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Here is a slightly more well defined question than in the title, which avoids counter examples using a field with only two elements: prove that the union of three subspaces of a vector space $V$ over a field $F$ ($|F|> 2$) is a subspace of $V$ if and only if one of the subspaces contains the other two.

A very similar question has already been asked and answered here. However the answer there states things such as:

Without loss of generality, we can assume the whole space $V$ is in fact $V_1+V_2+V_3$. Easily seen that in fact we must also have $V = V_1 \cup V_2 \cup V_3$.

and

take $a$ to be anything except $0$,$1$, and take $b=a−1$.

Why can we assume such things? Can anybody provide a slightly less terse explanation, that justifies statements such as the above which may be less obvious to a beginner.

Jony Pollon
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  • vectors outside $V_1 + V_2 + V_3$ are irrelevant to what we want to prove. 2) because we are assuming $|F| > 2$, there are elements of $F$ other than $0$ and $1$, so can choose an $a$ that is neither $0$ nor $1$.
    • – Rob Arthan Jun 14 '20 at 22:53
  • @RobArthan 1) But why is $V= V_1+V_2+V_3$ - how is it assumed that the sum of our sub-spaces are the original vector space - this might not be the case? 2) But what is $a$ is $0$ or $1$? Why can this be done without loss of generality? – Jony Pollon Jun 14 '20 at 23:06
  • The vectors outside $V_1 + V_2 + V_3$ are irrelevant to what we want to prove, so we can forget about them and assume that $V = V_1 + V_2 + V_3$. 2) $a$ is our choice: as $|F| > 2$, we can choose $a$ to be different from both $0$ and $1$.
    • – Rob Arthan Jun 14 '20 at 23:13
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    Consider the subspaces $V_1, V_2,$ and $V_3$ of $V.$ We wish to prove that $V_1 \cup V_2 \cup V_3$ is a subspace of $V$ if and only if $V_i, V_j \subseteq V_k$ for some distinct integers $i, j, k \in {1, 2, 3}.$ If $V_i, V_j \subseteq V_k,$ then we have that $V_1 \cup V_2 \cup V_3 = V_k$ is a subspace by hypothesis, so the interesting thing here is that if $V_1 \cup V_2 \cup V_3$ is a subspace, then $V_i, V_j \subseteq V_k.$ Consider the subspace $W = V_1 + V_2 + V_3.$ If $V_1 \cup V_2 \cup V_3$ is a subspace, then it is not difficult to check that $W = V_1 \cup V_2 \cup V_3.$ Start here. – Dylan C. Beck Jun 15 '20 at 15:40
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    Like Rob suggests, the point is to show the containment, so we may choose the scalars $a$ (other than $0$ or $1$) and $b = a - 1$ if that choice behooves us. – Dylan C. Beck Jun 15 '20 at 15:42