Let $A$ be a commutative Banach Algebra with unit and $M$ a maximal Ideal of $A$. Then $M$ is obviously a subspace.
But why must $M$ also be closed?
I don't see in which way a maximal ideal is connected to a closed subspace.
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The closure $\operatorname{cl}(M)$ of $M$ is also an ideal, as can be checked easily. As $M$ is maximal, we must have either $\operatorname{cl}(M) = M$ or $\operatorname{cl}(M) = A.$ We need show that the later doesn't happen; specifically, $1\not\in \operatorname{cl}(M).$
This follows from the fact that the set of invertible elements in a Banach algebra is open (see Prove that the set of invertible elements in a Banach algebra is open). Note that for a commutative Banach algebra we have that any element $x$ is not invertible exactly if it is contained in a maximal ideal.
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1$\begingroup$ Potentially $\operatorname{cl}(M)$ could be $A$. To argue that this doesn't happen one needs to prove that any element sufficiently close to $1$ is invertible, and therefore, not in $M$. $\endgroup$ Jan 3, 2020 at 16:42
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