Is the closure of a symmetric operator symmetric? Every symmetric operator $A$ can be closed to an operator. Are this closed extensions symmetric? Are there any conditions that make this true such as $Dom(A)$ is densly defined in some Hilbert space? 
 A: The symmetric condition for $A$ may be written as the equivalent condition
$$
                        \mathcal{G}(A)\perp_{H\times H} J\mathcal{G}(A),
$$
where the graph of $A$ is defined as $\mathcal{G}(A)=\{ (x,Ax) \in H\times H : x\in\mathcal{D}(A)\}$, and where $J : H\times H \rightarrow H\times H$ is the unitary map defined by $J(x,y)=(-y,x)$. Let $\overline{\mathcal{G}(A)}$ denote the closure of $\mathcal{G}(A)$ in $H\times H$. Because of joint continuity of the inner product on $H\times H$, and because of the isometric nature of $J$, it follows that
$$
     \overline{\mathcal{G}(A)} \perp_{H\times H} J\overline{\mathcal{G}(A)}.
$$
The closure $\overline{\mathcal{G}(A)}$ in $H\times H$ is $\mathcal{G}(\overline{A})$ where $\overline{A}$ is the closure of $A$. Hence, the above is equivalent to the symmetry of $\overline{A}$:
$$
              \mathcal{G}(\overline{A})\perp_{H\times H}J\mathcal{G}(\overline{A}).
$$
A: Here is a simpler more "analysis" proof.
Suppose $\psi,\phi\in$ Dom$(A^{cl})$. Then there are sequences $\psi_n\to\psi$ and $\phi_m\to\phi$ with $\psi_n,\phi_m \in$ Dom$(A)$ and $$\lim A\phi_n=A^{cl}\phi$$ $$\lim A\psi_m=A^{cl}\psi$$
Then by the continuity of the inner product, the following are equal $$<A^{cl}\phi,\psi>$$
$$<\lim A\phi_n,\psi>$$
$$\lim<A\phi_n,\psi>$$
$$\lim<A\phi_n,\lim \psi_m>$$
$$\lim_n\lim_m<A\phi_n,\psi_m>$$
$$\lim_n\lim_m<\phi_n,A\psi_m>$$
$$\lim<\phi_n,\lim A\psi_m>$$
$$\lim<\phi_n,A^{cl}\psi>$$
$$<\lim\phi_n,A^{cl}\psi>$$
$$<\phi,A^{cl}\psi>$$
So yes, the closure of a symmetric operator is symmetric.
