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."

For the sake of completeness, I will repeat parts of my earlier proof.

Assume that a counterexample exists: a one-to-one correspondence of plane P to plane P' such that every circle is mapped to a circle, but at least one line is not mapped to a line.

Without loss of generality (WLOG), assume that the y-axis is one such line.  Also, assume WLOG that (0,1) and (0,-1) are mapped to themselves.

Let C(x) be the set of points on the circle that intersects points (0,1), (0,-1), (x,0), where x is a positive real number.  Note that the union of all C(x) is the entire plane, excluding the y-axis, but including (0,1) and (0,-1).  Every circle C(x) on P must be mapped to some circle C(x') on P'.

Since this is a one-to-one correspondence, the leftover points must all be mapped to each other.  Therefore, the y-axis on P must be mapped to the union of the y-axis and all the leftover C(x').  Basically, this means it maps to a line and a nonzero number of circles (there must be at least one circle, since we assumed that this line does not map to a line).

Example: A crude ascii image of the sort of set on P' that might be mapped to the y-axis on P.  In the general case, we do not know how many circles there are, or even that there are a finite number of circles.

   |
 _-|-_
/  |  \
\_ | _/
  -|-
   |

Time to show the contradiction.

Consider the points (0,0) and (0,2) on P'.  They must map to some pair of points on the y-axis in P.  No matter what the pair is, there must exist a circle on P that passes through both.  Geometrically, this circle must pass through the y-axis exactly two times.

This circle must be mapped to a circle on P'.  The circle on P' passes through points (0,0) and (0,2), but no other points that map to the y-axis.  But this is geometrically impossible.

Since there is a contradiction, we must conclude our original assumption is false.  There is no counterexample.

Edited on December 15, 2006, 2:57 pm

Posted by Tristan on 2006-12-15 14:56:08