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This question is motivated by Distance between Unilateral shift and invertible operators., which proves that that presence of a non-unitary isometry is enough to guarantee that the set of invertibles is NOT dense. I was wondering whether the existence of a non-unitary proper partial isometry is enough to ensure this too.

I suppose I mean to restrict to the case where $A$ is infinite-dimensional, since the set of invertible elements in a finite-dimensional C$^{*}$-algebra is always dense.

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Any proper projection is a proper partial isometry, so there are lots of candidates. For example, if $A=L^\infty[0,1]$, or $UHF(2^\infty)$ to have a separable example, there are lots of projections. And both these algebras have real rank zero, so the selfadjoint invertibles are dense in the selfadjoints, which of course implies that the invertibles are dense.

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