Proof.

IfW1∪W2W1∪W2is a subspace, thenW1⊂W2W1⊂W2orW2⊂W1W2⊂W1.

IfW1⊂W2W1⊂W2orW2⊂W1W2⊂W1, thenW1∪W2W1∪W2is a subspace.

IfW1∪W2W1∪W2is a subspace, thenW1⊂W2W1⊂W2orW2⊂W1W2⊂W1.

(⟹)(⟹)Suppose that the unionW1∪W2W1∪W2is a subspace ofVV.Seeking a contradiction, assume thatW1⊄W2W1⊄W2andW2⊄W1W2⊄W1.This means that there are elements

x∈W1∖W2andy∈W2∖W1.x∈W1∖W2andy∈W2∖W1.

SinceW1∪W2W1∪W2is a subspace, it is closed under addition. Thus, we havex+y∈W1∪W2x+y∈W1∪W2.

It follows that we have either

x+y∈W1orx+y∈W2.x+y∈W1orx+y∈W2.

Suppose thatx+y∈W1x+y∈W1. Then we write

y=(x+y)−x.y=(x+y)−x.

Since bothx+yx+yandxxare elements of the subspaceW1W1, their differencey=(x+y)−xy=(x+y)−xis also inW1W1. However, this contradicts the choice ofy∈W2∖W1y∈W2∖W1.

Similarly, whenx+y∈W2x+y∈W2, then we have

x=(x+y)−y∈W2,x=(x+y)−y∈W2,

and this contradicts the choice ofx∈W1∖W2x∈W1∖W2.

In either case, we have reached a contradiction.Therefore, we have eitherW1⊂W2W1⊂W2orW2⊂W1W2⊂W1.

IfW1⊂W2W1⊂W2orW2⊂W1W2⊂W1, thenW1∪W2W1∪W2is a subspace.

(⟸)(⟸)I haveW1⊂W2W1⊂W2, then it yields thatW1∪W2=W2W1∪W2=W2and it is a subspace ofVV.

Similarly, ifW2⊂W1W2⊂W1, then we haveW1∪W2=W2W1∪W2=W2and it is a subspace ofVV.In either case, the unionW1∪W2W1∪W2is a subspace