Compact linear operator on a separable Hilbert space is approximately of finite dimension

Question

$X$ is a separable Hilbert space and $ A\in L(X,X)$ and compact. I need to prove that $A$ is approximately of finite dimension.

Answer 1

Given $\epsilon>0$, let $\mathcal N$ be a finite $\epsilon$-net in $A(B_X)$, where $B_X$ is the closed unit ball of $X$. Let $P$ be the orthogonal projection onto the linear span of $\mathcal{N}$. By construction, $\|Py-y\|\le \epsilon$ for all $y\in A(B_X)$. Hence, $\|PA-A\|\le \epsilon$. Since $PA$ has finite rank and $\epsilon$ was arbitrary, we are done.

As an aside, I prefer "is the norm limit of finite-rank operators" to "is approximately of finite dimension"; the former being more descriptive.