Set of Linear Subspaces is Closed under Intersection

Theorem
Let $\left({V, +, \circ}\right)_K$ be a $K$-vector space.

Let $\left({M_i}\right)_{i \in I}$ be an $I$-indexed collection of subspaces of $V$.

Then $M := \displaystyle \bigcap_{i \in I} M_i$ is also a subspace of $V$.

Proof
It needs to be demonstrated that $M$ is:
 * $(1): \quad$ a closed algebraic structure under $+$
 * $(2): \quad$ closed for scalar product $\circ$.

So let $a, b \in M$.

By definition of intersection, $a, b \in M_i$ for all $i \in I$.

As the $M_i$ are subspaces of $V$, $a + b \in M_i$ for all $i \in I$.

That is, by definition of intersection, $a + b \in M$.

It follows that $M$ is closed under $+$.

Now let $\lambda \in K, a \in M$.

By definition of intersection, $a \in M_i$ for all $i \in I$.

As the $M_i$ are subspaces of $V$, $\lambda \circ a \in M_i$ for all $i \in I$.

That is, by definition of intersection, $\lambda \circ a \in M$.

It follows that $M$ is closed under $\circ$.

Hence the result, by definition of subspace.