Is it necessary to show 0 is included in a set to show it is a vector subspace?

Question

To show a set is a vector subspace, I see it´s necessary to prove

a) An addition property: if $x$ and $m$ are both elements of the set, then $x+m$ must also be an element of the set for it to be a vector subspace.

b) A scalar multiplication property: if $x$ is an element of the set, then $kx$ is also defined in the set, where $k$ is a scalar.

However, many people insist that it´s also necessary to prove that the neutral element, $0$ is also defined in the set for it to be a vector subspace, independent of the other two properties.

Isn´t this property, however, implied by a) and b) as follows?:

Given the additive property, $x-m=0$, where $x=m$

or by the scalar property, $kx=0$, where $k=0$

Answer 1

You need to show that there is an element in your set.

The empty set satisfies the “addition property” as well as the “scalar multiplication property”, so you have to ensure the set is not empty.

Usually, asking for $0$ to be in the set is the first property to check, but just being not empty is sufficient, along with the other mentioned properties.

Indeed, if $x_0\in S$, then $(-1)x_0\in S$ (scalar multiplication property) and therefore $$ x_0+(-1)x_0\in S $$ But it follows from the vector space axioms that $x_0+(-1)x_0=0$, so we're done.

Perhaps more simply: the scalar multiplication property implies that $0x_0=0\in S$ (but you still need to have an element to start with).

Answer 2

Those properties rely on the existence of an element $0$ if it where not to exist how would you argue what is the value of $x-x$?

Answer 3

It is sufficient too prove that

1. For any u, v in the subset, u+ v is in the subset.
2. For any v in the subset and scalar a, av is in the subset.

That is, that the subset is closed under vector addition and closed under scalar multiplication.

Since the set of scalars is a field, it contains a 0 element and,for any vector v, 0v is the 0 vector.