What is the Closure of a Proper Ideal in a Unital Banach Algebra?

[Fredrik]

Homework Statement

Prove that the closure of a proper ideal in a unital Banach algebra is a proper ideal.

Homework Equations

The hint is to use the result of the previous exercise: If I is an ideal in a unital normed algebra A, and I≠{0}, we have

I=A $\Leftrightarrow$ I contains 1 $\Leftrightarrow$ I contains an invertible element

The Attempt at a Solution

x is in the closure of I if and only if there's a sequence (x_n) in I such that x_n→x.

It's easy to show that the closure of I is closed under linear combinations:

$$\|(ax_n+by_n)-(ax+by)\|\leq |a|\|x_n-x\|+|b|\|y_n-y\|<\varepsilon$$

(I don't feel like typing every word of the argument. I'm assuming that it's obvious what I have in mind.).

It's also easy to show that if i is in the closure of I and x is in A, xi is in the closure of I:

$$\|xi_n-xi\|\leq \|x\|\|i_n-i\|<\varepsilon$$

These results imply that the closure is an ideal. What I don't see is how to prove that it's a proper ideal, i.e. that it's neither {0} nor A. The most straightforward approach seems to try to prove that 1 can't be a member, but I don't see how to do that. I'm probably missing something really obvious as usual.

 

[ystael]

Recall that the invertible elements of $$A$$ form an open subset of $$A$$.

 

[Fredrik]

Thank you. That was all I needed to hear. That means that there's an open ball around 1 that's a subset of I^c, and that means that no sequence in I can have 1 as a limit. So 1 can't be a member of the closure.