it does not contain ${0_V}$, the zero vector). If it's non-empty, then there must exist at least one vector in the subspace. Take any vector ${u \in U}$. Then we have
$${0_F\times u = 0_V}$$
And since the subspace is closed under scalar multiplication we have that for any non-empty subset of ${V}$ that is closed under scalar multiplication and vector addition that ${0_V}$ exists within this subset. Hence the first condition that ${0_V}$ exists within the subspace can be replaced with it being non-empty. In this context, both statements are equivalent.
Ultimately - you cannot get away without defining ${0}$. If you don't define it - how do you then define additive invertibility?
Edit: As @ArturoMagidin has pointed out in the comments - it is possible to come up with a different set of axioms (by different I mean that you replace $4$) that don't include $0$, but that satisfies all other conditions. This is different from the standard vector space axioms, but is pretty cool - so you should check it out! :D
