I am self-studying the chapter of closed and closable operators. I have the following problem which I cannot find its proof.
Let $A$ be a closable operator and denote by $B$ a closed extension. We call $\bar{A}$ the closure of the (closable) operator $A.$ It is the smallest closed extension of $A$ in the sense that if $A\subset B$ and $B$ is closed, then $\bar{A}\subset B.$
Could anyone help with a proof? (I can guess this, but not sure what to prove...)