Closed ideals in $\mathbb B(H)$

Question

Let $\mathbb{H}$ be a non-separable Hilbert space.

If $\alpha$ is an countably many infinite cardinal number, let $I_{\alpha}=\{A\in \mathbb{B(H)}\:dim~ cl(ran A)\le \alpha\}.$

Show that $I_{\alpha}$ is a closed ideal in $\mathbb{B(H)}$.

I know that $I_\alpha$ is linear submanifold and is invariant under left and right multiplication. But how to prove that $I_\alpha$ is closed?

Answer 1

For any $x\in\mathbb H $ we have $Ax=\lim A_nx$, with $A_nx\in\text {ran }\,A_j $. Because a Hilbert space is metric, we can do this with sequences. Thus $$ \text {ran}\,A\subset \text {cl}\left (\bigcup_n\text {ran}A_j\right). $$