From: Dan Asimov <dasimov@earthlink.net> To: Eugene Salamin <gene_salamin@yahoo.com> Sent: Saturday, November 12, 2011 2:24 PM Subject: Re: [math-fun] Tiling puzzle

I wrote: << PUZZLE: Can the real line be partitioned into two congruent sets that are each dense?

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Yes, and not just 2, but any finite or countable number.  Define an equivalence relation on the real numbers, x == y if x-y is rational, and let S(0) be a set containing one number from each equivalence class.  S(0) can be chosen to be dense.  Let S(r) be the translate of S(0) by r.  For rational r, the countably many S(r) are disjoint and their union is all of the real line.  The S(r) give a countable dense partition of the real line.  For a finite partition, first let T[k] for 0<=k<n be a partition of the real line into n congruent sets, e.g. the union of k+mn <= x < k+mn+1 for all integers m.  Let S[k] for 0<=k<n be the union of S(r) for r in T[k].  Then these S[k] give a finite dense partition.

To show that the uncountable set S(0) can be chosen to be dense, let r[n] be the n-th rational, and choose real number x[n] in the interval I[n]:  r[n] - 1/2^n < x[n] < r[n] + 1/2^n, but such that x[n] =/= x[k] for k<n.  Let S(0) contain the x[n].  Now let x be a real number, let ε > 0, and let I be the open interval from x-ε to x+ε.  Since there are infinitely many rationals in I, some interval I[n] is contained within I.  Then x[n] is in I, so |x - x[n]| < ε, and thus S(0) is dense in the reals.

  --  Gene

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_______________________________________________> Yes, that's essentially what I had in mind -- nice solving.

Now for Part B: Is there a way to solve this puzzle (for 2 sets) so that one of the sets is a subgroup of R ?

--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.   --  Gene