Prove that $C[0, 1]$ is NOT Approximately Finite

Question

The following question is from $C^*$- Algebras by Example written by Kenneth R. Davidson. The original question is Problem III.6 in exercises after Chapter 3.

$\mathit{Definition}$: A $C^*$- Algebra $\mathfrak{A}$ is called approximately finite (or AF) iff it is the closure of an increasing union of finite dimensional subalgebras $\mathfrak{A}_k$.

Let $X$ be the Cantor set constructed by the traditional "middle-third" method. Say $J_0 = [0, 1]$, $J_1 = [0, \frac{1}{3}]\,\bigcup\,[\frac{2}{3}, 1]$ and $J_n$ be the $2^n$ disjoint intervals constructed in the same way. According to the textbook, define $\mathfrak{A}_n$ be the subalgebra of functions in $C(X)$ which are constant in $J_n$. Hence we have $C(X) = \overline{\bigcup_{n \geq 0}\,\mathfrak{A}_n}$. Here the topology is induced by the $\| \cdot \|_{\infty}$ norm and so is $C[0, 1]$. The question wants us to show $C[0, 1]$ can be embedded into $C(X)$ and the embedding image, as a subalgebra of $C(X)$ is not AF. Since $C(X) \subseteq C[0, 1]$, define $\mathcal{C}_n = \{f \in C[0, 1]\,\vert\,f$ is constant in each disjoint interval of $J_n \}$. Then I believe $\overline{\bigcup_{n \geq 0} \mathcal{C}_n} = C[0, 1]$ and let the embedding be $\iota: C[0, 1] \rightarrow C(X), f \rightarrow f \vert_X$. I can not see why the image fails to be AF (very likely the embedding is wrong ...).

One of the key characterization of AF-Algebra in the same book is:

$\mathit{Theorem\,III.4}\,$: A $C^*$- Algebra $\mathfrak{A}$ is AF iff $\mathfrak{A}$ is separable and: $$(\ast) \hspace{0.2cm} \forall\,\epsilon > 0\,\text{and}\,A_1, A_2, \dots, A_n \in \mathfrak{A} \hspace{0.2cm} \exists\,\text{a subalgebra}\,\mathcal{B} \leq \mathfrak{A}\,\text{with}\,dim[\mathcal{B}] < \infty \\ \text{such that}\,d(A_i, \mathcal{B}) < \epsilon\,\forall\,1 \leq i \leq n$$

$C[0, 1]$ is separable but I can not see why $(\ast)$ fails in $C[0, 1]$ either. Any hints will be appreciated.

Answer 1

There are continuous surjective maps $k: X\to [0,1]$. For example choose a ternary expansion $x=\sum_n \frac{x_n}{3^n}$ for every $x\in X$ ($x_n\in \{0,2\}$) and let $k(x) = \sum_n \frac{x_n/2}{2^n}$. By being a bit careful about the definition you can check elementarily that this can give you a continuous surjective map.

Now define $k^*: C([0,1])\to C(X)$, $f\mapsto f\circ k$. This is obviously a $*$-morphism. Further it is injective, since if $f(k(x))=0$ for all $x\in X$ clearly $f(y)=0$ for all $y\in [0,1]$ by $k$ being surjective. Now make use of the fact that an injective $*$-morphism between $C^*$ algebras is an isometry to see that $k^*(C([0,1]))\cong C([0,1])$.

Answer 2

The quoted theorem implies that every self-adjoint element in an AF C*-algebra can be approximated by a finite sum of orthogonal projections, i.e. such elements are dense. Evidently this is not true for $C[0,1]$.

An abelian C*-algebra is AF if and only if its spectrum is totally disconnected. So the existence of a non-AF subalgebra of an AF algebra is due to the existence of a surjection from a totally disconnected space (the Cantor set $X$) to a non-totally disconnected space (the unit interval).

A non-AF subalgebra cannot lie in a (dense, possibly) union of finite-dimensional subalgebras---otherwise it would be AF. So the map suggested in the question cannot be an embedding. No embedding of $C[0,1]$ into $C(X)$ can lie in the inductive system defining $C(X)$ (without its limit points).

This example also shows that the AF property is not preserved by C*-subalgebras. It is preserved, however, by hereditary subalgebras.