Suppose that $W_1, W_2, ..., W_n$ is a family of subspaces of V. Prove that the following set is a subspace of $V$: Suppose that $W_1, W_2, ..., W_n$ is a family of subspaces of V. Prove that the following set is a subspace of $V$:
$\cap_{i=1}^n W_i = ${$z | z \in W_i, 1 \leq i \leq n$}
The question counts for 4 marks. I'm just a bit confused as to how exactly this must be "proven"
My thoughts:
Since $W_1, ..., W_n$ are all subspaces, vector addition and scalar multiplication are closed under each of them.
Since any one of these subspaces is a vector space itself, the intersection of any of these subspaces must also be a vector space because vector addition and scalar multiplication are closed under all of them individually and, therefore, together as well.
$W_1 \cap W_2 \cap ... \cap W_n \subseteq V$ since they are a family of subspaces/subsets of $V$.
The set is a subspace of $V$ because it is a vector space under the vector addition and scalar multiplication defined on $V$

Your reasoning is correct but I guess you should provide more details for this task. So to prove for example that this intersection is closed under addition, a proper proof would look like this :
Take $x,y\in \cap_{i=1}^n W_i$, then for each $i$, we have $x,y\in W_i$. Since $W_i$ is a vector space, $x+y\in W_i$. Since this holds for each $i$, we have that $x+y\in \cap_{i=1}^n W_i$. Hence $\cap_{i=1}^n W_i$ is closed under addition.
