Decomposition of a compact Hausdorff $S$ space by a closed self-adjoint algebra in $C(S)$

Question

The following question is from Chapter IX in Linear Operaotrs edited by Dunford & Schwartz, Exercise 4.1

Let $S$ be a compact Hausdorff topological space and $C(S)$ be all complex continuous functions defined on $S$. For a $f \in C(S)$, let $f^*(x) = \overline{f(x)}$ be the involution. Now assume $\mathfrak{A}$ is a closed unital self-adjoint subalgebra of $C(S)$. Then show that there exists a decomposition of $S$ into a (disjoint) union of closed sets (say $\dot\bigcup_{\lambda \in \Lambda} F_{\lambda}$) on each of which all elements of $\mathfrak{A}$ will be constant($\forall\,f \in \mathfrak{A}\,\forall\,\lambda \in \Lambda, f\vert_{F_{\lambda}}$ is constant), and such that each continuous function that is constant on each $F_{\lambda}$ belongs to $\mathfrak{A}$.

It is proved that in a compact Hausdorff space $X$, each closed ideal $\mathcal{I}$ in $C(X)$ injectively correspond to a closed set $F \subset X$ such that $\mathcal{I} = \mathcal{I}_F = \{g \in C(X)\,\vert\,g\vert_F = 0\}$. In the above question, I do not know how to use this result. Should we partition $\mathfrak{A}$ by intersection of ideals?

Any hints will be appreciated.

Answer 1

Hint: Thinking in terms of decompositions into closed sets here is rather misleading. Rather, what you want to think about is the equivalence relation on $S$ defined by the partition of $S$ into the $F_\lambda$. Can you describe what that equivalence relation ought to be, in terms of $\mathfrak{A}$? (What equivalence relation does $\mathfrak{A}$ naturally define on $S$, such that each element of $\mathfrak{A}$ is constant on the equivalence classes?) Now use Stone-Weierstrass on the quotient of $S$ by this equivalence relation to show that it has the desired properties.

More details are hidden below.

Define an equivalence relation $\sim$ on $S$ by $x\sim y$ iff $f(x)=f(y)$ for all $f\in\mathfrak{A}$. Our partition $\{F_\lambda\}$, if it exists (and corresponds to a closed equivalence relation), must consist of the equivalence classes of this equivalence relation. So let $T$ be the quotient space $S/{\sim}$; since $\sim$ is a closed equivalence relation, $T$ is also compact Hausdorff. Note moreover that by the universal property of the quotient space, we can identify $C(T)$ with the subalgebra of $C(S)$ consisting of functions that are constant on each equivalence class of $\sim$. With this identification, the definition of $\sim$ immediately implies $\mathfrak{A}\subseteq C(T)$, and we wish to prove that $\mathfrak{A}=C(T)$.

To do this, we apply Stone-Weierstrass. Since $\mathfrak{A}$ is a closed self-adjoint unital subalgebra of $C(T)$, it suffices to show that it separates points of $T$. In other words, if $A$ and $B$ are two distinct equivalence classes of $\sim$, we wish to find a function $f\in\mathfrak{A}$ which takes different values on $A$ and $B$. But this is immediate from the definition of $\sim$: if no such $f$ existed, then any $a\in A$ would be equivalent to any $b\in B$, so they would not be distinct equivalence classes.

Answer 2

$\hspace{0.44cm}$According to @Sasha's ideas, because the inclusion mapping $\iota: \mathfrak{A} \rightarrow C(S)$ is a injective continuous mapping between $C^*$-Algebra, its induced mapping $\overset{\wedge}{\iota}$ from the spectrum of $\mathfrak{A}$, say $\mathcal{M}_{\mathfrak{A}}$, to the spectrum of $C(S)$ (which is homeomorphic to $S$) will be a surjective continuous mapping. Hence for each $\phi \in \mathcal{M}_{\mathfrak{A}}$ and each $g \in \mathfrak{A}, g \vert_{(\overset{\wedge}{\iota})^{-1}\,\{\phi\}} = \phi(g)$ and $\{(\overset{\wedge}{\iota})^{-1}\,\{\phi\}\}_{\phi \in \mathfrak{A}}$ is the desired decomposition. Call each set $S_{\phi}$

$\hspace{0.44cm}$ Given a function $h \in C(S)$ that is constant in each $S_{\phi}$, we have $\overset{\wedge}{h}(\phi) = h \vert_{S_{\phi}}$ is a well-defined element in $C[\mathcal{M}_{\mathfrak{A}}]$ and hence $h \in \mathfrak{A}$.