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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 ?

cpiegore
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samah Darweesh
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    No, your proof is not "true". The claim "but $x\not \in A\cup B$" has no proof. You say "since $A+B=\langle A\cup B\rangle$, but why then $x$ is not in $A\cup B$? – Dietrich Burde Dec 23 '21 at 17:59
  • Because A+B=⟨A∪B⟩ and A ∪ B ⊆ < A ∪ B > we can find an x such that x ∈ <A ∪ B > but x ∉ A ∪ B why is not true? – samah Darweesh Dec 23 '21 at 18:02
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    You have to explain why we can find such an $x$. Simply saying "because A+B=⟨A∪B⟩ and A∪B⊆<A∪B>" is not an explanation. Consider a vector space $V$ as a subspace of itself: then we have $V=\langle V\rangle$ and $V\subseteq V$ but that does not imply you can find an $x\in V$ but $x\not\in V$, so this reasoning alone is invalid. As a hint for how to proceed: your hypotheses tell you there exists an $x\in A\setminus B$ and a $y\in B\setminus A$, and elements of $A+B$ are sums of elements of $A$ and of $B$... – anon Dec 23 '21 at 18:15
  • What do you mean by $ < A \cup B > $? Is it the smallest subspace containing $ A \cup B $? – joy Dec 23 '21 at 18:17
  • @joy Yes. ${}{}$ – anon Dec 23 '21 at 18:22
  • @runway44 How does one find an $x$ such that $x\in V$ but $x\not\in V$ – cpiegore Dec 23 '21 at 18:22
  • @cpiegore You can't, obviously, which is my point. – anon Dec 23 '21 at 18:23

1 Answers1

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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.

joy
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  • What is the difference between $A \cup B$ and $\langle A \cup B \rangle$? – cpiegore Dec 23 '21 at 20:18
  • OP used the notation $ \langle A \cup B \rangle $ to mean the smallest subspace containing $ A \cup B $. – joy Dec 23 '21 at 20:23
  • The gap or the defect in her proof? – Mikasa Dec 24 '21 at 12:17
  • I said gap as in this case there indeed exists an $ x \in \langle A \cup B \rangle \cap (A \cup B)^c $, but its existence is not evident from just the fact that $ A \cup B \subseteq \langle A \cup B \rangle $. – joy Dec 24 '21 at 13:27