phr@ucbvax.ARPA (Paul Rubin) (05/22/85)
Construct a one-to-one, onto mapping between the open unit interval (0,1) and the closed interval [0,1]. This means that every point in (0,1) must be mapped to a unique point in [0,1] and vice versa. It's not too hard, once you figure out the right approach.
tanner@h-sc1.UUCP (jonathan tanner) (05/23/85)
> Construct a one-to-one, onto mapping between the open unit interval > (0,1) and the closed interval [0,1]. To make my formulas neater, I will construct a one-to-one, onto mapping between A = (-1,1) and B = [-1,1]. Compose this with the mappings x -> 2x-1 from (0,1) to (-1,1) and x -> (x+1)/2 from [-1,1] to [0,1], to solve the problem as stated. Define h : A -> B by h(x)=x, unless x=1/2^m or x=-1/2^m, for some integer m, in which case define h(x)=2x. That is, h(1/2)=1, h(-1/2)=-1, h(1/4)=1/2, h(-1/4)=-1/2, etc., and h fixes other points. Actually, 2 could be replaced by any number greater than 1. I solved this problem by following a proof of the Schroeder-Bernstein Theorem, which states that if there are one-to-one mappings f from A into B (in this case one can take f(x)=x) and g from B into A (e.g., g(x)=x/2) then there is a one-to-one mapping h from A ONto B. In other words, if card A <= card B and card B <= card A then card A = card B. Another solution is the map j : (0,1) -> [0,1] defined by j(x)=x, unless x=1/2^m, in which case define j(x)=4x, unless x=1/2, and define j(1/2)=0. This is my first posting to the net, so I hope I used proper etiquette. Sincerely, Jonathan -- Jonathan W. Tanner harvard!h-sc1!tanner harvard!h-ma1!tanner harvard!h-sc4!tanner Leverett B-24, Harvard College, Cambridge, MA 02138