# Regarding approximation by invertible operators [closed]

This post here states that if $$E$$ is an infinite dimensional space and if $$T$$ is an injective, bounded,non surjective opertor with closed range in $$E$$, then $$T$$ cannot be approximated in operator norm by invertible bounded linear operators on $$E$$. Can anyone tell why?

• We have $\|Tx\| \ge c \|x\|$ for some $c > 0$ and all $x \in E$ . Assume that there are invertible operators $T_n$ such that $T_n \to T$ wrt the operator norm. For all sufficiently large $n$ we then have $\|T_n x\| \ge c/2 \cdot \|x\|$ for all $x$, so $\|T_n^{-1}\| \le 2/c$. Thus, $(T_n^{-1})$ is a Cauchy sequence and hence convergent to an operator $S$. Now, one readily checks that $ST = TS = I$, so $T$ is invertible and hence surjective. Sep 30, 2018 at 12:47
• Why is $(T_n^{-1})$ Cauchy? Sep 30, 2018 at 14:45
• @M.González: We have $T_n^{-1} - T_m^{-1} = T_n^{-1} (T_m - T_n) T_m^{-1}$. Sep 30, 2018 at 15:31

If $$E$$ is a Banach space and $$T:E\to E$$ is an injective bounded operator with closed range $$R(T)$$, then there exists a number $$\delta>0$$ such that any bounded operator $$S:E\to E$$ with $$\|S-T\|<\delta$$ is also injective, has closed range and $$\dim \frac{E}{R(S)} =\dim\frac{E}{R(T)}$$.
The fact that $$S$$ is injective with closed range is easy: $$T$$ injective with closed range implies that $$T$$ is bounded below; i.e., there exists $$C>0$$ such that $$\|Tx\|\geq C\|x\|$$ for every $$x\in X$$. Thus if $$\|S-T\|, then $$S$$ is bounded below, hence injective with closed range.
To prove that $$R(S)$$ and $$R(T)$$ have the same codimension in $$E$$ is a bit more difficult. It can be seen as a consequence of the fact that the subset of semi-Fredholm operators $$SF(E)$$ is an open subset of the set of all bounded operators on $$E$$, and the index is constant in each connected component of $$SF(E)$$. See this link.