Let f be a one-to-one correspondence of the points in a plane. Prove or disprove the following statement:

"If f maps circles to circles, then it maps straight lines to straight lines."

(In reply to re(3): Full Proof by Bractals)

I think the proof does work, and the choice of (0,0), (0,2) is very relevant.  However, I think I would have explained it differently.

I would have tightened up the extra circle concept first and expained as follows:

We have the assumption that f(y-axis) not equal to y-axis
so there exists some y st f((0,y)) not on the y axis.  The circle defined by (0,1), (0,-1), and (0,y) in P' must be mapped to by the y-axis because it cannot have been mapped to by a circle that goes through (0,1), (0,-1) in P.

Further, the y-axis must be mapped to by the y-axis by similar logic (it was not mapped to by any circle that goes through (0,1) and (0,-1).

Now there must be at least 1 circle in P that includes f^-1(0,0) and f^-1(0,2)

The mapping of that circle is a circle in P' that includes (0,0), and (0,2).  Thus, it includes 2 points on the y axis and two points on the aforementioned circle defined yb (0,1), (0,-1), and f(0,y).  f^-1 of all four of those points must lie on y axis in P so the circle in P must intersect the y axis in 4 points which is impossible.

Basically, I think Tristan got the proof.

Posted by Joel on 2006-12-18 00:21:46