I know this question has been asked before, but my proof is different, although a bit complicated, and I want to make sure that I didn't miss anything. Also, I know that the statement is not precisely true, and that the cardinality of a field F must be greater than or equal the number of subsets for the statement to be true; however, this was not covered by the book yet so I didn't worry about it. I've written the proof very informally since it's "exhaustive" because it considers multiple cases and it would be difficult to write it more formally. Please let me know if a clarification is needed.
here is the exact phrasing from "Linear Algebra Done Right" by Axler:
Prove that the union of three subspaces of V is a subspace of V if and only if one of the subspaces contains the other two.
[This exercise is surprisingly harder than the previous exercise, possibly because this exercise is not true if we replace $\mathbf{F}$ with a field containing only two elements.]
$\mathbf{F}$ here is treated as $\Bbb{R}$ or $\Bbb{C}$ so I didn't worry about having a field of two elements in my proof.
Let U$_a$,U$_b$,U$_c$ be subspaces of the vector space V. Let a∈U$_a$, b∈U$_b$, c∈U$_c$ and Let U= U$_a$∪ U$_b$∪U$_c$. For U to be a subspace:
1.The identity vector must exist: it's there since the U is a union of subspaces.
2.Closed under addition and scalar multiplication:
$a,b,c \in U\rightarrow a+b,b+c,a+c,a+b+c \in U$
Now consider the following cases:
case $1$:
Let it be that for all the subsets
$x+y \in U_x$ or $x+y \in U_y$ where, $x,y in \{a,b,c\}$ and $x\neq y$. (I'll refer to this as "condition")
Let's take as an example:
$a+b \in U_a$, $b+c \in U_b$, $a+c \in U_c$
$a+b \in U_a$ $\rightarrow$ $b \in U_a$
$b+c \in U_b$ $\rightarrow$ $c \in U_b$
$a+c \in U_c$ $\rightarrow$ $a \in U_c$
Now $a+b+c$ can belong to any $U_i$ and that would imply that $a,b,c$ belong to that $U_i$ proving our proposition. (Or we can use the fact that the subsets are equivalent since they're contained in each other)
case 2:
Two of the subsets satisfy the previous condition but one does not.
Let's take as an example: $a+b \in U_a$, $a+c \in U_b$, $b+c \in U_c$
$a+b \in U_a$ $\rightarrow$ $b \in U_a$
$b+c \in U_c$ $\rightarrow$ $b \in U_c$
Now if $a+b+c \in U_a$ or $a+b+c \in U_c$ then our proposition is proved. However, if it belongs to $U_b$ then nothing is implied. But by the implications above, b belongs to both $U_a$ and $U_c$. Therefore:
$U_b$ $\subset$ $U_c$ and $U_b$ $\subset$ $U_a$ $\rightarrow$ $a+c \in U_a$ and $a+c \in U_c$. Proving our proposition.
case 3:
None of the subsets satisfy the previous condition so we have:
$b+c \in U_a$, $a+c \in U_b$, $a+b \in U_c$
consider $ai+bj+ck$ where $i,j,k \in \mathbf{F}$. $ai+bj+ck \in U$ so it must belong to one of subsets. However, if i=j or j=k or i=k, then it is obvious that it would belong to one of the subsets;
for example: if $i=j$ then $(a+b)i+ck \in U_c$ by addition and scalar closure.
So to take the argument further, suppose that $i\neq j \neq k$.
Then $ai+bj+ck$ belongs to one of the subsets. For the sake of the argument let that subset be $U_a$. Then, since $ai ∈ U_a$, and $(b+c) \in U_a$ we can do the following:
$ai+bj+ck - (ai + (b+c)j) = c(k-j)$. Therefore $c(k-j) \in U_a$ and this implies that $c \in U_a$.
Finally since $a,c,b+c \in U_a$ $\rightarrow$ $b \in U_a$ completing the argument. This argument can be replicated for when $ai+bj+ck \in U_b$ or $U_c$.
My Main question is regarding whether this covered all the cases and that there was not any wrong assumptions in my proof. I've reviewed multiple times but I still have my doubts. Also I'm open to any advices regarding my communication of the question and if I should've made something more clear.
Thanks.