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

#### SC0

1. The problem statement, all variables and given/known data

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.]

3. The attempt at a solution

Let Ua,Ub,Uc be subspaces of the vector space V. Let a∈Ua, b∈Ub, c∈Uc and Let U= Ua∪ Ub∪Uc. 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∈Ux or x+y∈Uy where, x,y∈{a,b,c} and x≠y. (I'll refer to this as "condition")
Let's take as an example: a+b∈Ua, b+c∈Ub, a+c∈Uc
a+b∈Ua → b∈Ua
b+c∈Ub → c∈Ub
a+c∈Uc → a∈Uc
Now a+b+c can belong to any Ui and that would imply that a,b,c belong to that Ui 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∈Ua, a+c∈Ub, b+c∈Uc
a+b∈Ua → b∈Ua
b+c∈Uc → b∈Uc
Now if a+b+c∈Ua or a+b+c∈Uc then our proposition is proved. However, if it belongs to Ub then nothing is implied. But by the implications above, b belongs to both Ua and Uc. Therefore:
Ub ⊂ Uc and Ub ⊂ Ua → a+c∈Ua and a+c∈Uc. Proving our proposition.
case 3:
None of the subsets satisfy the previous condition so we have:
b+c∈Ua, a+c∈Ub, a+b∈Uc
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∈Uc 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 Ua. Then, since ai∈Ua, and (b+c)∈Ua we can do the following:
ai+bj+ck−(ai+(b+c)j)=c(k−j). Therefore c(k−j)∈Ua and this implies that c∈Ua.
Finally since a,c,b+c∈Ua → b∈Ua completing the argument. This argument can be replicated for when ai+bj+ck∈Ub or Uc.
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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 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.

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#### fresh_42

Mentor
2018 Award
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∈Ux $\vee$ x+y∈Uy where, x,y ∈ {a,b,c} and xy.
What I mean by that is that if x ∈ Ua and y ∈ Ub 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

Staff Emeritus
Homework Helper
What I mean by that is that if x ∈ Ua and y ∈ Ub 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, Ua = {(r, 0, 0): r real}, and Ub = {(0, r, 0): r real}.

#### SC0

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

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

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