Is the set of all invertible linear operators dense in the set of all linear operators? Exercise 2 after $\S$ 91 from Paul R. Halmos's "Finite-Dimensional Vector Spaces" (second edition) invites to prove or disprove the following assertion.

For every (bounded) linear transformation $A$ (on an inner product space) there exists a sequence $(A_n)$ of invertible linear transformations such that $A_n \rightarrow A$.

The inner product space, say $\mathcal V$, of the assertion is not specified to be over the complex (or real) field, and is not said to be finite-dimensional or complete either. Also, for reference, $\S 91$ (from the book) identifies "$A_n \rightarrow A$" in inner product spaces by

$\Vert A_n - A \Vert \rightarrow 0$ as $n \rightarrow \infty$.

Further, $\S \ 87$ has the following definition for the norm $\Vert \cdot \Vert$ of a linear operator:

$\Vert A \Vert = \inf \ \big\{K: \Vert Ax \Vert \leq K \Vert x \Vert \text{ for all vectors } x\big\}.$

I am able to see why the assertion holds if $\mathcal V$ is finite-dimensional; see my "constructive" proof below. I am unable to imagine what happens in the general case however, and would appreciate a proof or a counterexample in infinite dimensional spaces. Would also appreciate an advice if my proof is found to be inaccurate! Thanks.

Proof (in finite dimensions): Let $\mathcal V$ be a $k$-dimensional inner product space ($0 \leq k < \infty$), and let $A$ be any linear operator on $\mathcal V$. If $\mathcal N(A)$ is the null-space of $A$, then $\mathcal N^\perp(A) \oplus \mathcal N(A) = \mathcal V$. Also, if the rank of $A$ is $m \ (\leq k)$, and if $\mathcal R(A)$ is the range of $A$, then $m = $ dim $\mathcal R(A)=$  dim $\mathcal N^\perp(A) = k-$dim $\mathcal N = k-$dim $\mathcal R^\perp(A)$. It is also true that $\mathcal R^\perp(A) \oplus \mathcal R(A) = \mathcal V$.
We now begin constructing a sequence $(A_n)$ of invertible linear operators such that $A_n \rightarrow A$. Let $B$ be the restriction of $A$ to the $m$-dimensional $\mathcal N^\perp(A)$. It is clear that $B$ maps $m$-dimensional $\mathcal N^\perp(A)$ onto $m$-dimensional $\mathcal R(A)$, and is invertible. Next, write $C_n x_i = \frac{1}{n}y_i$ for $n = 1, 2, \cdots$ and for $i = 1, \cdots, k-m$, where $\{x_1, \cdots, x_{k-m}\}$ and $\{y_1, \cdots, y_{k-m}\}$ are any bases in the $(k-m)$-dimensional subspaces $\mathcal N(A)$ and $\mathcal R^\perp (A)$ respectively. (It is clear that $C_n$ is a linear map with rank $k-m$, and is invertible.) Next, if, for every $z = (z_1+z_2)$ in $\mathcal V$, we write $A_n z = B z_1 + C_n z_2$ for $n = 1, 2, \cdots$, whenever $z_1$ and $z_2$ are in $\mathcal N^\perp(A)$ and $\mathcal N(A)$ respectively, then we find that $A_n$ is a linear mapping of $\mathcal V$ onto $\mathcal V$, and is invertible. Finally, because $A_n z - Az = (Az + C_nz_2) - Az = C_n z_2$, which implies that $(A_n - A)z \rightarrow 0$, it follows that $A_n \rightarrow A$.
 A: The assertion fails in the infinite dimensional case: For example consider the Hilbert space $V=l^2(\mathbb{N},\mathbb{R})$ and let $R,L$ be the right and left shift operator on $V$, that is for $x=(x_n) \in V$
$$
L(x)=(x_2,x_3, \dots), ~ R(x)=(0,x_1,x_2, \dots).
$$
Then $L$ is not invertible ($L$ is not injective), $R$ is a right inverse of $L$ ($LR =I_V$) and $\|L\|,\|R\| = 1$. Now let $A:V \to V$ be any linear and continuous operator with $\|A\|<1$. Since $\|RA\| \le \|R\| \|A\| < 1$ the
operator $(I+RA)$ is invertible (Neumann's series). Since
$$
L+A = L(I+RA)
$$
we have $L=(L+A)(I+RA) ^{-1}$. Thus $L+A$ can't be invertible (otherwise $L$ would be invertible). So, no sequence of invertible operators can have limit $L$.
A: Just to give a simpler proof in the finite dimensional case. Let $a_n$ be a sequence of numbers such that
(1) $a_n\rightarrow 0$ as $n\rightarrow\infty$
(2) $a_n$ is not an eigenvalue of $A$ for any $n$.
Then $A-a_n I_n$ is invertible and converges to $A$.
Note that this argument fails in infinite dimensional case for a simple reason: The spectrum of $A$ is not necessarily finite, therefore we may not be able to pick $a_n\rightarrow 0$ that avoids spec(${A}$). Indeed, the shift operators have the closed unit disk as its spectrum.
