TITLE: How does $\lim A_n$ being not invertible imply $\sup_n\|A_n^{-1}\|=\infty$?
QUESTION [7 upvotes]: Consider a sequence of operators $\{A_n\}_{n=1}^{\infty}\subset B(X,Y)$, where $X,Y$ are normed vector spaces and $B(X,Y)$ denotes the space of bounded linear operators from $X$ to $Y$. Assume that all $A_n$ are invertible (i.e., 1-1 and onto) and
  $$
\lim_{n\to\infty} A_n=A\ \ \text{in operator norm}
$$
  where $A:X\to Y$ is not invertible. Show that
  $$
\sup_n \|A_n^{-1}\|=\infty
$$

If $\sup_n\|A_n^{-1}\|=M<\infty$ for some $M\in{\Bbb R}$, then I'm trying to show that $A$ must be invertible. What I've done is showing that $A$ is 1-1:
Since $\|A_n-A\|\to 0$ as $n\to\infty$, we have
$$
\|A_nx-Ax\|\leq\|A_n-A\|\|x\|\to 0\ \ \text{as}\  n\to\infty
$$
for all $x\in X$. Hence, if $Ax=0$ then $\|A_nx\|\to 0$, i.e., for all $\varepsilon>0$, there exists $N>0$ such that
$$
\|A_nx\|\leq \varepsilon/M
$$
for all $n>N$. (One can show that $M>0$, so we have no problem with $1/M$). It follows that
$$
\|x\|=\|A_n^{-1}(A_nx)\|\leq \|A_n^{-1}\|\|A_nx\|\leq M\cdot \varepsilon/M=\epsilon.
$$
Thus $x=0$.
How can I show that $A$ is also onto?

[ADDED:]I suspect that one might need the assumption of completeness of $X$ and $Y$. But I don't see a quick counterexample when one of $X$ and $Y$ is not Banach space.

REPLY [7 votes]: We have to consider the problem over the completion $\overline X$ of $X$. But that does not hurt our reasoning: if $A$ is not invertible, neither will its extension to $\overline X$ be.
Assuming that $\|A_n^{-1}\|\leq M$ for all $n$, we have
$$
\|A_n^{-1}-A_m^{-1}\|=\|A_n^{-1}(A_m-A_n)A_m^{-1}\|\leq\,M^2\,\|A_m-A_n\|.
$$
As $\{A_n\}$ is Cauchy, we deduce that so is $\{A_n^{-1}\}$. 
Let $A'=\lim A_n^{-1}$. Then
$$
AA'=\lim_n A_nA_n^{-1}=I,\ \ A'A=\lim_nA_n^{-1}A_n=I,
$$
so $A$ is invertible. Note that we are using that the norm limit is multiplicative.