# Convergence of finite rank operators $(T_n)_n$: why finite?

Let $(T_n)_n$ be a sequence of compact operators such that $\Vert T_n-T\Vert\underset{n\to\infty}{\to} 0$ then I can prove that $T$ is a compact operator.

Which I don't understand is that in my course the theorem is

If a sequence $T_n$ a finite rank operators converge to $T$ then $T$ is compact.

It's strange because in the proof I don't see where we need that $(T_n)_n$ is finite rank.

The proof use the fact that if $T_n$ is compact the it's totally bounded so that $T_N(B_E)\subset \cup_{j=1}^J B(x_j,\varepsilon).$

am I missing something ?

• True. You only need that the operators $T_n$ are compact to infer that $T$ is. Finite rank operators are automatically compact. – Giuseppe Negro Dec 9 '17 at 15:59

## 1 Answer

A norm limit of compacts is compact. As far as I can tell, the proof is only saying that finite-rank operators are compact, which is an easy direct consequence of the fact that the closed unit ball in a finite-dimensional space is compact.

• ok that makes sense, thanks. – user331066 Dec 9 '17 at 16:06