[math-fun] Can a nonconstant continuous function on the reals
assume only rational values? Related puzzle: Find an x in closed form s.t. CantorStaircase[x] is irrational. --rwg Spoiler: "2/)]2/1 ,0 ,3[atehTcitpillE + 1-( == ]]3/1 ,0 ,3[atehTcitpillE + 1-[esacriatSrotnaC"
I'll bite on the first question: No. Suppose f: R —> R is a nonconstant continuous function taking only rational values, say at least the values p < q in Q. Let c be an irrational number such that p < c < q . Let finv(X) denote the inverse image by f of the set X. Then by assumption we have R = finv((-oo,c)) + finv((c,oo)) where + denotes disjoint union. This exhibits the connected set R of reals as the disjoint union of two nonempty open sets — contradiction. —Dan
On Feb 7, 2016, at 10:08 PM, Bill Gosper <billgosper@gmail.com> wrote:
assume only rational values? Related puzzle: Find an x in closed form s.t. CantorStaircase[x] is irrational. --rwg Spoiler: "2/)]2/1 ,0 ,3[atehTcitpillE + 1-( == ]]3/1 ,0 ,3[atehTcitpillE + 1-[esacriatSrotnaC"
Somehow "Can a nonconstant continuous function on the reals ..." led to this question; don't ask me why: For any irrational number c, let G(c) be the subgroup of the circle group R/Z defined by G(c) = {nc mod 1 | n in Z} This defines a circular ordering on the members of G(c): For any three distinct elements x, y, z of G(c), the (circular) sequence (x,y,z) is either clockwise or counterclockwise. Since the definition of G(c) provides a 1-1 correspondence K in Z <—> Kc mod 1 in S^1, where C denotes the circle of radius 1 given by S^1 = R/2piZ, which we also identify via [t] in R/2piZ <—> exp(it) with the unit circle in the complex plane. Now we can can think of this circular ordering O_c: G(c)^3 - diag^ —> {+1, -1}, ((( where diag^ = {(x,y,z) in G(c)^3 | xx(z-y) + yy(x-z) + zz(y-x) = 0} is the superdiagonal. ))) as applying to the integers: O_c(K,L,M)) = +1, (Kc mod 1, Lc mod 1 , Mc mod 1) is counterclockwise in G(c); = -1, otherwise. QUESTION: Given two irrationals b, c, consider sense-preserving bijections h (i.e., h preserves the circular orderings of all distinct triples) h: G(b) —> G(c) and let the set of all such be denoted H+(b,c) = {h: G(b) —> G(c) | O_b(K,L,M) = O_c(h(K), h(L), h(M))} Does there exist a real number p > 0 such that the inf over h in H+(b,c) of the expression disc(b,c) = Sum_{x in G(b)} |x - h(x)|^p is nonzero but finite? —Dan
Adam privately observes a second spoiler, CantorStaircase[(1 + EllipticTheta[3, 0, -1/3])/2] == (1 + EllipticTheta[3, 0, -1/2])/2 and remarks this is merely the problem of exhibiting an irrational point in Cantor's set. --rwg On Sun, Feb 7, 2016 at 10:08 PM, Bill Gosper <billgosper@gmail.com> wrote:
assume only rational values? Related puzzle: Find an x in closed form s.t. CantorStaircase[x] is irrational. --rwg Spoiler: "2/)]2/1 ,0 ,3[atehTcitpillE + 1-( == ]]3/1 ,0 ,3[atehTcitpillE + 1-[esacriatSrotnaC"
participants (3)
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Bill Gosper -
Dan Asimov -
Dan Asimov