A recent issue of *American Math. Monthly* has a paper that partitions
$\mathbb{R}$ into an arbitrary finite number of uncountable sets such
that every real number is a condensation point of all the sets in the
partition. In fact, we can find uncountably many uncountable
subsets $S\_\alpha$ of $\mathbb{R}$ that are (1) pairwise disjoint, (2)
have union $\mathbb{R}$, and (3) every real number is a condensation
point of all the sets. One way to do this is the following. For any
infinite binary sequence $\alpha=(a\_1, a\_2, \dots)$, let $S\_\alpha$ be
the set of all real numbers whose fractional part has the binary
expansion $0.b\_1b\_2\cdots$ such that the sequence
$(b\_2,b\_4,b\_6,\dots)$ differs from $\alpha$ in only finitely many
terms. When $\alpha=(1,1,1,\dots)$ we must remove from $S\_\alpha$ all
numbers whose binary expansion ends in infinitely many 1's, due to the
nonuniqueness of the binary expansion. The set of distinct
$S\_\alpha$'s then has the desired properties. This partition of
$\mathbb{R}$ also has the property that any two of the sets $S\_\alpha$
and $S\_\beta$ are translates of each other, except for
$S\_{1,1,1,\dots}$. This suggests the question: does there exist a
partition of $\mathbb{R}$ satisfying (1), (2), and (3) such that
*every* pair of sets in the partition are translates?



Nice partition of $\mathbb{R}$ into uncountably many uncountable sets
I think the following works:


Fix an uncountable subgroup $G$ of $(\mathbb{R}, +)$ - note that any such is dense in $\mathbb{R}$, and in fact intersects each nonempty open interval uncountably often - of uncountable index, and consider the set of cosets of $G$.


(Such a group can be produced by transfinite recursion using a well-ordering of $\mathbb{R}$: alternate between throwing reals into the group, and barring reals from entering the group.)




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Note that the argument above relies on the axiom of choice. However, this isn't necessary: [provably in ZF, there is a $\mathbb{Q}$-linearly independent set $S$ of reals of size continuum](https://mathoverflow.net/q/282832/8133). We can split this into two disjoint sets $S\_1, S\_2$ of size continuum (pick some appropriate real $a$ and consider the points to the left of $a$ versus the right of $a$). Now consider the $\mathbb{Q}$-vector space generated by $S\_1$.


