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Closely related question

Suppose that $V= P \ \cup Q \ \cup R $. Leaving out the trivial case where one subspace contains the other one, we consider the following case:

$P-Q \neq \emptyset, Q-P \neq \emptyset,Q-R\neq \emptyset, R-Q\neq \emptyset,P-R\neq \emptyset, R-P\neq \emptyset $, i.e. each subspace contains elements, unavailable to the other ones.

We take $p \in P\setminus Q\cup R, q\in Q\setminus P\ \cup R$. Now, we consider the elements $p+q$ and $p+(-q)$. Evidently the two aforementioned elements can neither be in $P$ nor in $Q$ [ $\because $ if in $P$, $(p+q)+(-p)=q \in P$, contradiction. Similar argument for the other one] . Since, the union of the three subspaces make up the entire vector space, both of them must be in $R$. Now, $(p+q)+(p+(-q))=2p\in R$. As $R$ is a real subspace, $\frac{1}{2}(2p)=p\in R$, a contradiction.

Therefore, one of them must be $V$.

Is the proof correct?

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    $\begingroup$ Compare with the proof at this duplicate. $\endgroup$ – Dietrich Burde Mar 16 '19 at 19:52
  • $\begingroup$ @DietrichBurde Almost similar. $\endgroup$ – Subhasis Biswas Mar 16 '19 at 19:53
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    $\begingroup$ Or see this duplicate, this is well-known. $\endgroup$ – Dietrich Burde Mar 16 '19 at 19:54
  • $\begingroup$ didn't read the rest of the answer on the first look, as the question differed. $\endgroup$ – Subhasis Biswas Mar 16 '19 at 19:54
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    $\begingroup$ We get $V\subseteq P\subseteq V$ by the duplicate, so equality (or with $Q$ or $R$). It is exactly the same question. $\endgroup$ – Dietrich Burde Mar 16 '19 at 19:55

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