The question I'm working on is as follows:
Suppose $U_1$ and $U_2$ be subspaces of a vector space $V$. Show that $U_1+U_2$ equals the intersection of all subspaces of $V$ that include $U_1 \cup U_2$.
My attempt:
Let $K = \bigcap_{i = 0}V_i$ be the intersection of all subspaces such that $U_1$ $\cup$ $U_2$ $\subseteq V_i$ for all $i$. Let $U = U_1+U_2 = \{x + y \in V : x \in U_1, y \in U_2\}$. We must show that $U \subseteq K$ and $K \subseteq U$.
Since the intersection of any subspaces results in another subspace, we know K is a subspace. Since subspaces are closed under addition we have that for $x,y$ in $K$, $x+y\in K$. Letting $x$ and $y$ be elements from $U_1 \cup U_2$, it is clear that $U \subseteq K$.
The span of a set $S$ is the intersection of all subspaces of a vector space $V$ that contain $S$. Hence $K = \operatorname{span}(U_1 \cup U_2)$. So elements from $K$ can be written as linear combinations from $U_1 \cup U_2$. Since $U$ can also be represented as linear combinations from $U_1 \cup U_2$ it follows that $K \subseteq U$.
This felt a little too easy so I'm thinking that I have missed something important. Any help here would be great.
Best Answer
It is really that simple! Suppose you have any subspace $W$ of $V$ such that $U_1\cup U_2\subseteq W$. Then we have, for any $u_1\in U_1, u_2\in U_2$, $u_1+u_2\in W$ because $W$ is a subspace. Thus, $U_1+U_2\subseteq W$. As $W$ was arbitrary, we know that this inclusion holds for all such subspaces, and therefore holds for their inclusion giving us $U_1+U_2\subseteq \bigcap_{U_1\cup U_2\subseteq W}W$. On the other hand, $U_1+U_2$ is itself a subspace of $V$ containing $U_1\cup U_2$ which gives the converse inclusion.