I am learning Kaplansky density theorem, there is a problem as following:
If $C$ is a unital $C^*$-subalgebra of $B(H)$ and $A$ is its SOT clousure, $ReA$ and $ReC$ is set of self adjoint elements in $A$ and $C$, ball$ReA$ and ball$ReC$ is the closed unit ball of $ReA$ and $ReC$.
And I find a proof in a book: the author firstly prove that $ReA$ is the SOT clousre of $ReC$, because $\forall x\in ReA$, there exist $\{x_i\}\subset C$ such that $x_i \rightarrow x$ in *-SOT by von Neumann double commutant theorem, thus also $x_i^* \rightarrow x$ in *-SOT, thus $\frac{1}{2}(x_i^*+x_i)\rightarrow x$ in SOT, thus $ReA \subset \overline{ReC}^{SOT}$
Now I want to ask that weather $\overline{ReC}^{SOT}\subset ReA $ ? i.e. the SOT limit of a sequence of self adjoint elements in the $C^*$-algebra is a self adjoint element or not ?
Yes, because sot implies wot. Note that you cannot use sequences in general, you have to use nets. If $\{a_j\}\subset C$ consists of selfadjoints and $a_j\to a$ sot, then $$ \langle a^*\xi,\xi\rangle=\langle \xi,a\xi\rangle=\lim_j\langle \xi,a_j\xi\rangle=\lim_j\langle a_j\xi,\xi\rangle=\langle a\xi,\xi\rangle. $$ As this works for all $\xi\in H$, you get that $a^*=a$.