I am self studying linear algebra by Linear Algebra by Kenneth M Hoffman but I am having a difficulty in proof checking because there is no one to check my solutions. I think there is no harm in asking for examining my solution here.
Here is the specific problem:
Let $W_1$ and $W_2$ be subspaces of a vector space $V$ such that the set theoretic union $W$ is also a subspace. Prove that one of the subspace contains the other.
Proof by Contradiction:
Suppose that it is false that one of the subspace contains the other. Then there exists two vectors $\alpha_1 \in W_1$ but $\alpha_1 \notin W_2 $ and $\alpha_2 \in W_2$ but $\alpha_2 \notin W_1 $.
$\alpha_1+\alpha_2 \in W=W_1 \cup W_2$ because $\alpha_1$ and $\alpha_2$ are two vectors in $W$ which is a subspace of $V$. This implies that $\alpha_1+\alpha_2 \in W_1$ OR $\alpha_1+\alpha_2 \in W_2$. Suppose that $\alpha_1+\alpha_2 \in W_1$,( If $\alpha_1+\alpha_2 \in W_2$, the proof follows by symmetry i.e replacing $W_1$ by $W_2$) then $(\alpha_1+\alpha_2)-\alpha_1= \alpha_2 \in W_1$ because $W_1$ is a subspace of $V$. This contradicts with the implication that 'there exists two vectors $\alpha_1 \in W_1$ but $\alpha_1 \notin W_2 $ and $\alpha_2 \in W_2$ but $\alpha_2 \notin W_1 $.' which followed by the assumption that 'it is false that one of the subspace contains the other', so this assumption must be false. Hence,it is true that one of the subspace contains the other.
Please verify my solution for correctness and tell me how can I make it more compact like the proofs the textbook provides.
Best Answer
Your proof is fine.
You could write the following passage more compact:
This can be shortened to "But this contradicts the assumption $\alpha_2 \notin W_1$".