Are Hilbert-Schmidt operators in non-separable Hilbert spaces compact?

Question

The definition of Hilbert-Schmidt operator should still be valid even when the Hilbert space is not separable: If $e_i$ for $i\in I$ is an orthonormal basis for a Hilbert space, and

$\mbox{Trace}(T)=\sum_{i\in I}\|Te_{i}\|^{2}<\infty$

Then $T$ is Hilbert-Schmidt. Of course if the sum is finite then there can only be countably many non-zero terms in the summation.

However, I am not sure how to show that such operators are compact.

Answer 1

Preword

The Hilbert dimension plays no role at all!

Problem

A trace class operator is Hilbert Schmidt. (Decomposition)

A Hilbert Schmidt operator is compact. ([Denseness][2])