Subspace criterion (zero vector?) This may be a poor question but it is one that I cannot wrap my head around.

The theorem I am presented with is as follows:


  (Subspace criterion) Suppose $V$ is vector space over $F$.
  A nonempty subset $U ⊆ V$ is a subspace if and only if $U$ is closed for addition
  and multiplication with scalars. 


When proceeding to worked examples, I noticed that the first property that is checked when proving a set is a subspace using the above criterion is whether the zero vector exists. However, this property is not mentioned in the criterion?

I assume that there is some connection between a set being closed under addition and it has a zero vector but I am failing to deduce it.

Any help is greatly appreciated.
Consider a subspace of $V$ of dimension $1$, $U$, where the entire subspace is spanned by scalar multiples of some element $\vec{u}$.

No matter what $\vec{u}$ is, we can always construct a sum of the form:

$$\vec{u} + -\vec{u}$$

This is equal to $\vec{0}$ by definition, so the zero vector needs to be in the subspace in order for the subspace to be closed with respect to vector addition.
Let 0'=additive identity of field F,now since vector subspace S is closed with respect to scalar multiplication,hence 0'(a)€S
0'(a)=(0'+0')a  {a€S}
0'(a)=0'a+0'a
0'(a)=0''€S{where 0''=zero of vector subspace},,Hence there is a connection between closed wrt scalar multiplication and existence of Zero of subspace