re:Interesting Question

From: Theorist (Alexey007_at_hotmail-dot-com.no-spam.invalid)
Date: 11/06/04 

Date: 6 Nov 2004 03:29:30 -0600

I figured out how to do it.
First you show that the interior of each S_t is contained in the
union, this is done by using the fact that a subset of an interval
that's order isomorphic to an interval is an interval.

Then you conclude that the union is open and consider it's boundary.
It's easy to show that every three points on the boundary have the
same circumradius 1, from where it's easy to complete the proof.

Anyway what I needed this for is to show that the plane without n
points (n>1) cannot be partitioned into (non-degenerate) circles.

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