Is the set of closed range operators open?

Question

As we know the set of surjective operators is open.

Let $T\in B(\mathcal{H})$ be a closed range operator. Is there any $\delta>0$ such that for every $T'\in B(T, \epsilon) $, we have $T'$ is closed range? If it is not true, please give me a counter example.

Answer 1

This is not true. One (boring) counterexample is that the zero operator $0$ has closed range, but if $T$ is an operator with non-closed range, then $\frac{1}{n}T\to 0$.