Is it true that range of $(I + T)$ closed whenever $T$ is compact? 
Let $\mathcal H$ be a Hilbert space and $T \in \mathcal L (\mathcal H)$ be a compact operator. Then is it true that range of $(I + T)$ is closed?

If the answer is in affirmative sense then how to prove it? Any help in this regard will be warmly appreciated.
Thanks for your time.
 A: Yes this is true. I present the outline of the proof I know, where for convenience I define $S:=I+T$:
First note that $Z:=\ker(S)$ is finite dimensional. This follows from the fact that $L(H)=K(H)$ if and only if $H$ is finite dimensional (These being the algebras of bounded and compact operators on $H$ respectively).
Using this fact we can write $H=Z\oplus Y$ for some closed subspace $Y$ of $H$. Furthermore $S(H)=S(Y)$. Clearly $S|_{Y}$ is a bounded operator, thus it will have a closed range if we can show that $S|_Y$ is bounded below. Suppose to the contrary that this isn't the case. There must thus exist a sequence $(y_n)\in Y$ with $\|y_n\|=1$ for all $n$ such that $\lim_{n\to\infty}\|Ty_n+y_n\|= 0$. As $T$ is compact there exists some subsequence $y_{n_k}$ so that $\lim_{k\to \infty} Ty_{n_k}=z\in Y$ (because $Y$ is closed). From this we see that $y_{n_k}$ has some limit $y\in Y$. Furthermore $\lim_{k} Ty_{n_k}+y_{n_k}=Ty+y$, but as $\lim_{k\to\infty}\|Ty_{n_k}+y_{n_k}\|= 0$ we see that $y\in \ker(S)$. Thus $y\in\ker(S)\cap Y$, so $y=0$, but this contradicts the fact that each $y_n$ is a unit vector. Thus indeed $S|_Y$ is bounded below, and so $S(H)$ is closed.
Note that nowhere did we use the Hilbert space structure in this argument, and also the fact that if we replace $I$ with $\lambda I$ for some nonzero $\lambda$ the arguments still go through. We thus have a close link with Fredholm theory.
