Today I have heard that statement but I can't find the reference.


Can somebody know a reference and/or a proof?



Is the numerable product of finite abelian groups a cantor set?
The idea is that a base for the product topology can be formed by sets of the following form:


$$U\_{\alpha\_0,....,\alpha\_{n}} = \{(x\_i : i \in \mathbb{N}) \in \prod\_{i \in \mathbb{N}} G\_i : x\_0 = \alpha\_0, x\_1 = \alpha\_1, x\_2 = \alpha\_2 ..., x\_k = \alpha\_k\}$$ 


where $\alpha\_0 \in G\_0$, $\alpha\_1 \in G\_1$, .... $\alpha\_k \in G\_k$. 


Sets of this form form a tree under the reverse containment relation.


For $\sigma = (\alpha\_0,....,\alpha\_k)$ and $\tau = (\beta\_0,...,\beta\_j)$, we have that $U\_{\sigma} \supseteq U\_{\tau}$ (i.e., $U\_{\tau}$ is an extension of $U\_{\sigma}$ in this tree of open sets) if and only if $\sigma$ is an initial segment of $\tau$.


The facts that a base for the topology is a finite branching tree and that every two points are separated by the topology together imply that the space is homeomorphic to Cantor space.


The idea of the proof is to construct a map $\varphi$ taking finite binary strings (representing basic open sets in Cantor space) to finite binary strings of elements of the groups $G\_i$ in the product. You just want to make sure of the following properties, and then this will give you a homoemorphism by telling you the preimages of all the basic open sets:


(1) If $\sigma \subseteq \tau$, then $\varphi(\sigma) \subseteq \varphi(\tau)$


(2) Every finite string $\tau$ of elements of the groups $G\_i$ has $\tau \subseteq \varphi(\sigma)$ for some finite binary string $\sigma$. 


(3) If $\sigma$ and $\tau$ are incomparable, then all sufficiently long extensions $\sigma' \supseteq \sigma$ and $\tau' \supseteq \tau$ have $\phi(\sigma')$ incomparable with $\phi(\tau')$


It is an argument by induction to show that such a map exists.


