Proof of Theorem TSS

(⇒) We have the hypothesis that $W$ is a subspace, so by Property Z we know that $W$ contains a zero vector. This is enough to show that $W\neq\emptyset$. Also, since $W$ is a vector space it satisfies the additive and scalar multiplication closure properties (Property AC, Property SC), and so exactly meets the second and third conditions. If that was easy, the other direction might require a bit more work.

(⇐) We have three properties for our hypothesis, and from this we should conclude that $W$ has the ten defining properties of a vector space. The second and third conditions of our hypothesis are exactly Property AC and Property SC. Our hypothesis that $V$ is a vector space implies that Property C, Property AA, Property SMA, Property DVA, Property DSA and Property O all hold. They continue to be true for vectors from $W$ since passing to a subset, and keeping the operation the same, leaves their statements unchanged. Eight down, two to go.

Suppose $\vect{x}\in W$. Then by the third part of our hypothesis (scalar closure), we know that $(-1)\vect{x}\in W$. By Theorem AISM $(-1)\vect{x}=\vect{-x}$, so together these statements show us that $\vect{-x}\in W$. $\vect{-x}$ is the additive inverse of $\vect{x}$ in $V$, but will continue in this role when viewed as element of the subset $W$. So every element of $W$ has an additive inverse that is an element of $W$ and Property AI is completed. Just one property left.

While we have implicitly discussed the zero vector in the previous paragraph, we need to be certain that the zero vector (of $V$) really lives in $W$. Since $W$ is nonempty, we can choose some vector $\vect{z}\in W$. Then by the argument in the previous paragraph, we know $\vect{-z}\in W$. Now by Property AI for $V$ and then by the second part of our hypothesis (additive closure) we see that \begin{equation*} \zerovector=\vect{z}+(\vect{-z})\in W \end{equation*}

So $W$ contains the zero vector from $V$. Since this vector performs the required duties of a zero vector in $V$, it will continue in that role as an element of $W$. This gives us, Property Z, the final property of the ten required. (Sarah Fellez contributed to this proof.)