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

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.

• Your definition of $\mathcal C_n$ does not result in a finite dimensional algebra (because any $f\in \mathcal C_n$ can vary in any way it pleases outside of $J_n$). Commented Jun 7, 2020 at 11:56
• A simple way to see hat $C([0,1])$ cannot be an AF-algebra is to notice that every finite-dimensional commutative $C^\ast$-algebra of dimension at least $2$ has a non-trivial projection, while $C([0,1])$ does not have any non-trivial projections. Commented Jun 7, 2020 at 12:02
• Another angle is that the spectrum of any element in $C([0,1])$ is connected. I think the only finite dimensional $C^*$ algebra where every element has a connected spectrum is $\Bbb C$, which seems like a statement that admits an elementary proof. (This is pretty much the same comment as what @MaoWao said, just in another dress) Commented Jun 7, 2020 at 13:09
• @s.harp Thank you for your input. According to both your and MaoWao's reasons I believe $C[0, 1]$ alone do not have non-trivial finite dimensional subalgebras. What I really need to show is $\iota(C[0, 1])$, as a subalgebra in $C(X)$, which contains non-trivial projection, is not AF. I do need to edit the question because the part about $\mathcal{C}_n$ is not very clear. Commented Jun 7, 2020 at 13:44
• The image of $C([0,1])$ under the restriction map is all of $C(X)$, because any continuous function on $X$ can be extended to a continuous function on $[0,1]$. Commented Jun 7, 2020 at 14:14

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 $$k^*(C([0,1]))\cong C([0,1])$$.

• Thank you for your answer. Now it suffices to show $C[0, 1]$ alone is not AF. Commented Jun 7, 2020 at 18:31

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.

• This could be another good exercise. If the image of the embedding is just a hereditary subalgebra instead of an ideal, then the image, as a subalgebra, is again AF. Commented Jun 9, 2020 at 3:47