# Maximal Subalgebra in Normed Algebra is Closed

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## Theorem

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

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

Let $B$ be a subalgebra of $A$ that is maximal with respect to set inclusion.

Then $B$ is closed.

## Proof

From Closure of Subalgebra in Normed Algebra is Subalgebra, the closure $B^-$ of $B$ is a subalgebra with $B \subseteq B^-$.

Since $B$ is maximal with respect to set inclusion, we have that $B = B^-$.

From Set is Closed iff Equals Topological Closure, we conclude that $B$ is closed.

$\blacksquare$