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From: tanner@h-sc1.UUCP (jonathan tanner)
Newsgroups: net.math
Subject: Re: a new (old) head-scratcher -- solution
Message-ID: <376@h-sc1.UUCP>
Date: Thu, 23-May-85 10:34:36 EDT
Article-I.D.: h-sc1.376
Posted: Thu May 23 10:34:36 1985
Date-Received: Fri, 24-May-85 22:18:20 EDT
References: <7368@ucbvax.ARPA>
Organization: Harvard Univ. Science Center
Lines: 34

> 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