How to show a compact, closed-range operator on an infinite-dimensional Hilbert space has finite rank, without using the open-mapping theorem? If $H$ is an $\infty$-dimensional Hilbert space and $T:H\to{H}$ is a compact operator with closed range, how do I show that $T$ has finite rank, without using the open-mapping theorem? (The open-mapping theorem is not in my lecture notes).
The definitions I have in my lecture notes are:
(Let $B(H)$ denote the space of all bounded operators mapping $H\to{H}$, $K(H)$ denote the space of all compact operators mapping $H\to{H}$, $R(H)$ denote the space of all finite rank operators mapping $H\to{H}$).


*

*$T\in{B(H)}$ is compact if the closure of $T(B(0,1))$ is a compact set.

*$T\in{B(H)}$ has finite rank if $Range(T)=T(H)$ is finite-dimensional.


I'm not sure how to do the proof, but I think that the following propositions in my lecture notes could be useful:


*

*$T\in{R(H)}$ iff $T\in{B(H)}$ is the norm limit of a sequence of finite rank operators, i.e. $K(H)$ is the closure of $R(H)$.

*Let $T\in{R(H)}$. Then there is an orthonormal set $\{e_1,...,e_L\}$  s.t.
$$Tu=\sum\limits_{i,j=1}^{L}{c_{ij}(u,e_j)e_i}$$
where $c_{ij}$ are complex numbers.


Thank you in advance.
 A: Sketch: Find a way to write the range of $T$ as a countable union of compact sets $K_i$.  Then the Baire category theorem will guarantee that one of the $K_i$ has nonempty interior (relative to the range of $T$).  This means the range of $T$ is locally compact, hence finite dimensional.
A: The key thing to prove (or know) is that if $V$ is a Hilbert space whose closed unit ball is compact, then $V$ is finite-dimensional.
(The same is true with the word Hilbert replaced by the word Banach, but the Hilbertian case is simpler to prove, because every Hilbert space has an orthogonal basis.)
I assume that you have seen this result in your notes; if not, it is not too hard to prove once one recalls that in a metric space the notions of compactness and sequential compactness coincide.
Now you say you want to avoid the open mapping theorem for Banach spaces. I think this is OK because soft algebraic arguments tell us that $T:H\to H$ factors as
$H \to H/\ker(T) \to \overline{{\rm Im} T} \to H$ where each arrow is continuous and linear, and the second arrow is a bijection. So now apply the result I originally mentioned with $V= \overline{{\rm Im} T}$.
Filling in the remaining details seems to me like an instructive exercise, so I leave it to you.
