Countable separating set of a von Neumann algebra

Question

I studied the following proposition , but I have a question in the proof of this proposition.

Let $A$ be a von Neumann algebra. If $A$ possesses a countable separating set then show that $A$ is $\sigma- $ finite.

Proof: Let $M$ be a countable separating set for $A$. Let $\{E_i\}_{i\in I}$ be a pairwise disjoint family of projections of $A$. Fr every $x\in M$, we have $E_ix=0$ for all but countably many of the indices $i$...

My question: How can we say for countably many of the indices $i$, $E_ix\neq 0$ ?

Answer 1

Because $$ \|x\|^2\geq\|\sum_iE_ix\|^2=\sum_{i,j}\langle E_ix,E_jx\rangle=\sum_i\langle E_ix,e_i\rangle=\sum_i\|E_ix\|^2. $$ Now you just use the fact that a convergent series can only have countably many nonzero terms.