Which of the following sets are subspaces of V? Let V be a vector space and let U, W, X $\subset$ V be sub-vector spaces of V.
Which of the following sets are again subspaces of V?
a) U $\triangle$ W
b) U $\cap$ W
c) U $\cup$ W
d) v + U for a v $\in$ V
*So a subspace of a Vector V is a subset U that satisfies three Axioms(Conditions):
(UV0) The zero Vector is in U
(UV1) U is closed under addition
(UV2) U is closed under multipication of scalars*
-For a) I don`t know how to prove if it is a vector subspace of V or not! Need some help
-For b) I have proved this way but I´m not sure: We have to check if the three Axioms are satisfied:
(UV0): Is 0 $\in$ U and 0 $\in$ W, and because that both of them are subspaces $\Rightarrow$ 0 $\in$ U $\cap$ W $\Rightarrow$ U $\cap$ W $\neq$ $\varnothing$
(UV1): Let u,v $\in$ U $\cap$ W, and because that U and W are subspaces, follows:
(u,v $\in$ U $\Rightarrow$ u+v $\in$ U) and (u,v $\in$ W $\Rightarrow$ u+v $\in$ W) $\Rightarrow$ u+v $\in$ U $\cap$ W
(UV2): Let u $\in$ U $\cap$ W and $\lambda \in$ K $\Rightarrow$ (u $\in$ U, $\lambda$ $\in$ K $\Rightarrow$ $\lambda$u $\in$ U ) and (u $\in$ W, $\lambda$ $\in$ K $\Rightarrow$ $\lambda$u $\in$ W )
$\Rightarrow$ $\lambda$u $\in$ U $\cap$ W.
-For c) I need an Idea aswell and
-For d) I think it should be proved with to cases 1.Fall v $\notin$ U and 2.Fall v$\in$U, I´m not sure if this is the right Idea and if it is how to prove it
Thank you very much:)
 A: Proposition: $\;U\cup V\;$ is a vectorial subspace iff $\;U\subset V\;$ or $\;V\subset U\;$
Hint for the proof: Direction $\;\Longleftarrow\;$ is trivial. For the other diection, suppose that both $\;U\not\subset V\;$ and also $\;V\not\subset U\;$ . Then there exists $\;u\in U\setminus V\;$ nad $\;v\in V\setminus U\;$ ...well, your turn
A: All your ideas are right. The proof is right. Adapt it to prove that $v+U$ is a subspace when $v \in U$. Can $0 \in v+U$ if $v \notin U$?
For union think of lines in the plane for a counterexample.
For $U \Delta V$, can $0$ be in it?
Proof for $v+U$. If $v \notin U$, then suppose for a contradiction that $v+U$ is a subspace. Then $0 \in v+U$ i.e. there is an $u \in U$ such that $0 = v + u$. But then $v=-u \in U$, which is a contradiction.
If $v \in U$, then $\forall u \in U$, we have that $v+u \in U$. So $v+U \subseteq U$. Moreover, $\forall u \in U$ we have $(-v)+u \in U$, so $u = v + (-v) + u \in v + U$. Hence, also $U \subseteq v + U$.
By double inclusion, $v + U = U$, and thus is a subspace.
