direct proof of $U_1\cup U_2$ is subspace of $V$ $\implies$ $U_1\subset U_2$ or $U_2\subset U_1$

Question

I'm trying to prove the following by direct proof (I've already got a proof by contradiction).

The union of two subspace of $V$ is a subspace of $V$ implies one of the subspaces is contained in the other.

Here's how far I've gotten: Suppose $U_1\cup U_2$ is a subspace of $V$. Let $u_1 \in U_1$ and $u_2 \in U_2$. Then, by closure of vector addition in $U_1\cup U_2$, $u_1 + u_2 \in U_1\cup U_2$. This implies that $u_1 + u_2 \in U_1$ or $u_1 + u_2 \in U_2$. WLOG, suppose $u_1 + u_2 \in U_1$. Then $u_2 = (u_1 + u_2) - u_1 \in U_1$.

At this point, I'd like to be able to say that because $u_2$ was chosen arbitrarily, this implies that $U_2 \subset U_1$. But I also made the choice that $u_1 + u_2 \in U_1$, so for some other choice of $u_2$ it could be the case that $u_1 + u_2 \in U_2$ which would imply $u_1 \in U_2$.

Is there any way to complete this proof directly or do I have to stick with the proof by contradiction for this one?

Answer 1

You don't need to use contradiction really, but mere inspection of cases. Just prove that if $U_1$, say, is not contained in $U_2$ i.e. if there's a $u_1\in U_1\smallsetminus U_2$, then $U_2\subset U_1$.

Answer 2

The idea is OK, go by contradiction.

Suppose neither inclusion holds, but $U_1 \cup U_2$ is a subspace. So we have $u_1 \in U_1 ,u_1 \notin U_2$ and $u_2 \in U_2, u_2 \notin U_1$ to witness the non-inclusions.

But $u_1 + u_2 \in U_1 \cup U_2$, say in $(u_1 + u_2) \in U_i$. But then $(u_1 + u_2 ) - u_i \in U_i$, while it equals $u_j$ where $j \neq i$. Contradiction, as $u_j \notin U_i$ when $j \neq i$.

Answer 3

Your suspect is right: “without loss of generality” is improper.

For instance, if you chose $u_1=0$, then saying that “without loss of generality $u_1+u_2\in U_1$” assumes that $U_2\subseteq U_1$.

There is no proof by contradiction involved.

If there exists $u_2\in U_2$ such that, for all $u_1\in U_1$, we have $u_1+u_2\in U_2$, then $U_1\subseteq U_2$, because $u_1=(u_1+u_2)-u_2$.

Assume that $U_1\not\subseteq U_2$. Then, for every $u_2\in U_2$, there exists $u_1\in U_1$ with $u_1+u_2\in U_1$. In particular, $u_2=(u_1+u_2)-u_1\in U_1$. Hence $U_2\subseteq U_1$.