Let $\mathbb{R}$ be the set of real numbers. Given a subset $S$ of $\mathbb{R}$, let $\mathcal{T}\_S$ be the translation-invariant topology generated by $S$. That is, $\mathcal{T}\_S$ is the topology with a subbasis consisting of all translates of $S$. Suppose $A$ is a subset of $\mathbb{R}$ such that for every nonempty subset $S$ of $A$, we have that $\mathcal{T}\_S$ is disconnected. Does it follow that $\mathbb{R}$ cannot be written as a finite union of translates of $A$?


Some preliminary thoughts on this question: If we replace $\mathbb{R}$ by $\mathbb{Z}$, then the answer is no. For example, we can take $A$ to be the set of even integers, and one can verify that it has the above property. But $\mathbb{Z}$ equals the union of $A$ and $A+1$. So far, for $\mathbb{R}$ the only sets $A$ I’ve been able to come up with that have the required property are subsets of finite unions of cosets of a proper subgroup of $\mathbb{R}$. But the real numbers, unlike the integers, cannot be written as a finite union of cosets of a proper subgroup. So one approach to this problem might be to show that those are the only such $A$. However, I have no idea whether that’s true.



Subsets of $\mathbb{R}$, every nonempty subset of which generates a disconnected translation-invariant topology
This example is inspired by an example given op page 13 in [J. van Mill, Homogeneous subsets of the real line, Compositio Mathematica, 46 (1982) no. 1, pp. 3-13](http://www.numdam.org/item/?id=CM_1982__46_1_3_0).


Let $H$ be a Hamel base for $\mathbb{R}$ over $\mathbb{Q}$ such that
$1\in H$.
For $x\in\mathbb{R}$ let $q(x)$ denote the coefficient of~$1$ in its expression
as a linear combination of members of $H$.


Let $G=\{x:q(x)=0\}$ and let
$I=\mathbb{Q}\cap\bigcup\{(\pi+2n,\pi+2n+1):n\in\mathbb{Z}\}$.
We set $A=G+I$, so $A$ is the union of the cosets $G+q$ where $q\in I$, or $A=\{x:q(x)\in I\}$.
Notice that $A+\pi=\mathbb{R}\setminus A$, so in the translation-invariant
topology generated by $A$ the set $A$ itself is clopen, and $\mathbb{R}$
is the union of two translates of $A$.


We show that if $S$ is a nonempty subset of $A$ then $A$ is open in
the translation-invariant topology $\tau\_S$ generated by $S$, so that
$\tau\_S$ is not connected.


To begin note that $I+2\mathbb{Z}=I$,
so that $G+S+2\mathbb{Z}\subseteq A$ and the set $G+S+2\mathbb{Z}$
can be written as $G+T+2\mathbb{Z}$ for some $T\subseteq Q\cap(\pi,\pi+1)$.


Now consider the translation-invariant topology $\tau\_T$ on $\mathbb{Q}$ generated by $T$.
This topology contains nonempty sets of arbitrarily small diameter.
Let $\varepsilon>0$ and take $p,q\in T$ such that $p < \inf T+\varepsilon/2$
and $q>\sup T-\varepsilon/2$.
Then $q\in T\cap(T+(q-p))\subseteq(q-\varepsilon/2,q+\varepsilon/2)$.
It follows that $(\pi,\pi+1)\cap\mathbb{Q}$ belongs to $\tau\_T$.


But then
$$
A=G+\bigl((\pi,\pi+1)\cap\mathbb{Q}\bigr)+2\mathbb{Z}
$$
belongs to the translation-invariant topology generated by
$G+T+2\mathbb{Z}$, and hence to~$\tau\_S$.


