Question in proof that $\dim(U_1+U_2)=\dim U_1+\dim U_2-\dim(U_1 \cap U_2)$ In the proof that Dim$(U_1+U_2)=DimU_1+DimU_2-Dim(U_1 \cap U_2)$
In "Linear Algebra Done Right" by Sheldon Axler:

Theorem 2.18: 
If $U_1$ and $U_2$ are subspaces of a finite dimensional vector space,
  then:
$\dim(U_1+U_2)=\text{dim}(U_1)+\text{dim}(U_2)-\text{dim}(U_1 \cap U_2)$

PROOF (in short): 
Let $(u_1,u_2,...)$ be a basis of $U_1\cap U_2$. This can be extended to a basis $(u_1,u_2,...,u_m,v_1,v_2,...,v_j)$ of $U_1$. Also, it can be extended to a basis $(u_1,u_2,...,u_m,w_1,w_2,...,w_k)$ of $U_2$.
Clearly $\text{span}(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$ is $U_1 + U_2$. To show that this list is a basis of $U_1+U_2$ we just need to show that it is linearly independent. 
I cannot understand why $\text{span}(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$ is $U_1 + U_2$. Can someone help explain this to me?
 A: This is an instance of a more general fact:

$$\operatorname{span}(A\cup B)=\operatorname{span}(A) + \operatorname{span}(B).$$

which may be proven by noting that if $v_a\in \operatorname{span}(A)$ and $v_b\in \operatorname{span}(B)$ then clearly $v_a+v_b$ can be written as a linear combination of elements in $A\cup B$ by simply expanding $v_a$ as a combination in $A$ and $v_b$ as a combination in $B$. Conversely, if $v\in \operatorname{span}(A \cup B)$, then $v$ can be written as a linear combination of elements in $A\cup B$, which can be split as a sum of a linear combination of elements of $A$ plus a combination of elements of $B$.
Here, we could take $A$ as the basis of $U_1$ and $B$ as the basis of $U_2$ to see that $\operatorname{span}(A\cup B)= U_1+U_2$.
A: In very simple terms:


*

*by construction every vector in $U_1$ is a linear combination of $(u_1,...,u_m,v_1,...,v_j)$ and every vector in $U_2$ is a linear combination of $(u_1,...,u_m,w_1,...,w_k)$.

*By definition, every vector in $U_1+U_2$ is of the form $u+u^\prime$ with $u\in U_1$ and $u^\prime\in U_2$.
Putting things together every vector in $U_1+U_2$ can be written as a linear combination of $(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$.
