vector space question thats driving me nuts?

Question

let V be a vector space and let m be any set of linear subspaces of V. let A be the union of the subspaces in m, that is, A = U m = (x e V: xeM for some M e m)
e=element of

show that there exists a smallest linear subspace of V that contains every M e m

tough isnt it!

Answer 1

Nope. Consider that for any set A⊆V whatsoever, there is a smallest subspace containing it as a subset - specifically, the intersection of all subspaces of V that contain A, which is also equal to the algebraic span of A (this is sometimes the definition of span (A), in which case the statement becomes that it is equal to the set of all finite linear combinations of elements of A).

The proof of this is straightforward -- let S be the intersection of all subspaces of V containing A. Obviously, if it is a subspace, then it is a subset of every subspace of V containing A, and thus is the smallest one. Therefore, we just need to show that S is a subspace, which means we have to show 0∈S, (u∈S ∧ v∈S) → u+v∈S, and for any scalar k, u∈S → ku∈S:

1) Let W be an arbitrary subspace of V containing A. Since W is a subspace, 0∈W. Thus every subspace containing A contains 0, it follows that their intersection contains 0 -- i.e. 0∈S

2) Suppose u∈S and v∈S. Let W be an arbitrary subspace of V containing A. Since S is the intersection of all such subspaces, S⊆W, thus u∈W and v∈W. Since W is a subspace, it is closed under addition, therefore u+v∈W. Since this is true of an arbitrary subspace of V containing A, it must be true of all of them, therefore u+v is in the intersection of all such W, so u+v∈S

3) Let k be any scalar and suppose u∈S. Let W be an arbitrary subspace of V containing A. Since S is the intersection of all such subspaces, S⊆W, thus u∈W. Since W is a subspace, it is closed under scalar multiplication, ku∈W. Since this is true of an arbitrary subspace of V containing A, it must be true of all of them, therefore ku is in the intersection of all such W, so ku∈S.

Note that the assumption that A was a union of subspaces was not required for the proof.

Answer 2

A itslef is not linear subspace, because taking two elements
x,y in A, x-y is not necessarily in A
What you need to take is the subspace generated by all elements in A. The subspace generated by A is composed of
sums of the form x_1+x_2+...+x_n, where x_i are in some M_i, M_i in m

Equivalently, you can take the intersection of all linear subspaces that include A. The intersection is nonempty since V itself includes A.

I let you prove that the sets above satisfy the axioms of a linear subspace