$V$ is some vector space over a field $F$.
$W$ is subspace of $V$.
$(V\setminus W) \cup \{0\}$ is also a subspace of $V$ (that is all vectors in $V$ but not in $W$, and $0$ vector).
I need to prove that either $V=W$ or $W =\{0\}$ - just contains the zero vector and nothing more.
Supposing $V=W$ then $V=W$ and we are done, but if they are not equal how do I find that every vector in $W$ is the zero vector?
On
As i understand the above answer: If W=V then its trivial. assume W≠V.
Take v∈V∖W and 0≠w∈W.
if v+w ∈W and W is a subspace then (v+w)−w ∈ W which means that v∈W.
which contradicts v∈V∖W{0} unless v=0.
if v+w ∈ V∖W∪{0} and V∖W∪{0} is a subspace then (v+w)-v ∈V∖W∪{0}. which means that w∈V∖W.
which contradicts w∈W, unless w=o.
thus any vector in W must be zero --> W = {0}.
Suppose that $v \in V\setminus W $ and $0\neq w \in W$. Then if $v+w \in W$, we get the contradiction $v+w-w\in W$. So $v+w \in V\setminus W\cup \{0\}$ which is supposed to be a subspace. Hence the contradiction $v+w-v=w\in V\setminus W$.
That is too much of contradictions... and conclude the proof.