I came up with this proof which involves interpreting a truth table. I omitted the implication in the backwards direction because I'm more interested in reviewing the technique used in the first part. I wanted to post this here to get some feedback and to clarify any conceptual misunderstandings. If you know of any ways of improving or simplifying my proof I'd love to hear it. I know this question has been posted on here before but I think this is a pretty conclusive way of solving the problem in a way which actually shows you why the answer must be.
Let $W_1,W_2$ be subspaces of $V$. Prove the union $W_1 \cup W_2$ is also a subspace of $V$ if and only if $W_1 \subset W_2$, $W_2 \subset W_1$, or $W_1 = W_2$.
If $W_1 \cup W_2$ is a subspace of $V$, then $x,y \in W_1 \cup W_2 \iff x+y \in W_1 \cup W_2$.
By definition of union and closure under vector addition, $x+y \in W_1 \cup W_2 \equiv (x+y \in W_1) \lor (x+y \in W_2) \equiv (x \in W_1 \land x\in W_1)\lor (x \in W_2 \land x\in W_2).$
Clearly, this statement is only true when:
- Both $x$ and $y$ are in both $W_1$ and $W_2$.
- $x$ is in $W_1$ and $W_2$ but $y$ is only $W_1$ or only in $W_2$.
- $y$ is in $W_1$ and $W_2$ but $x$ is only $W_1$ or only in $W_2$
- $x$ and $y$ are only chosen from either $W_1$ or $W_2$ but not both.
These four cases can be true or false depending on the values of $x$ and $y$. However, only one of these cases may be true considering all possible $x,y \in W_1 \cup W_2$. Working backwards, we have $W_1 = W_2$ with (1), $W_1 \subset W_2$ or $W_2 \subset W_1$ with (2) and (3), and (4) can be discarded as it would necessitate that we cannot choose elements from one of the subspaces, which is to say one of the subspaces is empty, which is a contradiction.
We can also work backwards from here to show that if (1) or (2)/(3) are true, so it must be that $W_1 \cup W_2$ is a subspace of $V$ $\blacksquare$
Best Answer
Welcome to MSE! There's a few misunderstandings here.
First, your criterion
is not true. The forward direction is (that's more or less the definition of a subspace) but the backwards direction is not. For instance, if $v$ is any element not in $W_1 \cup W_2$, then $v + (- v) = 0 \in W_1 \cup W_2$, despite $v, -v \not \in W_1 \cup W_2$!
As another point, I'm not sure how you're making step $\star$ in
Judging by your truth table below, I think you meant to write
As we said earlier, though, knowing that $x+y \in W_1$ (resp. $W_2$) doesn't actually tell you that $x \in W_1 \land y \in W_1$ (resp. $W_2$)!
You then go on to make some other conclusions based on this faulty logic. These are also somewhat confusing, but I can't tell if they're just worded in a way I'm not used to or if there's another conceptual misunderstanding.
In particular, I'm not at all sure what you mean by
There is, of course, a cleaner approach -- you should argue by contradiction (Or by contrapositive if you prefer to think about it that way).
Say $W_1 \subseteq W_2$ and $W_2 \subseteq W_1$ are both false.
Then we can find an element $x_1 \in W_1 \setminus W_2$ and $x_2 \in W_2 \setminus W_1$.
Can you show that $x_1 + x_2 \not \in W_1 \cup W_2$? If you do, you'll have shown that $W_1 \cup W_2$ is not a subspace (which is what we want to show, to complete the contrapositive).
You'll need to use your (good!) idea from earlier that
$$x_1 + x_2 \in W_1 \cup W_2 \equiv (x_1 + x_2 \in W_1) \lor (x_1 + x_2 \in W_2)$$
you can then argue in a similar fashion to what you were doing earlier. It will be helpful to remember that $-x_1 \in W_1$ and $-x_2 \in W_2$, since they are both subspaces. Do you see how to take it from here?
I hope this helps ^_^