You have an ink stamp that is so amazingly precise that, when inked and pressed down on the plane, it makes every circle whose radius is an irrational number (centered at the center of the stamp) black.
Is it possible to use the stamp three times and make every point in the plane black?
If it is possible, where would you center the three stamps?
(In reply to
re(9): Brian's solution by SilverKnight)
Sounds like a miscommunication. Sorry, my fault. In my original post I used the terms "white" and "black" to describe subsets of R^6 and I should have used different terms. Here is a somewhat different restatement, which I think captures Larry's point, and answers both you and DJ:
Claim: Let S(A,B) [A and B in the plane] be the set of points C in the plane such that placing DJ's stamp at A,B, and C will leave a white point somewhere in the plane. Then if A and B are distinct, S(A,B) has measure 0.
Lemma: Let x be any point in the plane, and Px be the set of points where DJ's stamp can be placed and leave x white. Then Px has measure 0.
Proof of lemma: Px is clearly the set of circles, centered at x, with rational radii. This is a countable collection of sets of measure 0 (circles have measure 0 in the plane), and hence has measure 0.
Proof of claim: Choose A, B distinct. As Larry showed, after stamping at A and B there are only countably many points still white. Call this set X. S(A,B) = {Union of all Px: x is in X}. By the lemma, each Px has measure 0. So again we have a countable collection of sets of measure 0, so S(A,B) has measure 0. End of proof of claim.
So, since S(A,B) has measure 0, it has a non-empty complement. Any point D chosen from the complement of S(A,B) will give you three points (A,B,D) where you can stamp and cover the plane. Since A and B were arbitrary (provided they are distinct), and S(A,B) has measure 0, then Larry was correct in his assertion that picking three points at random from the plane, the odds of covering all the points int the plane is 1.
Edited on December 9, 2003, 3:05 pm
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