Boundary of invertibles in a normed algebra A student and I are reading the book Introduction to Banach Spaces and Algebras, by Allan, and we're stuck.  Exercise 4.5 says:

Let $A$ be a normed algebra with unit sphere $S$.  Let $a\in A$.  Then $a$ is a topological divisor of 0 if
  $$ \inf\{\|ab\|+\|ba\|:b\in S \}=0. $$
  Prove that every element in the frontier of $G(A)$ is a topological divisor of $0$.

Here $G(A)$ is the collection of invertible elements of $A$.  I assume that the question really means to say that $A$ is a unital normed algebra.  Then the book already essentially proves this result for Banach algebras (Corollary 4.13).
So if $B$ is the completion of $A$, and if $a$ is still in the frontier of $G(B)$, then we're done (the infimum obviously doesn't change if we replace $S$ by the unit sphere of $B$).
Conversely, if there is an example of $a\in\partial G(A)$ with $a\in G(B)$, then we have a counter-example to the exercise.  So my question is:

If $a\in\partial G(A)$ and $B$ is the completion of $A$, then is $a\in\partial G(B)$?

Edit: Embarrassingly, I think I can now answer this!
Let $A$ be the complex polynomials, interpreted as an algebra of continuous functions on the interval $[0,1]$.  A little bit of algebra shows that $G(A)$ consists of just the constant polynomials.  So $G(A)$ is actually closed (not open, which would be the case if $A$ were Banach).  So being careful about what "frontier" means, I guess $G(A)$ is its own frontier.  But then the exercise is trivially false, as the frontier of $G(A)$ contains invertibles.
So the exercise seems wrong.  But somehow my counter-example seems cheap.  So a new question:

Can the frontier of $G(A)$ contain a non-invertible element which is invertible in $B$?  Are there examples where $G(A)$ is open?

 A: I think I can answer the first part of the revised question, though I could just be doing something stupid.
Let $A=\ell^1(F_2)$ sitting inside $B=C_r^*(F_2)$. There exists $a\in A$ which is self-adjoint but whose spectrum in $A$, call it $S$, is not contained in ${\mathbb R}$.
(We can take $a$ to have finite support: I forget the exact formula which works, but it can be found in Palmer Vol. II in one of the sections on "hermitian" groups, the point being that $F_2$ is not hermitian.)
Take a sequence $a_n=a-\lambda_n I$ for scalars $\lambda_n\notin S$ with $\lambda_n\to \lambda$ for some $\lambda \in S \setminus {\mathbb R}$. Then we have a sequence of invertible elements in $A$, which converge in $B$ to an element that is invertible in $B$ but not in $A$. I guess that by taking bicommutants we can get a commutative example from this one, in particular $B=C(X)$ for some $X\subseteq {\mathbb R}$.
Of course this doesn't answer your second question, where (if I understand correctly) you want $G(A)$ to be open in $A$ with respect to the norm from $B$.
