[math-fun] Question on real functions
Let f be a continuous bijection between two real intervals, which remains a bijection when restricted to the rationals. The only examples of such f that I could construct are piecewise of the form y = (ax+b)/(cx+d), with a, b, c, d integer, cd != 0. Are there other examples?
Well, any order-preserving permutation P: Q_d -> Q_r of the rationals induces a homeomorphism of the reals of which P is the restriction to the rationals (where Q_d = Q_r = Q, to distinguish between domain and range). You can construct a P as above based on countably many countable choices: Let P^: Q_r -> Q_d denote the inverse function of P. Consider any enumeration of the rationals Q = {q_j | j in Z+}. 1. First set P(q_1) = r_1, some arbitrary rational. 2. Let q_j be the rational of lowest index j such that P^(q_j) is still undefined. Pick P^(r_2) arbitrarily to satisfy the condition that P^ preserves order where defined. 3. Let q_k be the rational of lowest index k such that P(q_k) is still undefined. Pick P(q_k) arbitrarily to satisfy the condition that P preserves order where defined. 4. Alternate between steps 2. and 3. countably many times. This clearly results in an order-preserving bijection P: Q_d -> Q_r. I.e., an order-preserving permutation P: Q -> Q. P now extends uniquely to a homeomorphism of the reals. ((( 5. It is easy to use additional conditions to limit the arbitrary choices in steps 2. and 3., so that it is impossible for P: Q -> Q to be linear-fractional on any interval. Using almost any reasonable definition of probability, one can show that the probability of ending up with a linear fractional P is 0. ))) --Dan On 2013-03-30, at 7:42 AM, David Wilson wrote:
Let f be a continuous bijection between two real intervals, which remains a bijection when restricted to the rationals.
The only examples of such f that I could construct are piecewise of the form y = (ax+b)/(cx+d), with a, b, c, d integer, cd != 0.
Are there other examples?
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Dan Asimov -
David Wilson