Equivalence classes of points of R^2

Question

Let $A$ and $B$ be two sets of points in $\Bbb R^2$. We define an equivalence relation on the powerset of $\Bbb R^2$, by saying that $R(A,B)$ iff there is a translation $f$ on $\Bbb R^2$ such that the image set of $A$ under $f$ is $B$. It is easy to check that this $R$ is indeed an equivalence relation. The equivalence class of the empty set has cardinality $1$, and so does the equivalence class of the set $\Bbb R^2$. My conjecture is that the equivalence class of every nonempty proper subset of $\Bbb R^2$ has cardinality of the continuum. Is this true or not?

Answer 1

Perhaps surprisingly, it is not.

Let $\mathcal{H}$ ba a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ (that is, a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$) one of whose elements is $1$, and let $S$ be the set of all points $(x, y)$ such that the coefficient of $1$ in the expression for $x$ using $\mathcal{H}$ is $0$.

Then there are countably many elements in the equivalence class of $S$; namely, for each rational $a\in \mathbb{Q}$, the set $S_a$ of all $(x, y)$ such that the $1$-coefficient of $x$ is $a$.