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.
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.
