What is the intersection of the closures of left invertible operators and right invertible operators? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:08:27Z http://mathoverflow.net/feeds/question/99815 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99815/what-is-the-intersection-of-the-closures-of-left-invertible-operators-and-right-i What is the intersection of the closures of left invertible operators and right invertible operators? Qingping Zeng 2012-06-17T03:04:37Z 2012-06-19T11:02:03Z <p>From Douglas Zare's answer (see <a href="http://mathoverflow.net/questions/99777/does-x-embed-in-y-and-y-embed-in-x-always-imply-that-x-isomorphic-on" rel="nofollow">http://mathoverflow.net/questions/99777/does-x-embed-in-y-and-y-embed-in-x-always-imply-that-x-isomorphic-on</a>), one know that $$\overline{G_{l}(X,Y)} \bigcap \overline{G_{r}(X,Y) } = \overline{G(X,Y)}$$ does not hold in general, where $G_{l}((X,Y),G_{r}((X,Y)$ and $G(X,Y)$ denote the set of left invertible operators, right invertible operators and invertible operators. (We say an operator $T$ left invertible, if $ST=I$ for some operator $S$.) But does this equality holds when $X=Y$? If not, for what kinds of $X$, this equality holds? Furthermore, does this equality holds when $B(X)$ is replaced by a Banach algebra $A$ with an identity? </p>