Let $V$ be a $\mathbb K$ vector space, $M_1$ and $M_2$ subsets of $V$. ($[M] =$ linear span of $M$) Show that:
a) $[M_1 \cup M_2] = [[M_1 \cup M_2]]$ and specially $[[M]] = [M]$.
What I've tried: I guess once I've proven that $[[M]] = [M]$, it follows that $[M_1 \cup M_2] = [[M_1 \cup M_2]]$, as $[M_1 \cup M_2]$ is nothing more than a normal subset. Can I just say that by definition the linear span is already all linear combinations of vectors from $M$? Then getting again all linear combinations from all linear combinations isn't getting me anything new. Or is there a math way to write this?
b) $[M_1 \cap M_2] \subset [M_1] \cap [M_2]$
What I've tried: Playing with properties like the following:
$M \subset [M]$
$M_1 \subset M_2 \Rightarrow [M_1] \subset [M_2]$
But to no avail. So I have $M_1 \cap M_2 \subset [M_1 \cap M_2]$, $M_1 \subset [M_1]$ and $M_2 \subset [M_2]$, but couldn't prove anything with this.
c) no equality is valid in b).
What I've tried: I suppose they want me to show that $[M_1 \cap M_2] =[M_1] \cap [M_2]$ is false? I fail to understand how this is not true.
Many many thanks in advance!