Why is identity map on a separable Hilbert space not compact? False proof.
Let $e_n$ be the orthonormal basis. Then the projection map onto $H$ is defined by $\sum(x,e_k)e_k$. What is stopping us from taking a finite part of this series. This gives us a compact operators, that converge to the identity map. Clearly something is wrong here not all Hilbert spaces are finite dimensional.
Best Answer
Beacause that sequence of operators does not converge to the identity: if $n\in\Bbb N$,$$\left\lVert e_{n+1}-\sum_{j=1}^n\langle e_{n+1},e_k\rangle e_k\right\rVert=\lVert e_{n+1}\rVert=1$$and therefore$$\left\lVert\operatorname{Id}-\sum_{k=1}^n\langle\cdot,e_k\rangle e_k\right\rVert\geqslant1.$$