In what sense are self adjoint elements of C*-algebra "closed"? Let $\mathcal A$ be a unital C*-algebra. 
I read the fact, that the self-adjoint elements $\mathcal S:=\{A\in \mathcal A \vert A^*=A\}$ are closed in $\mathcal A$.
In what sense is "closed" meant here? Topologically closed, i.e. every convergent sequence in $\mathcal S$ has a limit in $\mathcal S$ or algebraically closed? What is to prove for algebraic closure, is it closure under vector addition, under algebra multiplication and under scalar multiplication?
 A: If the set of self-adjoint elements was stable under multiplication, we would have $AB=(AB)^*=B^*A^*=BA$, i.e. any two self-adjoint elements would commute. Writing any element $C=\frac{C+C^*}{2}+i\frac{C-C^*}{2i}$ as a linear combination of two self-adjoint elements, we see that this would imply that $\mathcal A$ be commutative. The converse is clear. Hence the set of self-adjoint elements is stable under multiplication if and only if $\mathcal{A}$ is commutative.
Note that $iA$ is not self-adjoint if $A$ is nonzero self-adjoint, as $(iA)^*=-iA$. So that's not a complex subspace. But for $s,t$ real scalars, and $A,B$ self-adjoint, then
$$
(sA+tB)^*=\overline{s}A^*+\overline{t}B^*=sA+tB
$$
is still self-adjoint. So algebraically, that's a real vector subspace. 
Topologically, it is norm closed. Indeed, the function $A\longmapsto A^*$ is an isometry, so it is norm-continuous, as well as $f:A\longmapsto A-A^*$. Hence the set of self-adjoint elements $f^{-1}(0)$ is closed in the norm topology.
If $A$ sits in $B(H)$, meaning it is a norm closed $*$-subalgebra of $B(H)$, then the set of self-adjoint elements is even closed for the wot and the sot (weak and strong operator topology). That's because $A\longmapsto A^*$ is wot-continuous.
