Help with proving a set is a spanning set I'm trying to prove that a vector space $V \cong U \oplus (V/U)$ where $U \subset V$. I want to do this by proving that $\{u_i\}_{i \in I} \oplus \{v_j +U\}_{j \in J}$ is a basis for $V$ where $\{u_i\}_{i \in I}$ is a basis for $U$ and $\{v_j\}_{j \in J}$ is a set in $V$ such that $\{v_j+U\}_{j \in J}$ forms a basis for $V/U$. I think I've proved that the set is linearly independent but I'm struggling with the spanning proof.
So far I've got, for $v \in V$;
\begin{eqnarray*}
v &=& v+(u-u),\qquad u \in U \\
 &=& -u +v+u \\
 &=& -\sum_{i \in I}a_iu_i+\sum_{j \in J}b_j(v_j+U)\\
\end{eqnarray*}
but I'm not at all confident in the last step as I think I'm mixing up cosets and elements of cosets. Any advice?
 A: It is in general not true that $V=U\oplus(V/U)$, even as sets, because $V/U$ is a set of subsets of $V$. However, it is true that $V\cong U\oplus(V/U)$, even as vector spaces. How to prove this depends on what you already know. Starting from scratch, you could prove that:


*

*If $\mathcal{B}_{V/U}$ is a basis for $V/U$, then for any $u\in U$ the set
$$\mathcal{B}_{V/U}':=\{v_i+u\mid v_i+U\in\mathcal{B}_{V/U}\},$$
is a linearly independent subset of $V$.

*If $\mathcal{B}_U$ is a basis for $U$, then 
$$\mathcal{B}:=\mathcal{B}_U\cup\mathcal{B}_{V/U}',$$
is a linearly independent subset of $V$.

*The set $\mathcal{B}$ above spans $V$.

*The above bijection between $\mathcal{B}_U\cup\mathcal{B}_{V/U}$ and $\mathcal{B}$ gives rise to an isomorphism $V\cong U\oplus(V/U)$.



In your proof of step 3, multiple things go wrong:


*

*In the first expression, this $u\in U$ needs some quantification. Do you mean to say that these equalities hold for all $u\in U$, or for some, or for a specific one?

*Your second equality is between elements and subsets of $V$, which is generally problematic.

*Note that it is in general not true that an element of $V$ of the form $v+u$ is in the span of $\{v_j\}_{j\in J}$.


In stead, consider the following approach:
As $\{v_j+U\}_{j\in J}$ spans $V/U$, for every $v\in V$ there exist $\{b_j\}_{j\in J}$ (with finite support) such that
$$v+U=\sum_{j\in J} b_j(v_j+U)=(\sum_{j\in J} b_jv_j)+\sum_{j\in J} b_j U=(\sum_{j\in J} b_jv_j)+U.$$
It follows that $v-\sum_{j\in J}b_jv_j\in U$. So there exist $\{a_i\}_{i\in I}$ (again with finite support) such that
$$v-\sum_{j\in J}b_jv_j=\sum_{i\in I}a_iu_i\qquad\text{ so }\qquad v=\sum_{i\in I}a_iu_i+\sum_{j\in J}b_jv_j,$$
which shows that that $\{u_i\}_{i\in I}\cup\{v_j\}_{j\in J}$ spans $V$.
