So, I'm trying to prove the following result.
Let $V$ be a vector space over $F$. Let $U_1$ and $U_2$ be subspaces of $V$. Then, the following is true:
$U_1 \cup U_2 = V \implies [(U_1 = V) \lor (U_2 = V)]$
Proof Attempt:
Since $U_1 \cup U_2$ is a vector space, we have previously shown that $U_1 \subset U_2 \lor U_2 \subset U_1$. Hence:
$[U_1 \cup U_2 = U_1] \lor[ U_1 \cup U_2 = U_2] \implies (U_1 = V) \lor (U_2 = V)$.
Where we have made use of the fact that $A \subset B \implies A \cup B = B$. This proves the result.
Can someone check my proof above and see if it's correct? Also, how would I prove this without having proved the statement right at the beginning? Like, how might I prove this from first principles?
Best Answer
\begin{equation} U_1 \cup U_2 = V \iff ( \textbf{v} \in U_1 \cup U_2 \iff \textbf{v} \in V) \iff ((\textbf{v} \in U_1 \lor \textbf{v} \in U_2) \iff \textbf{v} \in V) \iff (\textbf{v} \in U_1 \iff \textbf{v} \in V) \lor (\textbf{v} \in U_2 \iff \textbf{v} \in V) \iff (U_1 = V) \lor (U_2=V) \end{equation}