Linear Subspaces Closed under Intersection

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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.

$\blacksquare$