What is the quotient space of $\mathbb{Z}$-indexed $\{0,1\}$-sequences with resprect to shifts? Let $V=\{0,1\}^{\mathbb{Z}}$ be equipped with product topology. For $k\in\mathbb{Z}$ let $T^k:V\rightarrow V$ be the shift-by-$k$-operator, so
$$T^k((x_j)_{j\in\mathbb{Z}}):=(x_{j+k})_{j\in\mathbb{Z}}.$$
Two elements $x,y\in V$ are called equivalent (denoted by $x\sim y$) iff there is a $k\in\mathbb{Z}$ such that $T^k(x)=y$. What is the quotient space of $V$ with respect to $\sim$, what is it homeomorphic to? 
 A: This is not an answer, but it does give some insight into the space.
For $x\in V$ let $[x]$ be the $\sim$-equivalence class of $x$, and let $\pi:V\to V/{\sim}:x\mapsto[x]$ be the quotient map. Suppose that $B$ is a basic open set in $V$, i.e., one that specifies the values at some finite set of indices. Then
$$\pi^{-1}\big[\pi[B]\big]=\bigcup_{k\in\Bbb Z}T^k[B]\;,$$ 
which is open in $V$, so $\pi$ is an open map.
Let $x,y\in V$, and suppose that $x$ has a finite substring that is not a substring of $y$. Let $B$ be the basic open nbhd of $x$ determined by this substring. Then no shift of $y$ is in $B$, so $y\notin\pi^{-1}\big[\pi[B]\big]$, and hence $[y]\notin\pi[B]$. That is, $\pi[B]$ is an open nbhd of $[x]$ that does not contain $y$.
Now suppose that every finite substring of $x$ is also a finite substring of $y$, and let $U$ be an open nbhd of $[x]$. There is a basic open nbhd $B$ of $x$ such that $\pi[B]\subseteq U$. This $B$ is determined by some finite substring of $x$, so by hypothesis $T^k(y)\in B$ for some $k\in\Bbb Z$, $y\in\pi^{-1}\big[\pi[B]\big]$, and hence $[y]\in\pi[B]\subseteq U$. That is, every open nbhd of $[x]$ contains $[y]$. We now have the following result:

For any $x,y\in V$, $[x]$ has an open nbhd that does not contain $[y]$ if and only if $x$ has a finite substring that does not appear in $y$.

Clearly $V/{\sim}$ is not $T_1$, and in fact it is not even $T_0$. Let $x\in V$ be any bisequence that has every finite binary string as a subsequence, and let $y$ be the complementary bisequence: $y_n=1-x_n$ for each $n\in\Bbb Z$. Clearly $x$ and $y$ have exactly the same finite substrings. Suppose that $x\sim y$; then there is a $k\in\Bbb Z$ such that $y=T^k(x)$, so that $y_i=x_{i+k}$ for each $i\in\Bbb Z$. By hypothesis there is an $n\in\Bbb Z$ such that $x_i=0$ for $i=n,\ldots,n+k$. But then 
$$1=1-x_n=y_n=x_{n+k}=0\;,$$
which is absurd. Thus, $[x]\ne[y]$, and neither $[x]$ nor $[y]$ has an open nbhd separating it from the other.
