# A real vector space $V$ is the union of three subspaces $P$, $Q$, $R$ of $V$. Prove that one of $P$,$Q$, $R$ is $V$

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?

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