The relation between two subspaces [duplicate]

Prove or disprove

İf $V$ is a vector space over a field $F$, $A$ and $B$ are subspaces of $V$ such that $A ⊄ B$ and $B ⊄ A$ then $∃ x ∈ A+B$ such that $x ∉ A ∪ B$

Edit : Since $A + B = < A ∪ B >$ and since $A ∪ B ⊆ < A ∪ B >$ this mean $∃ x ∈ <A ∪ B >$ but $x ∉ A ∪B$ and since $A + B = <A ∪ B >$ then the result is true. Is my proof true ?

Solution 1:

Since $ A \not\subset B $ and $ B \not\subset A $, you have $ A \cap B^c $ and $ B \cap A^c $ non-empty. Take $ u \in A \cap B^c $ and $ v \in B \cap A^c $. Then, $ (u + v) \in A + B $. Suppose, if possible, $ (u + v) \in A $. But then $ v = (u + v) - u \in A $, since $ u \in A $ and $ A $ is a subspace. But this contradicts the assumption $ v \in B \cap A^c $. So, $ (u + v) \not\in A $. Arguing similarly, we get $ (u + v) \not\in B $. Hence $ (u + v) \not\in A \cup B $.

The gap in your proof is that $ A \cup B \subseteq \langle A \cup B \rangle $ does not imply the existence of $ x \in \langle A \cup B \rangle \cap (A \cup B)^c $. This is apart from the fact that the proof of $ A + B = \langle A \cup B \rangle $ is relatively more involved than the simpler proof of your problem statement.