The union of three subspaces of V is a subspace of V

[SC0]

Homework Statement

This is the exact phrasing form 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 F with a field containing only two elements.]

The Attempt at a Solution

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∈U→a+b,b+c,a+c,a+b+c∈U
Now consider the following cases:
case 1:
Let it be that for all the subsets
x+y∈U_x or x+y∈U_y where, x,y∈{a,b,c} and x≠y. (I'll refer to this as "condition")
Let's take as an example: a+b∈U_a, b+c∈U_b, a+c∈U_c
a+b∈U_a → b∈U_a
b+c∈U_b → c∈U_b
a+c∈U_c → a∈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∈U_a, a+c∈U_b, b+c∈U_c
a+b∈U_a → b∈U_a
b+c∈U_c → b∈U_c
Now if a+b+c∈U_a or a+b+c∈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 ⊂ U_c and U_b ⊂ U_a → a+c∈U_a and a+c∈U_c. Proving our proposition.
case 3:
None of the subsets satisfy the previous condition so we have:
b+c∈U_a, a+c∈U_b, a+b∈U_c
consider ai+bj+ck where i,j,k∈F. ai+bj+ck∈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∈U_c by addition and scalar closure.
So to take the argument further, suppose that i≠j≠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)∈U_a we can do the following:
ai+bj+ck−(ai+(b+c)j)=c(k−j). Therefore c(k−j)∈U_a and this implies that c∈U_a.
Finally since a,c,b+c∈U_a → b∈U_a completing the argument. This argument can be replicated for when ai+bj+ck∈U_b or U_c.
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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 know that the statement is not precisely true, and that the cardinality of a field F must be greater than or eqU_al 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.

 

[fresh_42]

Can you structure your proof in a readable manner?
a) Avoid double usage of the same signs.
b) Say which direction of the two ($\Longleftarrow ,\Longrightarrow$) you refer to by what you write?
c) Use a better emphasis on the quantifiers you use, i.e. does "where $x,y \in \{\,a,b,c\,\}$" mean for all pairs or only for a special pair?
d) Use LaTeX to type the few formulas, see https://www.physicsforums.com/help/latexhelp/

You defined the three subsets as $U_a,U_b,U_c$ and their union as $U$. Fine. I personally would have preferred numerical indices in order to avoid confusion with the vectors of the same name, which is a potential source for cheating, but o.k. Now what I do not get is: Which direction do you want to prove before you defined these cases? Or do you prove equivalence in each of them? What do you mean by "Two of the subsets satisfy the previous condition but one does not." Which "previous condition"?

Sorry, but it takes more effort to guess what you might have meant, than it takes to prove the statement. At least as far as I am concerned.

 

[SC0]

Yeah, I apologize for the mess. I found it very difficult to organize my thoughts regarding this question and I ended up writing it in an unclear way.

What I'm trying to prove is this:
(Union of three subspaces of V is a subspace of V) $\Rightarrow$ (One of the subspaces contains the other two)
As regarding the "previous condition" and every usage of the word "condition" in my post, it refers to this:
Let it be that for all the subsets:
x+y∈U_x $\vee$ x+y∈U_y where, x,y ∈ {a,b,c} and x≠y.
What I mean by that is that if x ∈ U_a and y ∈ U_b then x+y belongs to either one of them, as opposed to some other subset.

Please let me know wherever clarification is needed, and thanks for the help, this problem has been bugging me for a while now.

 

[vela]

What I mean by that is that if x ∈ U_a and y ∈ U_b then x+y belongs to either one of them, as opposed to some other subset.

This doesn't hold. For example, consider the case where V = R³, U_a = {(r, 0, 0): r real}, and U_b = {(0, r, 0): r real}.

 

[SC0]

This doesn't hold. For example, consider the case where V = R³, U_a = {(r, 0, 0): r real}, and U_b = {(0, r, 0): r real}.

I know, this is only one case of three other cases. I'm trying to prove, exhaustively, that for the union of three subspaces to be a subspace one of them must contain both of them.

The first case I consider is if this condition (x ∈ U_a and y ∈ U_b then x+y belongs to either one of them, as opposed to some other subset.) happens to apply to all of the subsets.
The second case is if this condition applies only to two of the subsets.
The third case is when the condition does not apply to any of them.