From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, November 12, 2011 9:52 AM Subject: [math-fun] Tiling puzzle

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

--Dan

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] for0<=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