The intersection of an infinite number of subspaces is a subspace

Question

Let $V$ be a finite dimensional vectorspace over a field $\mathbb{ F}$. It's easy to show that if $U$ and $V$ are subspaces of $V$ then $U \cap V$ is a subspace. But what if there are an infinite number of them? I have seen sometimes in pure maths that you take an arbitrary finite collection of sets and show the statement you are trying to prove is true for them to deduce it is true for the whole infinite collection. Why is this technique allowed? Can I do it for this case? Why?

Answer 1

If $V_1\cap V_2$ is a subspace, then $V_1 \cap V_1 \cap V_3=(V_1\cap V_2)\cap V_3$ which is again the intersection of two subspaces. We can use induction to show that if this works for $n $ subspaces, then it works for $n+1$ subspaces and hence any finite number of subspaces.

Now for the infinite case. Since they are all subspaces, they all contain the zero vector. It remains to show that this intersection is closed under addition and scalar multiplication.

We choose a $v $ and $w $ in their intersection. Since they are all subspaces, they are all closed under addition and $v+w$ is in each subspace and hence in their intersection. Similarly for scalar multiplication, if we pick a $v $ in their intersection, and a $c\in \mathbb {F}$, we must have that $cv $ is in each subspace since they are all closed under scalar multiplication, and hence $cv $ is also in their intersection. Since this intersection has satisfied the subspace test, it is a subspace.

But this is assuming the existence of an infinite number of sets for a given vector space over a field.

Answer 2

You don't need anything special to deduce that an intersection of arbitrarily many subspaces is a subspace. The fact is that any two vectors in the intersection are contained in any particular one of the subspaces. Since they are contained in that particular subspace, so is any linear combination of them. Since the subspace was arbitrary, any linear combination of the two vectors is contained in every subspace, hence in the intersection of all of the subspaces. (You also need to know the intersection contains 0, which is obvious.)