# If the union of two subspaces is the vector space, then one of the subspaces is the vector space Proof

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?

• You meant $(U_1=V) \lor (U_2=V)$ instead of $(U_1=V) = (U_2=V)$ , right? Otherwise, the proof seems fine to me. Feb 13, 2020 at 9:28

$$$$U_1 \cup U_2 = V \iff ( \textbf{v} \in U_1 \cup U_2 \iff \textbf{v} \in V) \iff ((\textbf{v} \in U_1 \lor \textbf{v} \in U_2) \iff \textbf{v} \in V) \iff (\textbf{v} \in U_1 \iff \textbf{v} \in V) \lor (\textbf{v} \in U_2 \iff \textbf{v} \in V) \iff (U_1 = V) \lor (U_2=V)$$$$
Prove by contradiction. Suppose $$U_1 \neq V$$ and $$U_2 \neq V$$. Then there exist vectors $$x \in V \setminus U_1$$ and $$y \in V \setminus U_2$$. Consider $$x+y$$. Either $$x+y \in U_1$$ or $$x+y \in U_2$$. In the first case $$y \notin U_1$$ [because $$y \in U_1$$ would imply that $$x=(x+y)-y\in U_1$$]. But then $$y$$ is neither in $$U_1$$ nor in $$U_2$$ which is a contradiction.
In the second case $$x \notin U_2$$ [because $$x \in U_2$$ would imply that $$y=(x+y)-x\in U_2$$]. But then $$y$$ is neither in $$U_1$$ nor in $$U_2$$ which is a contradiction.