When Does $\operatorname{Span}(A+B)=\operatorname{Span}(A)+\operatorname{Span}(B)$? let there be $A,B \in V$.

$$A+B=\{a+b: a\in A,b \in B\}$$
Does $\operatorname{Span}(A+B)=\operatorname{Span}(A)+\operatorname{Span}(B)$?
I think it is right, even if there are vectors that are both in $A$ and $B$.
 A: Your notation is unclear. You probably mean $A$ and $B$ are subsets of $V$. If we denote by $A+B=\{a+b: a\in A, b\in B\}$, then a vector $v$ belongs to $\def\span{\operatorname{span}}\span(A+B)$ if it can be written as
$$
v=\alpha_1(a_1+b_1)+\alpha_2(a_2+b_2)+\dots+\alpha_n(a_n+b_n)
$$
for scalars $\alpha_1,\alpha_2,\dots,\alpha_n$ and $a_1,a_2,\dots,a_n\in A$ and $b_1,b_2,\dots,b_n\in B$. In this case
$$
v=(\alpha_1a_1+\alpha_2a_2+\dots+\alpha_na_n)+
(\alpha_1b_1+\alpha_2b_2+\dots+\alpha_nb_n)
$$
and this proves $\span(A+B)\subseteq\span(A)+\span(B)$.
If the vector space $V$ is finite dimensional, in order to have the reverse inclusion the condition is
\begin{align}
\dim\span(A+B)&=\dim(\span(A)+\span(B))\\
&=\dim\span(A)+\dim\span(B)-\dim(\span(A)\cap\span(B))
\end{align}
This is not true in general; an easy example is $A=\{u\}$, $B=\{-u\}$, where $u\ne0$. Then $A+B=\{0\}$, while $\span(A)=\span(B)=\langle u\rangle$ and, clearly,
$$
\{0\}=\span(A+B)\ne\span(A)+\span(B)=\langle u\rangle.
$$
A: To show they are equal, you have to show span{$ A + B $} $\subseteq$ span{$ A$} + span{$B $}, and span{$ A$} + span{$B $} $\subseteq$ span{$ A + B $}.
For span{$ A + B $} $\subseteq$ span{$ A$} + span{$B $}, let $x \in$ span{$ A + B $}.  Then $x = c(a + b)$ for some scalar $c$ and elements $a \in A, b \in B$.  But then that means $x = ca + cb$, and $ca \in$ span{$A$}.  Similarly, $cb \in$ span{$B$}.  So $ca + cb \in $ span{$A$} + span{$B$}.
So we know as we just proved that span{$ A + B $} $\subseteq$ span{$ A$} + span{$B $}.  But the other way does not hold.  Here is a counter example:
Suppose $V = \mathbb{R}^{2}$.  Let $A = \left \{ \begin{bmatrix}1 \\ 0 \end{bmatrix} \right \}$ and $B = \left \{ \begin{bmatrix}0 \\ 1 \end{bmatrix}  \right \}$.  Then span{$A$} $ = \left \{ \begin{bmatrix}c \\ 0 \end{bmatrix} \mid c \in \mathbb{R} \right \}$.  Similarly, span{$B$} $ = \left \{ \begin{bmatrix}0 \\ d \end{bmatrix} \mid d \in \mathbb{R} \right \}$.  Then span{$A$} + span{$B$} $= \left \{ \begin{bmatrix}c \\ d \end{bmatrix} \mid c,d \in \mathbb{R} \right \} = \mathbb{R}^{2}$.  But span{$A + B$} $= \left \{  \begin{bmatrix}c \\ c \end{bmatrix} \mid c \in \mathbb{R} \right \}$.  This clearly is not all of $\mathbb{R}^{2}$, since it only contains vectors where the $x$ and $y$ coordinate are equal.  In particular, $\begin{bmatrix} 2 \\ 3 \end{bmatrix} \in \mathbb{R}^{2} = $ span{$A$} + span{$B$}, but $\begin{bmatrix} 2 \\ 3 \end{bmatrix} \not \in $ span{$A + B$}.
