why is the infinite product of the discrete two point space with itself, a topological homogeneous space? Why is the infinite product of the discrete two point space with itself, a topological homogeneous space?
 A: Let $X=\{0,1\}^\omega$, the product of countably infinitely many copies of the discrete two-point space. Let $x$ and $y$ be arbitrary points in $X$. Define $$d:\omega\to\{0,1\}:n\mapsto|x_n-y_n|\;.$$ In other words, $d_n=0$ when $x_n=y_n$, and $d_n=1$ when $x_n\ne y_n$. For each $z\in X$ let $\hat z\in X$ be defined as follows: $\hat z_n=z_n+d_n$, where the addition is carried out modulo $2$. In other words, $$\hat z_n=\begin{cases}z_n,&\text{if }d_n=0\\1-z_n,&\text{if }d_n=1\;.\end{cases}$$
It’s easy to see that $\hat x=y$ and $\hat y=x$. It’s also clear that $\hat{\hat z}=z$ for every $z\in X$, so the map $h:X\to X:z\mapsto\hat z$ is a bijection.
Let $\sigma=\{s_0,\dots,s_n\}$ be a finite sequence of $0$’s and $1$’s, and define $$B(\sigma)\triangleq\{z\in X:z_k=s_k\text{ for }k=0,\dots,n\}\;.$$
The collection $\mathscr{B}$ of all such sets $B(\sigma)$ is a base for $X$. You should have no trouble checking that $$h[B(\sigma)]=B(\hat\sigma)\tag{1}$$ for every $B(\sigma)\in\mathscr{B}$, where $\hat\sigma$ is defined in the obvious way: for $k=0,\dots,n$,
$$\hat s_k=\begin{cases}z_k,&\text{if }d_k=0\\1-z_k,&\text{if }d_k=1\;.\end{cases}$$
It’s an immediate consequence of $(1)$ that $h$ is a homeomorphism.
Intuitive Description: The map $h$ simply ‘flips’ each factor on which $x$ and $y$ disagree, thereby interchanging $x$ and $y$. ‘Flipping’ the factor like this is just switching the names of the points in that factor space; this doesn’t change the topology in any way.
Added: If you think of $X$ as a topological group, as Nate Eldredge suggested in the comments, $h$ is simply translation by $d$. Even if you’ve not dealt before with topological groups, it should be plausible that a translation is an autohomeomorphism.
