Closure of Subalgebra in Normed Algebra is Subalgebra

Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$.

Let $B$ be a subalgebra of $A$.

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

Proof

From Closure of Subspace of Normed Vector Space is Subspace, $B^-$ is a vector subspace of $A$.

Now let $x, y \in B^-$.

From the definition of a closed set in a normed vector space, there exists sequences $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ valued in $B$ such that:

$x_n \to x$ and $y_n \to y$.

From Product Rule for Sequence in Normed Algebra, we have:

$x_n y_n \to x y$

From the definition of closure, we have $x y \in B^-$.

So $B^-$ is a vector subspace of $A$ that is closed under multiplication.

Hence $B^-$ is a subalgebra of $A$.

$\blacksquare$