Dan, you are correct. My construction failed to consider the case of x = 1/4. -- Gene
________________________________ From: Dan Asimov <dasimov@earthlink.net> To: Eugene Salamin <gene_salamin@yahoo.com> Sent: Saturday, November 12, 2011 7:30 PM Subject: To Gene Re: [math-fun] Tiling puzzle
Gene,
I'm not sure if or where you may have erred, but I thought I had an argument making it impossible that one of the two sets is a subgroup of R:
----- Let x |-> x + t be the translation taking subgroup H bijectively onto H'.
If H is a subgroup, then it contains 0 and x in H implies -x in H.
And so the subgroup has a self isometry via x |-> -x.
But this means there is also a map of the form x |-> t - x taking H bijectively onto H'. But this is impossible since x -> t-x always has a fixed point at x = t/2. -----
Or have I made a mistake?
--Dan
<< It suffices to find an additive subgroup H of the reals of index 2, as then the two cosets H and H' are translates of each other. Start with H[0] = {0}, H'[0] empty, and let R be well ordered. H and H' will be built up by transfinite induction.
For α a successor ordinal, let x[α] be the smallest real in R - H - H'. Make H[α] by appending to H[α-1] all reals (y + n x[α]), with y in H[α-1] and n an integer. Make H'[α] by appending to H'[α-1] all reals (y' + (n+1/2) x[α]), with y' in H'[α-1]. Then H[α] and H[α] + H'[α] are subgroups of R and the former has index 2 in the latter.
For β a limit ordinal, let H[β] be the union of all H[α] for α < β, and let H'[β] be the union of all H'[α] for α < β. Continuing until the reals are exhausted yields the desired subgroup H.
Even though kleptomaniacs can't help themselves, they do.