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Let $W_1$ and $W_2$ be subspaces of a vector space $V$. Prove that $W_1 \cup W_2$ is subspace of $V$ if and only if $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$

Attempt:

$\boxed {p\Leftarrow q}$

(Almost trivial): Assume $W_1 \subseteq W_2$ then $W_1 \cup W_2=W_2$ and since $W_{1,2}$ are subspaces proof of this side is over.

$\boxed {p\Rightarrow q}$ I couldnot do the direct proof so I wanted to assume otherwise, here is where trouble begins.

I don't know how to take negative statement of the $p\Rightarrow q$ as follows:

Otherwise statement of the $\boxed {p\Rightarrow q}$ is $W_1 \cup W_2$ is subspace $V$ but $\boxed{\boxed{W_1 \subseteq W_2\; or\; W_2 \subseteq W_1}}$ is wrong.

How can I do the proof in the assumming otherwise, and can you give me hint to prove the theorem (hints or answers, of course, can be other methods.)

Micheal Brain Hurts
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1 Answers1

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If say we have $x\in W_1\setminus W_2$ and $y\in W_2\setminus W_1$, so $x,y \in W_1\cup W_2$ but we can't say $x+y\in W_1\cup W_2$.

nonuser
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