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Why is the infinite product of the discrete two point space with itself, a topological homogeneous space?

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Think of the two point space as the cyclic group of order 2. Then the infinite product is also a group. You can check that the product topology makes it a topological group, so it's certainly homogeneous. –  Nate Eldredge Feb 19 '12 at 2:05
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No need for groups: a 2 point discrete space is homogeneous. Any product of homogeneous spaces is homogeneous: work per coordinate and a product of homeomorphisms is a homeomorphism. Done. Products "create" homeogeneity too: $[0,1]^{\mathbb{N}}$ is homogeneous even when $[0,1]$ is not, and there are stronger theorems of this kind for zero-dimensional spaces too. –  Henno Brandsma Feb 21 '12 at 21:24

1 Answer 1

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.

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