Equivalence of these propositions Let $V$ be a $ \Bbb{K}$-vector space. Let $S$ be a subset of $V$, the $S$ is a subspace of $V$ if and only if:
1) $0_V \in S$
2) $v, u \in S \implies v+u\in S$
3) $c \in \Bbb{K}, v \in S \implies c\cdot v \in S$
My question is: can I summarize these 3 propositions in the following one? (or at least 2 and 3).
$c \in \Bbb{K}, v,u \in S \implies cv+u \in S$ 
If so, how do I prove the equivalence?
 A: 1) isn't necessary, so long as $S$ is non-empty. For if so, and 2) and 3) hold, then since $1 \in \Bbb K$, and $\Bbb K$ is a field, $-1 \in \Bbb K$ (since fields are abelian groups under addition).
Thus 3) says if $u \in S$, then $(-1)u \in S$, and it is easy to see that:
$(-1)u = -u$, since $(-1)u + u = (-1)u + (1)u = (-1+1)u = 0_{\Bbb K}u = 0_V$.
And now 2) says that since $u,-u \in S$, that $u + -u = 0_V \in S$.

To prove your proposition implies 2) and 3) for a NON-EMPTY set (the reverse implication is trivial), assume it holds.
Then take $c = -1$, and $u = v$, so we have that $-u + u = 0_V \in S$.
Therefore, we may take $c$ abitrary, $v \in S$, and $u = 0_V$, and your proposition then implies 2).
Finally, we can take $c = 1$, and $u,v \in S$, and your proposition then clearly implies 2).
The "usual" one-step check for subspaces (on non-empty sets-this is important) is "similar" to yours:
$a,b \in \Bbb K$, and $u,v \in S \implies au + bv \in S$. It should not be surprising that a vector space should be closed under finite linear combinations of its elements.
