Intersection of Subalgebras is Subalgebra

Theorem

Let $K$ be a field.

Let $\struct {A, +, \cdot, \circ}_K$ be an algebra over $K$.

Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of subalgebras of $A$.

Let:

$\ds B = \bigcap_{\alpha \mathop \in I} A_\alpha$

Then $B$ is a subalgebra of $A$.

Proof

From Set of Linear Subspaces is Closed under Intersection, $\struct {B, +, \cdot}_K$ is a vector subspace of $\struct {A, +, \cdot}_K$.

Now let $x, y \in B$.

That is, $x, y \in A_\alpha$ for each $\alpha \in I$.

Since $A_\alpha$ is a subalgebras of $A$ for each $\alpha \in I$, we have:

$x y \in A_\alpha$ for each $\alpha \in I$.

So $x y \in B$.

So $B$ is a subalgebra of $A$.

$\blacksquare$