union of subset and span proof I have to show the following for two subsets $M$ and $N$ of a Vector Space $V$:
$$span(M\cup span(N)) = span(M \cup N)$$
I do not quite know how to do that. I tried to show that both are subsets of one another, so I chose an arbitrary vector $f$ $$f= \sum_{i=1}^nx_iv_i+\sum_{i=1}^ny_iw_i$$ whereas $v_i \in M, w_i \in span(N)$. Then I tried to, instead of writing $w_i$ saying that, since $w_i$ is an element of the span of $N$ it can be written out as a linear combination so I end up with:
$$f=\sum_{i=1}^nx_iv_i+\sum_{i=1}^ny_i\sum_{l\in N}z_nl$$ Am I done here with the first part? How do I show that this is a subset of $span(M\cup N)$? 
 A: In order to show that $\operatorname{span}(A)=\operatorname{span}(B)$, you need to show that


*

*every element of $A$ belongs to $\operatorname{span}(B)$

*every element of $B$ belongs to $\operatorname{span}(A)$
On the other hand, if $B\subseteq A$, the second statement is obviously verified, because

if $B\subseteq A$, then $\operatorname{span}(B)\subseteq\operatorname{span}(A)$

(Prove this.)
Since $N\subseteq\operatorname{span}(N)$, we also have
$$
M\cup N\subseteq M\cup\operatorname{span}(N)
$$
Thus we just need to show that each element $v\in M\cup\operatorname{span}(N)$ belongs to $\operatorname{span}(M\cup N)$.
This is clear if $v\in M$; so, suppose $v\in\operatorname{span}(N)$. Then, as $N\subseteq M\cup N$, we have $\operatorname{span}(N)\subseteq\operatorname{span}(M\cup N$) and therefore $v\in\operatorname{span}(M\cup N)$
A: Suppose $M= \{v_1,...,v_m\}$ and $N = \{u_1,...,u_n \}$.
Pick $v \in span ( M \cup span(N) ) $, then 
$$ v = \sum_{i=1}^m \alpha_i v_i + \gamma \sum_{i=1}^n\beta_i u_i $$
for escalars $\alpha_i, \beta_i, \gamma $. Putting $\delta_i = \gamma \beta_i$, one has 
$$ v = \sum \alpha_i v_i + \sum \delta_i u_i $$
So, $v$ is a linear combinations of vectors in $M \cup N$, which means it lies in $span( M \cup N)$
