Let $\mathcal H$ be a Hilbert space. Let $T \in \mathcal L(\mathcal H)$ and $K$ be a compact operator on $\mathcal H.$ If $\text {ran} (T)$ is a closed subspace of $\mathcal H$ then show that $\text {ran} (T+K)$ is also closed.
Let $\{(T + K)x_n\}_{n \geq 1}$ be a sequence in $\text {ran} (T + K)$ converging to $x \in \mathcal H.$ If $\{x_n\}_{n \geq 1}$ is a bounded sequence then $\{Kx_n\}_{n \geq 1}$ will have a convergent subsequence say $\left \{Kx_{n_k}\right \}_{k \geq 1}$ converging to $y$ (say). Then $Tx_{n_k} \to (x - y).$ But since $\text {ran} (T)$ is closed it follows that there exists $z \in \mathcal H$ such that $(x - y) = Tz.$ So $x = y + Tz.$ Now I get stuck and couldn't proceed further. Because $y$ might not necessarily be in the range of $K.$ Instead what I can say is that there exists a closed ball $B$ of some positive radius such that $\overline {K(B)}$ contains $y.$ Also I have no idea how to proceed when $\{x_n\}_{n \geq 1}$ is unbounded. Some help is very much required at this moment.
Thanks for your time for reading.
The result if FALSE. For a counter-example, take $T$ to be the zero operator and $K$ any compact operator with a non-closed range.