Maximal ideal space of an $L^\infty$ space is extremely disconnected

Question

I'm reading the book An Introduction to Operator Algebras by Kehe Zhu, in a such book all the algebras are assumed to be unital and associative. Once said that, the autor proves that if $A$ is a commutative $C^*$-algebra then the Gelfand transform is a $C^*$-isomorphism from $A$ onto $C(M_A)$, being $M_A$ the maximal ideal space of $A$, that is, the set of non-trivial multiplicative linear functionals.

Next, the following theorem is stated: if $A=L^\infty(X)$ (for some measure space $(X,S,\mu)$) with its natural $C^*$-algebra structure, that is, pointwise operations and involution defined by $f^*=\overline{f}$, then $A$ is isomorphic as a $C^*$-algebra to $C(K)$, where $K$ is an extremely disconnected compact Hausdorff space.

Of course, the natural candidate is $K=M_A$, since it is already known that $K$ is a compact Hausdorff space it only remains to prove that $K$ is extremely disconnected, that is, for any open set $U \subset K$, $\overline{U}$ is open. This is exactly what the autor does.

In order to do so, given a fixed open set $U$ the following set is considered: $$ F=\left\{{f \in C(K): 0\leq f \leq 1, f|_{K-U}}=0\right\} $$ If $\Gamma:A \rightarrow C(K)$ is the Gelfand transform, then $\Gamma$ is an isomorphism of $C^*$-algebras (that is, it is a bijective algebra homomorphism which preserves involutions, actually it is also an isometry). Therefore, $\Gamma^{-1}(F)$ is a bounded set in $A$ (the autor also says that $\Gamma^{-1}$ "preserves" the order structure, but I guess that means that if $||f||\leq||g||$ then $||\Gamma^{-1}f||\leq||\Gamma^{-1}g||$, otherwise I have no idea what he means). From this, the autor concludes that $F$ has a least upper bound $f_0$ (again, I guess that means $||f||\leq||f_0||$ for all $f\in F$ and that if $g \in C(K)$ also satisfies this last relation then $||f_0||\leq||g||$ ).Next, the following is claimed: for every $x \in U$ there exist $f \in F$ such that $f(x)=1$ and therefore $f_0|_{U}=1$.

Why is this last claim true? Since $||f||\leq||f_0||$ we can assure that $||f_0||=1$ but unless we know that $f(x) \leq f_0(x)$ for all $x \in K$ and for all $f \in F$ (which doesn't seem true unless I have misunderstood something on the way) I don't see why we could ensure that $f_0|_{U}=1$. As you can see, I am having trouble in order to understand this proof. I would appreciate your help in understanding the proof as well as correcting any errors I may have made.

In advance thank you.

Answer 1

Edit: I asked a related question about the part of the proof that involves von Neumann algebras. It turns out that the argument requires the measure to be $\sigma$-finite (more properly, localizable); otherwise there is not guarantee that $L^\infty(X)$ is a von Neumann algebra, and there are examples of bounded sets of real-valued functions in $L^\infty(X)$ where the supremum does not exist.

(next comes the original answer)

The norm provides no order, the same way that the absolute value does not give you an order in $\mathbb C$. The order one considers in a C$^*$-algebra is among selfadjoint elements (analog to how in $\mathbb C$ you only have an order between real elements), and $a\leq b$ means that $b-a$ is positive. The Gelfand transform, being a $*$-homomorphism, preserves this order. In the particular case where the C$^*$-algebra is $C(K)$, $f\leq g$ means that $f,g$ are real valued that $f(t)\leq g(t)$ for all $t$.

In your case, because $\Gamma$ is a $*$-isomorphism, any $a\in \Gamma^{-1}(F)$ will satisfy $a=\Gamma^{-1}(f)$ with $0\leq f\leq 1$, and so $0\leq a\leq 1$ since $\Gamma(0)=0$, $\Gamma(1)=1$.

Now since $A=L^\infty(X)$ then $\Gamma^{-1}(F)$ has a least upper bound (see explanation at the bottom), and hence $F$ has a supremum because it's an isomorphic image of a set with a supremum.

Fix $u\in U$. Because $K$ is compact Hausdorff, by Urysohn's Lemma there exists $g\in C(K)$, with $0\leq g\leq 1$, $g(u)=1$ and $\operatorname{supp}g\subset U$. So $g\in F$ and so $1\geq f_0(u)\geq g(u)=1$ and $f_0(u)=1$. As this can be done for all $u\in U$, we get by continuity that $f_0|_{\overline U}=1$.

Now probably the proof continues by noticing that $f_0|_{K\setminus \overline U}=0$, for otherwise it would not be the least upper bound. And then $K\setminus\overline U=f_0^{-1}(\{0\})$ is closed.

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Why $\Gamma^{-1}(F)$ has a least upper bound? Briefly, for all finite subsets of $\Gamma^{-1}(F)$ the maximum is again in $L^\infty(X)$. This allows us to construct a bounded monotone increasing net of selfadjoint elements in $L^\infty(X)$, which converges since we are in a von Neumann algebra. So there exists $\eta\in L^\infty(X)$ which is the least upper bound of $\Gamma^{-1}(F)$. Hence $f_0=\Gamma(\eta)$ is a least upper bound for $F$. Because of Urysohn's Lemma, for each $u\in U$ there exists $g\in F$ with $g(u)=1$; this forces $f_0(u)=1$. And since $f_0$ is continuous, we get that $f_0\geq 1_{\overline U}$. Now, if $f_0(v)>0$ for some $v\in iK\setminus \overline U$, we could use Urysohn's Lemma again to find a function $h\in C(K)$ with $0\leq h\leq1$, $h|_{\overline U}=1$, and $h(v)=\frac12$. Then $hf_0\leq f_0$ with strict inequality at $v$, and $hf_0$ is still an upper bound for $F$, a contradiction. So $f_0|_{K\setminus \overline U}=0$.