Proposition 32.4.4.label Let $A$ be a unital Banach algebra and $I \subset A$ be a proper ideal, then:
- (1)
$I$ contains no invertible elements.
- (2)
$\ol I$ is also a proper ideal.
- (3)
$I$ is contained in a maximal ideal.
- (4)
If $I$ is maximal, then $I$ is closed.
Proof, [Fol16]. (1): If $I$ contains an invertible element, then $1 \in I$ and thus $A = I$.
(2): By Proposition 32.2.3, $G(A)$ is open. Since $G(A) \cap I = \emptyset$, $G(A) \cap \ol I = \emptyset$ as well.
(3): By Zorn’s lemma.
(4): By (2).$\square$
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