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.
$\blacksquare$