TITLE: Invertible operator with countable spectrum
QUESTION [6 upvotes]: Let $H$ be  a separable  Hilbert space  and $A$ is  an invertible  bounded operator on $H$. Can we approximate  $A$ with  an invertible operator $B$ such that $sp(B)$ is  a countable set?
Motivation:
If the answer is yes, this  would give's us  an alternative  proof  of  connected ness of $GL(H)$. This  alternative proof is identical to a short and interesting proof  of  connectedness  of  $GL(n,\mathbb{C})$, in page  19 of "Introduction to the  Baums  Connes conjecture" by Alain Valette

REPLY [8 votes]: The question has already been answered above (by Mike Jury). Here is another way of arguing: Any operator with countable spectrum is in the closure of the invertible operators. So are their limits. But again, the unilateral shift provides an example of an operator not in the closure of the invertibles. For if $\|S-X\|<1$ then $\|I-S^*X\|<1$, and so $S^*X$ is invertible. $X$ cannot be invertible, since otherwise $S^*$ would also be invertible. So no translate $S+\lambda I$ can be approximated by operators with countable spectrum.