AF C*-Algebras and Continuous Functions on Totally Disconnected Sets

Question

Why do the continuous functions on a totally disconnected set, such as the Cantor set, form an AF C$^*$-algebra? Conversely, why do commutative AF C$^*$-algebras consist of continuous complex functions on a totally disconnected compact metrizable space? I am new to operator theory and have only begun reading papers in the subject. Both claims appear in different papers and I have not been able to find references for these results.

Answer 1

The Cantor set is a projective limit of finite spaces (remember its basis is made of clopens, so just take a finite clopen partition and take the limit over the refinements).

Dually, the algebra of its continuous functions will be an inductive limit of algebras of functions over finitely many points, so finite dimensional algebras.

This remark answers both your questions.

Answer 2

Let $\mathcal{A}$ be an unital commutative $C^{*}$-algebra. Suppose that $\mathcal{A}$ is Approximately Finite-dimensional. By the Gelfand Theorem, we can suppose that $\mathcal{A}=C(X,\mathbb{C})$, where $X$ is a compact Hausdorff topological space.

A commutative finite-dimensional $C^{*}$-álgebra is generated by a finite number of projections (idempotents and self-adjoints elements). Since our $C^{*}$-algebra $C(X,\mathbb{C})$ is AF, it must be generated by a sequence of projections, that is, there exists a sequence $(p_{n})$ of projections of $C(X,\mathbb{C})$ such that the closed linear span generated by the $p_{n}$'s coincides with $C(X,\mathbb{C})$.

Now, if $f\in C(X,\mathbb{C})$ is a projection, we have $f^{2}=f$, and so, $f=\chi_{A}$ (the characteristic funcion of $A$), where $A=f^{-1}(\{1\})$. Since $f$ is continuous, the sets $A=f^{-1}(\{1\})$ and $X\setminus A=f^{-1}(\{0\})$ are closed subsets of $X$, that is, $A$ is a clopen (i.e, open and closed) subset of $X$. Conversely, the characteristic functions of clopen subsets of $X$ are continuous functions on $X$. Therefore, the projections of the $C^{*}$-algebra $C(X,\mathbb{C})$ are the characteristic functions of the clopen subsets of $X$

Therefore, our $C^{*}$-algebra $C(X,\mathbb{C})$ coincides with the closed linear span of a family of characteristic functions $\{\chi_{A_{n}}:n\in\mathbb{N}\}$, where $A_{n}$ is a clopen subset of $X$ for each $n\in\mathbb{N}$. In particular, the function constantly equal to $1$ on $X$ can be uniformly approximated by linear combinations of the $\chi_{A_{n}}$'s, and then, $X=\bigcup\limits_{n}A_{n}$ (if there existed an element $x\in X$ that does not belong to any $A_{n}$, any linear combination of the $\chi_{A_{n}}$'s would take the value $0$ in $x$, and then, we could not approximate the function constantly equal to $1$ on $X$)

Now, let us see that $X$ is totally disconnected.

We will use the following fact: a compact Hausdorff space $Y$ is totally disconnected if and only if for all $x, y\in Y$ with $x\neq y$, there exists a clopen subset $A$ of $Y$ such that $x\in A$ and $y\notin A$ (the reverse implication is almost immediate. For the other implication, see this Any two points in a Stone space can be disconnected by clopen sets)

We proceed by contradiction. Let us suppose that $X$ is not totally disconnected. Then, there exists $x, y\in X$, $x\neq y$ such that every clopen subset of $X$ that contains $x$, also contains $y$. Since the $A_{n}$'s are a covering of $X$, there exists a minimum $n_{0}\in\mathbb{N}$ such that $x\in A_{n_{0}}$, and then, $A_{n_{0}}$ also contains $y$. Since the singletons $\{x\}$ and $\{y\}$ are closed and $X$ is a compact Hausdorff space (in particular, it is a normal topological space), by the Urysohn's Lemma, there exists a continuous function $\varphi:X\longrightarrow [0,1]$ such that $\varphi(x)=1$ and $\varphi(y)=0$. Since $C(X,\mathbb{C})$ coincides with the closed linear span of the $\chi_{A_{n}}$'s, there exists $N\in\mathbb{N}$ and scalars $a_{1},\dots,a_{N}$ such that $\|\varphi-\sum_{i=1}^{N}a_{i}\chi_{A_{i}}\|_{\infty}<1/2$.

Denote by $J=\{1\leq m\leq N: x,y\in A_{m}\}$. We know that $J=\{1\leq m\leq N: x\in A_{m}\}$. Also, $J=\{1\leq m\leq N: y\in A_{m}\}$ (if we had $1\leq m\leq N$ such that $y\in A_{m}$ and $x\notin A_{m}$, then $X\setminus A_{m}$ is a clopen subset which contains $x$, but it does not contain $y$, and we would have a contradiction).

Evaluating at $x$ and $y$, we obtain respectively $|1-\sum_{i\in J}a_{i}|<1/2$ and $|\sum_{i\in J}a_{i}|<1/2$, which is absurd.

Therefore, $X$ is totally disconnected