This may be a standard question answered in a book, or article. I don't know. I know that there exist related results with $\lim^1$-sequences (Rosenberg and Schochet).


What is


$KK(C\_0(X),\mathbb{C})$


for $X =$ Cantor set?


What is it more generally when $C\_0(X)$ is an abelian AF-algebra?


My only question is, are these homology groups mostly UNCOUNTABLE?


Note that, for example,


$KK(\oplus\_{n=1}^\infty \mathbb{C} , \mathbb{C}) = \prod\_{n=1}^\infty \mathbb{Z}$


is uncountable.



K-homology of Cantor set and abelian AF-algebras
As David Handelman already pointed out, by continuity of $K$-theory we obtain that $K\_0(C(X)) \cong \bigoplus\_{\mathbb{N}} \mathbb{Z}$ and $K\_1(C(X)) = 0$. Since $C(X)$ is commutative, it lies in the Bootstrap class of $KK$-theory (see Blackadar, "Operator algebras", section V.1.5.2, p. 414). In particular, the universal coefficient theorem holds for $C(X)$ (see V.1.5.8, p. 416 in Blackadar), i.e.
$$
 0 \to Ext^1\_{\mathbb{Z}}(K\_\*(C(X)), K\_\*(\mathbb{C})) \to KK\_\*(C(X),\mathbb{C}) \to Hom(K\_\*(C(X)),K\_\*(\mathbb{C})) \to 0
$$
is exact and the map $\delta$ from Ext to KK has degree $1$, the map $\gamma$ from KK to Hom has degree $0$. Since $K\_\*(C(X))$ is free, we obtain that $\gamma$ is an isomorphism. This yields
$$
KK(C(X),\mathbb{C}) = KK\_0(C(X),\mathbb{C}) \cong Hom(K\_0(C(X)), \mathbb{Z}) = \prod\_{\mathbb{N}} \mathbb{Z} ,
$$
which is uncountable.


If $A$ is a commutative AF-algebra, then we have $K\_1(A) = 0$ and $A$ also lies in the Bootstrap class (because it is commutative), so we still obtain an isomorphism 
$$
KK\_0(A,\mathbb{C}) \cong Hom(K\_0(A), \mathbb{Z})
$$
in this case. Since $K\_0(A)$ is the direct limit over countable abelian groups, it should itself be countable again (feel free to correct me if I am wrong here). Therefore $KK(A,\mathbb{C})$ should be at most uncountable.


