Closure of Proper Ideal in Unital Banach Algebra is Proper Ideal

Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.

Let $I$ be a proper ideal of $A$.

Then $I^-$ is a proper ideal of $A$.

Proof

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

Let $x \in I^-$ and $y \in A$.

We need to show that $x y \in I^-$.

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

$x_n \to x$

Since $I$ is an ideal, we have $x_n y \in I$ for each $n \in \N$.

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

$x_n y \to x y$

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

So $I^-$ is an ideal.

It remains to show that it is proper.

Let $\map G A$ be the group of units of $A$.

From Ideal of Unit is Whole Ring, we have:

$J \cap \map G A = \O$

From Group of Units in Unital Banach Algebra is Open, $\map G A$ is open.

Hence from Open Set Disjoint from Set is Disjoint from Closure, we have that:

$J^- \cap \map G A = \O$

From Ideal of Unit is Whole Ring, we conclude that $I^-$ is a proper ideal of $A$.

$\blacksquare$