Question on Stone's Theorem : skewadjoint operator and closedness

Question

So, I'm studying at the moment the following theorem and I can't understand how the closedness of $A$ follows. I can see why the adjoint is closed but how the fact that the operator is skewadjoint yields the closedness of $A$?

Any help or hint is much appreciated.

Thanks in advance

Answer 1

If the operator is skewadjoint, $A=-A'$. Since $A'$ is closed the closedness of $A$ follows.

If $B$ is closed then also $-B$ is closed.